Text Book Back Questions and Answers - Chapter 7 Thermodynamics 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Thermodynamics
Textual EvaluationSolved
I. Choose the correct answer from the following:
Question 1.
The amount of heat exchanged with the surrounding at constant temperature and pressure is called
(a) $\Delta \mathrm{E}$
(b) $\Delta \mathrm{H}$
(c) $\Delta \mathrm{S}$
(d) $\Delta \mathrm{G}$
Answer:
(b) $\Delta \mathrm{H}$
Question 2.
All the naturally occurring processes proceed spontaneously in a direction which leads to
(a) decrease in entropy
(b) increase in enthalpy
(c) increase in free energy
(d) decrease in free energy
Answer:
(d) decrease in free energy
Question 3.
In an adiabatic process, which of the following is true?
(a) $q=w$
(b) $\mathrm{q}=0$
(c) $\Delta \mathrm{E}=\mathrm{q}$
(d) $\mathrm{P} \Delta \mathrm{V}=0$
Answer:
(a) $q=0$
Question 4.
In a reversible process, the change in entropy of the universe is '
(a) $>0$
(b) $>0$
(c) $<0$
$(d)=0$

Answer:
(d) $=0$
Question 5.
In an adiabatic expansion of an ideal gas
(a) $\mathrm{w}=-\Delta \mathrm{u}$
(b) $w=\Delta \mathrm{u}+\Delta \mathrm{H}$
(c) $\Delta \mathrm{u}=0$
(d) $w=0$
Answer:
(a) $w=-\Delta u$
Question 6.
The intensive property among the quantities below is
(a) mass
(b) volume
(c) enthalpy
(d) mass/volume
Answer:
(d) mass/volume
Question 7.
An ideal gas expands from the volume of $1 \times 10^3 \mathrm{~m}^3$ to $1 \times 10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure at $1 \times 10^5 \mathrm{Nm}^{-2}$. The work done is
(a) $-900 \mathrm{~J}$
(b) $900 \mathrm{~kJ}$
(c) $270 \mathrm{~kJ}$
(d) $-900 \mathrm{~kJ}$
Answer:
(a) $-900 \mathrm{~J}$
Hint:
$\mathrm{w}=-\mathrm{P} \Delta \mathrm{V}$
$\mathrm{w}=-\left(1 \times 105 \mathrm{Nm}^{-2}\right)\left(1 \times 10^{-2} \mathrm{~m}^3-1 \times 10^{-3} \mathrm{~m}^3\right)$
$\mathrm{w}=-10^5\left(10^{-2}-10^{-3}\right) \mathrm{Nm}$
$\left.\mathrm{w}=-10^5(10-1) 10^{3-}\right) \mathrm{J}$
$\mathrm{w}=-10^5\left(9 \times 10^{-3}\right) \mathrm{J}$
$\mathrm{w}=-9 \times 10^2 \mathrm{~J}$
$\mathrm{w}=-900 \mathrm{~J}$

Question 8.
Heat of combustion is always
(a) positive
(b) negative
(c) zero
(d) either positive or negative
Answer:
(b) negative
Question 9.
The heat of formation of $\mathrm{CO}$ and $\mathrm{CO}_2$ are $-26.4 \mathrm{kcal}$ and $-94 \mathrm{kcal}$, respectively. Heat of combustion of carbon monoxide will be
(a) $+26.4 \mathrm{kcal}$
(b) $-67.6 \mathrm{kcal}$
(c) $-120.6 \mathrm{kcal}$
(d) $+52.8 \mathrm{kcal}$
Answer:
(b) $-67.6 \mathrm{kcal}$
Hint:
$
\begin{aligned}
& \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} \\
& \Delta \mathrm{H}_{\mathrm{C}}^0(\mathrm{CO})=\left[\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CO}_2\right)-\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CO})+\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{O}_2\right)\right] \\
& \Delta \mathrm{H}_{\mathrm{C}}^0(\mathrm{CO})=-94 \mathrm{KCal}-[-26.4 \mathrm{KCal}+0] \\
& \Delta \mathrm{H}_{\mathrm{C}}^0(\mathrm{CO})=-94 \mathrm{KCal}+26.4 \mathrm{KCal} \\
& \Delta \mathrm{H}_{\mathrm{C}}^0(\mathrm{CO})=-67.4 \mathrm{KCal}
\end{aligned}
$
Question 10.
$\mathrm{C}$ (diamond) $\rightarrow \mathrm{C}$ (graphite), $\Delta \mathrm{H}=-\mathrm{ve}$, this indicates that
(a) graphite is more stable than diamond
(b) graphite has more energy than diamond
(c) both are equally stable
(d) stability cannot be predicted

Answer:
(a) graphite is more stable than diamond
Question 11.
The enthalpies of formation of $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{Cr}_2 \mathrm{O}_3$ are $-1596 \mathrm{~kJ}$ and $-1134 \mathrm{~kJ}$, respectively. $\Delta \mathrm{H}$ for reaction $2 \mathrm{Al}+\mathrm{Cr}_2 \mathrm{O}_3 \rightarrow 2 \mathrm{Cr}+\mathrm{Al}_2 \mathrm{O}_3$ is
(a) $-1365 \mathrm{~kJ}$
(b) $2730 \mathrm{~kJ}$
(c) $-2730 \mathrm{~kJ}$
(d) $-462 \mathrm{~kJ}$
Answer:
(d) $-462 \mathrm{~kJ}$
$
\begin{aligned}
& 2 \mathrm{~A} 1+\mathrm{Cr}_2 \mathrm{O}_3 \rightarrow 2 \mathrm{Cr}+\mathrm{Al}_2 \mathrm{O}_3 \\
& \Delta \mathrm{H}_{\mathrm{I}}^0=\left[2 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{Cr})+\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Al}_2 \mathrm{O}_3\right)\right]-\left[2 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{Al})+\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Cr}_2 \mathrm{O}_2\right)\right] \\
& \Delta \mathrm{H}_{\mathrm{I}}^0=[0+(-1596 \mathrm{~kJ})]-[0+(-1134)] \\
&
\end{aligned}
$
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{r}}^0=-1596 \mathrm{~kJ}+1134 \mathrm{~kJ} \\
& \Delta \mathrm{H}_{\mathrm{r}}^0=-462 \mathrm{~kJ}
\end{aligned}
$
Question 12.
Which of the following is not a thermodynamic function?
(a) internal energy
(b) enthalpy
(c) entropy
(d) frictional energy
Answer:
(d) frictional energy

Question 13.
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then
(a) $\Delta \mathrm{H}>\Delta \mathrm{U}$
(b) $\Delta \mathrm{H}-\Delta \mathrm{U}=0$
(c) $\Delta \mathrm{H}+\Delta \mathrm{U}=0$
(d) $\Delta \mathrm{H}<\Delta \mathrm{U}$
Answer:
(d) $\Delta \mathrm{H}<\Delta \mathrm{U}$
Question 14.
Change in internal energy, when $4 \mathrm{~kJ}$ of work is done on the system and $1 \mathrm{~kJ}$ of heat is given out by the system is ...........
(a) $+1 \mathrm{~kJ}$
(b) $-5 \mathrm{~kJ}$
(c) $+3 \mathrm{~kJ}$
(d) $-3 \mathrm{~kJ}$
Answer:
(c) $+3 \mathrm{~kJ}$
Hint:
$\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$
$\Delta \mathrm{U}=-1 \mathrm{~kJ}+4 \mathrm{~kJ}$
$\Delta \mathrm{U}=+3 \mathrm{kJc}$
Question 15.
The work done by the liberated gas when $55.85 \mathrm{~g}$ of iron (molar mass $55.85 \mathrm{~g} \mathrm{~mol}^{-1}$ ) reacts with hydrochloric acid in an open beaker at $25^{\circ} \mathrm{C}$.
(a) $-2.48 \mathrm{~kJ}$
(b) $-2.22 \mathrm{~kJ}$
(c) $+2.22 \mathrm{~kJ}$
(d) $+2.48 \mathrm{~kJ}$
Answer:
(a) $-2.48 \mathrm{~kJ}$
Hint:

$
\mathrm{Fe}+2 \mathrm{HCl} \rightarrow \mathrm{FeCl}_2+\mathrm{H}_2
$
1 mole of Iron liberates 1 mole of hydrogen gas $55.85 \mathrm{~g}$ Iron $=1$ mole Iron
$
\begin{aligned}
& \therefore \mathrm{n}=1 \\
& \mathrm{~T}=25^{\circ} \mathrm{C}=298 \mathrm{~K} \\
& \mathrm{w}=-\mathrm{P}\left(\frac{\mathrm{nRT}}{\mathrm{P}}\right) \\
& \mathrm{w}=-\mathrm{nRT} \\
& \mathrm{w}=-1 \times 8314 \times 298 \mathrm{~J} \\
& \mathrm{w}=2477.57 \mathrm{~J} \\
& \mathrm{w}=-2.48 \mathrm{k} \mathrm{J}
\end{aligned}
$
Question 16.
The value of $\mathrm{AH}$ for cooling 2 moles of an ideal mono atomic gas from $125^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$ at constant pressure will be [given $C_P=\frac{5}{2} \mathrm{R}$ ]
(a) $-250 \mathrm{R}$
(b) $-500 \mathrm{R}$
(c) $500 \mathrm{R}$
(d) $+250 \mathrm{R}$
Answer:
(b) $-500 \mathrm{R}$
Hint:
$
\begin{aligned}
& \mathrm{T}_{\mathrm{i}}=125^{\circ} \mathrm{C}=398 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=25^{\circ} \mathrm{C}=298 \mathrm{~K} \\
& \Delta \mathrm{H}=\mathrm{nC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right) \\
& \Delta \mathrm{H}=2 \times \frac{5}{2} \mathrm{R}(298-398) \\
& \Delta \mathrm{H}=-500 \mathrm{R}
\end{aligned}
$
Question 17.
Given that $\mathrm{C}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=\mathrm{akJ} ; 2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=-\mathrm{b} \mathrm{kJ}$;
Calculate the $\Delta \mathrm{H}^{\circ}$ for the reaction $\mathrm{C}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}$
(a) $\frac{b+2 a}{2}$
(b) $2 \mathrm{a}-\mathrm{b}$
(c) $\frac{2 a-b}{2}$
(d) $\frac{b-2 a}{2}$
Answer:
(d) $\frac{b-2 a}{2}$
Hint:
$
\begin{aligned}
& \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \Delta \mathrm{H}^{\circ}=-\mathrm{a} \mathrm{kJ} \ldots \ldots \ldots(1) \\
& 2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2 \Delta \mathrm{H}^{\circ}=-\mathrm{b} \mathrm{kJ} \ldots \ldots . .
\end{aligned}
$

$
\mathrm{C}+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CO} \Delta \mathrm{H}^{\circ}=?
$
(1) $x(2)$

$
\begin{aligned}
& 2 \mathrm{C}+2 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2 \Delta \mathrm{H}^{\circ}=-2 \mathrm{a} \mathrm{KJ} \\
& \text { Reverse of equestion (2) will be } \\
& 2 \mathrm{CO}_2 \rightarrow 2 \mathrm{CO}+\mathrm{O}_2 \Delta \mathrm{H}^{\circ}=+\mathrm{b} \mathrm{KJ} \\
& (3)+(4) \\
& 2 \mathrm{C}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO} \Delta \mathrm{H}^{\circ}=\mathrm{b}-2 \mathrm{a} \mathrm{KJ} \\
& (5) \div 2 \\
& \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO} \Delta \mathrm{H}^{\circ}=\frac{b-2 a}{2} \mathrm{KJ}
\end{aligned}
$
Question 18.
When 15.68 litres of a gas mixture of methane and propane are fully combusted at $0^{\circ} \mathrm{C}$ and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat released from this combustion in $\mathrm{kJ}$ is $\left(\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)=-890 \mathrm{~kJ} \mathrm{~mol}^{-1}\right.$ and $\Delta \mathrm{H}_{\mathrm{C}}$ $\left.\left(\mathrm{C}_3 \mathrm{H}_8\right)=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
(a) $-889 \mathrm{~kJ}$
(b) $-1390 \mathrm{~kJ}$
(c) $-3180 \mathrm{~kJ}$
(d) $-653.66 \mathrm{~kJ}$
Answer:
(d) $-653.66 \mathrm{~kJ}$
Hint:
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)=-890 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_3 \mathrm{H}_8\right)=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Let the mixture contain $\mathrm{x}$ lit of and lit of propane.

Volume of oxygen consumed $=2 x+5(15.68-x)=32$ lit
$
\begin{aligned}
& 2 \mathrm{x}+78.4-5 \mathrm{x}=321 \\
& 78.4-3 \mathrm{x}=32 \\
& 3 \mathrm{x}=46.41 \\
& \mathrm{x}=15.471
\end{aligned}
$
Given mixture contains 15.47 liters of methane and 0.213 liters of propane. Hence,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{C}}=\left[\frac{\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)}{22.4 \text { lit }} \times(\mathrm{x}) \text { lit }\right]+\left[\frac{\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_2 \mathrm{H}_4\right)}{22.4 \text { lit }} \times(3.67-\mathrm{x}) \text { lit }\right] \\
& \Delta \mathrm{H}_{\mathrm{C}}=\left[\frac{-890 \mathrm{~kJ} \mathrm{~mol}^{-1}}{22.4 \text { lit }} \times 1.23 \text { lit }\right]+\left[\frac{-1423}{22.4 \text { lit }} \times(3.67-1.23) \text { lit }\right]
\end{aligned}
$
$
\begin{aligned}
\Delta \mathrm{H}_{\mathrm{C}} & =\left[-614.66 \mathrm{~kJ} \mathrm{~mol}^{-1}\right]+\left[-20.81 \mathrm{~kJ} \mathrm{~mol}^{-1}\right] \\
\Delta \mathrm{H}_{\mathrm{C}} & =635.47 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$

Question 19.
The bond dissociation energy of methane and ethane are $360 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $620 \mathrm{Id} \mathrm{mol}^{-1}$
(a) $170 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(b) $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(c) $80 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(d) $220 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Answer:
(c) $80 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Hint:
$
\begin{aligned}
& 4 \mathrm{E}_{\mathrm{C}-\mathrm{H}}=360 \mathrm{~kJ} \mathrm{~mol}-1 \\
& \mathrm{E}_{\mathrm{C}-\mathrm{H}}=90 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{C}}+6 \mathrm{E}_{\mathrm{C}-\mathrm{H}}=620 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{C}}+6 \mathrm{x} 9 \mathrm{O}=62 \mathrm{O} \mathrm{kJ} \mathrm{mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{C}}+540=620 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{C}}=80 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 20.
The correct thermodynamic conditions for the spontaneous reaction at all temperature is (NEET phase $-\mathrm{I})$
(a) $\Delta \mathrm{H}<0$ and $\Delta \mathrm{S}>0$
(b) $\Delta \mathrm{H}<0$ and $\Delta \mathrm{S}<0$
(c) $\Delta \mathrm{S}>0$ and $\Delta \mathrm{S}=0$
(d) $\Delta \mathrm{H}<0$ and $\Delta \mathrm{S}>0$
Answer:
(a) $\Delta \mathrm{H}<0$ and $\Delta \mathrm{S}>0$
Question 21.
The temperature of the system, decreases in an
(a) isothermal expansion
(b) isothermal compression
(c) adiabatic expansion
(d) adiabatic compression
Answer:
(c) adiabatic expansion
Question 22 .
In an isothermal reversible compression of an ideal gas the sign of $\mathrm{q}, \mathrm{AS}$ and $\mathrm{w}$ are respectively
(a),,+--

(b),,-+-
(c),,+-+
(d),,--+
Answer:
(d),,--+
Hint:
During compression, energy of the system increases, in isothermal condition, to maintain temperature constant, heat is liberated from the system. Hence $\mathrm{q}$ is negative. During compression entropy decreases. During compression work is done on the system, hence $w$ is positive.
Question 23.
Molar heat of vaporization of a liquid is $4.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the entropy change is $16 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. the boiling point of the liquid is
(a) $323 \mathrm{~K}$
(b) $27^{\circ} \mathrm{C}$
(c) $164 \mathrm{~K}$
(d) $0.3 \mathrm{~K}$
Answer:
(b) $27^{\circ} \mathrm{C}$
Hint:
$
\begin{aligned}
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{\mathrm{T}_{\mathrm{b}}} \\
& \mathrm{T}_{\mathrm{b}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{\Delta \mathrm{S}_{\mathrm{V}}}=\frac{4800 \mathrm{~J} \mathrm{~mol}^{-1}}{16 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}=300 \mathrm{~K}=27^{\circ} \mathrm{C}
\end{aligned}
$

Question 24.
$\Delta \mathrm{S}$ is expected to be maximum for the reaction
(a) $\mathrm{Ca}_{(\mathrm{S})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CaO}(\mathrm{S})$
(b) $\mathrm{C}_{(\mathrm{S})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO} 2(\mathrm{~g})$
(c) $\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}(\mathrm{g})$
(d) $\mathrm{CaCO}_{3(\mathrm{~S})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})}$
Answer:
(d) $\mathrm{CaCO}_{3(\mathrm{~S})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})}$
Hint:
In $\mathrm{CaCO}_{3(\mathrm{~S})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})}$ entropy change is positive in (a) and (b) entropy change is negative; in (c) entropy change is zero.

Question 25.
The values of $\mathrm{H}$ and $\mathrm{S}$ for a reaction are respectively $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and loo $\mathrm{kJ} \mathrm{mol}^{-1}$. Then the temperature above which the reaction will become spontaneous is ............
(a) $300 \mathrm{~K}$
(b) $30 \mathrm{~K}$
(c) $100 \mathrm{~K}$
(d) $20^{\circ} \mathrm{C}$
Answer:
(a) $300 \mathrm{~K}$
Hint:
$
\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}
$
At $300 \mathrm{~K}$.
$\Delta \mathrm{G}=30000 \mathrm{~J} \mathrm{~mol}^{-1}-300 \mathrm{~K} \mathrm{x} 100 \mathrm{JK} \mathrm{mol}^{-1}$
$\Delta \mathrm{G}=0$
above $300 \mathrm{~K}$;
$\Delta \mathrm{G}$ will be negative and reaction becomes spontaneous.
II. Answer these questions briefly.
Question 26.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics states that "the total energy of an isolated system remains constant though it may change from one form to another"
(or)
Energy can neither be created nor destroyed, but may be converted from one form to another.
Question 27.
Define Hess's law of constant heat summation.
Answer:
Hess's law:
The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps.

Question 28.
Explain intensive properties with two examples.
Answer:
The property that is independent of the mass or size of the system is called as intensive property.
e.g., Refractive index and surface tension.
Question 29.
Define the following terms:
(a) isothermal process
(b) adiabatic process
(c) isobaric process
(d) isochoric process
Answer:
(a) isothermal process:
It is defined as one in which the temperature of the system remains constant, during the change from its initial to final states.
(b) Adiabatic process:
It is def'ined as one in which there is no exchange of heat (q) between the system and surrounding during operations.
(c) Isobaric process:
It is defined as one in which the pressure of the system remains constant during its change from the initial to final state.
(d) Isochoric process:
It is defined as one in which the volume of system remains constant during its change from initial to final state of the process.
Question 30.
What is the usual definition of entropy? What is the unit of entropy?
Answer:

1. Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
2. For a reversible change taking place at a constant temperature $(T)$. the change in entropy
3. of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature $(\mathrm{T})$.
$
\Delta \mathrm{S}_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{\mathrm{T}}
$
4. SI unit of entropy is $\mathrm{J} \mathrm{K}^{-1}$
Question 31.
Predict the feasibility of a reaction when
1. both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ positive
2. both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ negative
3. AH decreases but $\Delta \mathrm{S}$ increases
Answer:
1. When both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive, the reaction is not feasible.
2. When both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are negative, the reaction is not feasible.
3. When $\Delta \mathrm{H}$ decreases but $\Delta \mathrm{S}$ increases, the reaction is feasible.
Question 32 .
Define Gibb's free energy.
Answer:
Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
where $\mathrm{G}=$ Gibb's free energy
$\mathrm{H}=$ enthalpy
$\mathrm{T}=$ temperature
$\mathrm{S}=$ entropy

Question 33.
Define enthalpy of combustion.
Answer:
Enthalpy of combustion of a substance is defined as "the change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen". it is denoted as AH.
Question 34.
Define molar heat capacity. Give its unit.
Answer:
Molar heat capacity is defined as "the amount of heat absorbed by one mole of a substance in raising the temperature by I Kelvin". It is denoted as $\mathrm{C}_{\mathrm{m}}$
Unit of Molar heat capacity: SI unit of $\mathrm{C}_{\mathrm{m}}$ is $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
Question 35 .
Define the calorific value of food. What is the unit of calorific value?
Answer:
- The calorific value of food is defined as the amount of heat produced in calories (or Joules) when one gram of food is completely burnt.
- Unit of calorific value (a) $\mathrm{Cal} \mathrm{g}^{-1}$ (b) $\mathrm{J} \mathrm{Kg}^{-1}$
Question 36.
Define enthalpy of neutralization.
Answer:
The enthalpy of neutralization is defined as the change in enthalpy of the system when one gram equivalent of an acid is neutralized by one gram equivalent of a base or vice versa in dilute solution.
$
\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}_{(1)}=57.32 \mathrm{~kJ}
$
Question 37.
What is lattice energy?
Answer:
Lattice energy is defined as the amount of energy required to completely separate one mole of a solid ionic compound into gaseous constituent.

Question 38 .
What are state and path functions? Give two examples.
Answer:
- The variables like P. V, T and ' $n$ ' that are used to describe the state of the system are called as state functions. e.g.. pressure, volume, temperature, internal energy, enthalpy and free energy.
- A path function is a thermodynamic property of the system whose value depends on the path or manner by which the system goes from its initial to final state. e.g., work (w) and heat (q).
Question 39.
Give Kelvin statement of second law of thermodynamics.
Answer:
Kelvin-Planck statement:
It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.
Question 40.
The equilibrium constant of a reaction is 10 , what will be the sign of $\Delta \mathrm{G}$ ? Will this reaction be spontaneous?
Answer:
$
\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-2.303 \text { RT } \log \mathrm{K}_{\mathrm{eq}} \\
& \mathrm{K}_{\mathrm{eq}}=10 \\
& \therefore \Delta \mathrm{G}^{\circ}=- \text { ve value. }
\end{aligned}
$
So the reaction will be spontaneous.
Question 41.
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
Answer:
1. Enthalpy of neutralization of a strong acid by a strong base is always a constant and it is equal to $-57.32 \mathrm{~kJ}$ irrespective of which acid or base is used.
2. Because strong acid or strong base means it is completely ionized in solution state. For e.g., $\mathrm{NaOH}$ (strong base) is neutralized by $\mathrm{HCl}$ (strong acid), due to their complete ionization, the net reaction take place is only water formation.
So the enthalpy of neutralization is always constant for strong acid by a strong base.
$
\begin{aligned}
& \mathrm{H}^{+} \mathrm{Cl}^{-}+\mathrm{Na}^{+} \mathrm{OH}^{-} \rightarrow \mathrm{Na}^{+} \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{H}^{+} \mathrm{NO}_3{ }^{+}+\mathrm{K}^{+} \mathrm{OH}^{-} \rightarrow \mathrm{K}^{+} \mathrm{NO}_3{ }^{+}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
(Net reaction) $\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-57.32 \mathrm{~kJ}$

Question 42 .
State the third law of thermodynamics.
Answer:
It states that the entropy of pure crystalline substance at absolute zero is zero.
(or)
It can be stated as "it is impossible to lower the temperature of an object to absolute zero in a $\lim$
finite number of steps". Mathematically $T \rightarrow 0$
Question 43.
Write down the Born-Haber cycle for the formation of $\mathrm{CaCl}_2$
Answer:
Born - Haber cycle for the formation of $\mathrm{CaCl}_2$
$\mathrm{Ca}_{(\mathrm{S})}+\mathrm{Cl}_{2(\mathrm{l})} \rightarrow \mathrm{CaC}_{2(\mathrm{~S})} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}$
Sublimation : $\mathrm{Ca}(\mathrm{S}) \rightarrow \mathrm{Ca}(\mathrm{S}) \Delta \mathrm{H}_1^{\circ}$
Ionization : $\mathrm{Ca}_{(\mathrm{g})} \rightarrow \mathrm{Ca}^{2+}{ }_{(\mathrm{g})}+2 \mathrm{e}^{-}=\Delta \mathrm{H}_2{ }^{\circ}$
Vapourisation: $\mathrm{Cl}_{2(1)} \rightarrow \mathrm{Cl}_{2(\mathrm{~g})}=\Delta \mathrm{H}_3^{\circ}$
Dissociation : $\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{Cl}_{(\mathrm{g})}=\Delta \mathrm{H}_4^{\circ}$
Electron affinity : $2 \mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-2(\mathrm{~g})}(\mathrm{g})=\Delta \mathrm{H}_5 \circ$
Lattice enthalpy : $\mathrm{Ca}^{2+}{ }_{(\mathrm{g})}+2 \mathrm{Cl}^{-}{ }_{(\mathrm{g})} \rightarrow \mathrm{CaCl}_{2(\mathrm{~S})}=\Delta \mathrm{H}_6{ }^{\circ}$ $\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=\Delta \mathrm{H}_1{ }^{\circ}+\Delta \mathrm{H}_2{ }^{\circ}+\Delta \mathrm{H}_3^{\circ}+\Delta \mathrm{H}_4^{\circ}+\Delta \mathrm{H}_5{ }^{\circ}+\Delta \mathrm{H}_6{ }^{\circ}$
Question 44.
Identify the state and path functions out of the following
(a) Enthalpy
(b) Entropy
(c) Heat
(d) Temperature
(e) Work

(f) Free energy.
Answer:
State function : Enthalpy, entropy, temperature and free energy.
Path function: Heat and work.
Question 45 .
State the various statements of second law of thermodynamics.
Answer:
1. Entropy statement:
Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the universe".
2. Kelvin-Planck statement:
it is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir.
3. Efficiency statement:
Even an ideal, frictionless engine cannot convert $100 \%$ of its input heat into work.
Efficiency $=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}$
$\%$ Efficiency $=\left[\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}\right] \times 100$
$\%$ Efficiency $=\left[\frac{\text { Output }}{\text { Input }}\right] \times 100$
$\%$ Efficiency $<100 \%$
4. Clausius statement:
Heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to spend some work.
Question 46.
What are spontaneous reactions? What are the conditions for the spontaneity of a process?
Answer:

1. A reaction that does occur under the given set of conditions is called a spontaneous reaction.
2. The expansion of a gas into a evacuated bulb is a spontaneous process, the reverse process that is gathering of all molecules into one bulb is not spontaneous. This example shows that processes that occur spontaneously in one direction cannot take place in opposite direction spontaneously.
3. Increase in randomness favours a spontaneous change.
4. If enthalpy change of a process is negative, then the process is exothermic and occurs spontaneously. Therefore $\Delta \mathrm{H}$ should be negative.
5. if entropy change of a process is positive, then the process occurs spontaneously, therefore $\Delta \mathrm{S}$ should be positive.
6. If free energy of a process is negative, then the process occurs spontaneously, therefore $\Delta \mathrm{G}$ should be negative.
7. For a spontaneous. irreversible process. $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0, \Delta \mathrm{G}<0$. i.e., $\Delta \mathrm{H}=-\mathrm{ve}, \Delta \mathrm{S}=+$ ve and $\Delta \mathrm{G}=-\mathrm{ve}$.
Question 47.
List the characteristics of internal energy.
Answer:
- Internal energy of a system is an extensive property. It depends on the amount of the substances present in the system.
- Internal energy of a system is a state function. It depends only upon the state variables $(T$, P, V. n) of the system.
- The change in internal energy is as $\Delta \mathrm{U}=\mathrm{U}_2-\mathrm{U}_1$
- In a cyclic process, there is no energy change. $\Delta \mathrm{U}_{\text {(cyclic) }}=0$.
- If the internal energy of the system at final state $\left(U_f\right)$ is less than the internal energy of the system at its initial state $\left(\mathrm{U}_{\mathrm{i}}\right)$, then $\Delta \mathrm{U}$ would be negative.
- if $\mathrm{U}_{\mathrm{f}}<\mathrm{U}_{\mathrm{i}} \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\mathrm{ve}$ if $\mathrm{U}_{\mathrm{f}}>\mathrm{U}_{\mathrm{i}} \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=+\mathrm{ve}$
Question 48.
Explain how heat absorbed at constant volume is measured using bomb calorimeter with a neat diagram.
Answer:
1. For chemical reactions, heal absorbed at constant volume, is measured in a bomb calorimeter.
2. Description of the apparatus and procedure:
The inner vessel and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws. A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess
$
\mathrm{P}_{\mathrm{ext}}=\frac{\text { Force }(\mathrm{F})}{\operatorname{Area}(\mathrm{A})}
$
oxygen. The bomb is lowered in water, which is placed inside the calorimeter. A stirreris placed in the bomb to stir the water uniformly. The reaction is started by striking the substance through electrical heating.
3. During burning, the exothermic heat generated inside the bomb raises the temperature of the surrounding water bath. Temperature change can be measured accurately using Beckmann thermometer. Since the bomb calorimeter is sealed, its volume does not change, so the heat measurements in this case corresponds to the heat of reaction at constant volume. 

4. In a bomb calorimeter experiment, a weighed sample of benzoic acid (w) is placed in the bomb which is then filled with excess oxygen and sealed. Ignition is brought about electrically. The rise in temperature (AT) is noted. Water equivalent or calorimetry equivalent of the calorimeter is known from the standard value of enthalpy of combustion of benzoic acid.
5. $\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)=-3227 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$
\omega_{\mathrm{e}}=\frac{\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)}{\Delta \mathrm{T}} \times \frac{W}{M_2}
$
6. By knowing o value, the enthalpy of combustion of any other substance is determined adopting the similar procedure and using the substance in place of' benzoic acid. By this experiment, the enthalpy of combustion at constant volume $\left(\mathrm{AU}_{\mathrm{C}}{ }^{\circ}\right)$ is known, $\Delta \mathrm{U}_{\mathrm{C}}{ }^{\circ}=\omega_{\mathrm{e}} \cdot \Delta \mathrm{T}$
7. Enthalpy of combustion at constant pressure of the substance is calculated from the equation $\Delta \mathrm{U}^{\circ} \mathrm{C}$ (pressure $)=\Delta \mathrm{U}^{\circ} \mathrm{C}($ volume $)+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$
Question 49.
Calculate the work involved in expansion and compression process.
Answer:
1. The essential condition of expansion or compression of a system is, there should be difference between external pressure $\mathrm{P}_{\text {ext }}$ and internal pressure $\left(\mathrm{P}_{\text {int }}\right)$.
2. If the volume of the system is increased against the external pressure. the work is done by the system. By convention work done by the system is given a negative sign (-w).

3. If the volume of the system decreased, the work is done on the system. By convention work done on the system is given a positive sign $(+\mathrm{w})$.
4. For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is $\mathrm{V}_{\mathrm{i}}$ and pressure of the gas inside is $\mathrm{P}_{\text {int }}$
5. If external pressure is $\mathrm{P}_{\text {ext }}$ which is greater than $\mathrm{P}_{\text {int }}$ piston is moved inward till the pressure inside becomes equal to $P_{\text {int }}$ It is achieved in a single step and the final volume be $\mathrm{V}_{\mathrm{f}}$.
6. During this compression, piston moves a distance $\mathrm{x}$ ) and is cross-sectional area of the piston is $\mathrm{A}$, then, Change in volume $=\mathrm{xA}=\Delta \mathrm{V}=\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}} \ldots \ldots \ldots \ldots(1)$
$
\begin{aligned}
& \mathrm{P}_{\mathrm{ext}}=\frac{\operatorname{Force}(F)}{\operatorname{Area}(A)} \\
& \mathrm{F}=\mathrm{P}_{\mathrm{ext}} \cdot \mathrm{A}
\end{aligned}
$
7. if work is done by the system by pushing out the piston against external pressure $\left(P_{\text {ext }}\right)$ then according to the equation,
$
\begin{aligned}
& -\mathrm{w}=\mathrm{F} \cdot \mathrm{x} \ldots \ldots \ldots(3) \\
& -\mathrm{w}=\mathrm{P}_{\mathrm{ext}} \cdot \mathrm{A} \cdot \mathrm{x} \ldots \ldots \\
& -\mathrm{w}=\mathrm{P}_{\mathrm{ext}} \cdot \Delta \mathrm{V} \ldots \ldots \ldots \ldots \\
& -\mathrm{w}=\mathrm{P}_{\mathrm{ext}} \cdot\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right)
\end{aligned}
$
Simply $w=-\mathrm{P} \Delta \mathrm{V}$
8. From the above equation, we can predict the sign of work (w).
9. During expansion, work is done by the system, since $\mathrm{V}_{\mathrm{f}}>\mathrm{V}_i$ the sign obtained for work will be negative.
10. During compression, work is done on the system, since $\mathrm{V}_{\mathrm{f}}<\mathrm{V}_{\mathrm{i}}$ the sign obtained for work will be positive.
Question 50.
Derive the relation between $\Delta \mathrm{H}$ and $\Delta \mathrm{U}$ for an ideal gas. Explain each term involved in the equation.
Answer:
1. When the system at constant pressure undergoes changes from an initial state with $H_1, U_1$, $\mathrm{V}_1$ and $\mathrm{P}$ parameters to a final state with $\mathrm{H}_2, \mathrm{U}_2, \mathrm{~V}_2$ and $\mathrm{P}$ parameters, the change in enthalpy $\triangle \mathrm{H}$, is given by $\mathrm{AH}=\mathrm{U}+\mathrm{PV}$
2. At initial state $\mathrm{H}_1=\mathrm{U}_1+\mathrm{PV}_1 \ldots \ldots . .(1)$
At final state $\mathrm{H}_1=\mathrm{U}_1+\mathrm{PV}_1 \ldots \ldots . .(2)$
(2) $-(1) \Rightarrow\left(\mathrm{H}_2-\mathrm{H}_1\right)=\left(\mathrm{U}_2-\mathrm{U}_1\right)+\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)$
$\Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V} \ldots \ldots \ldots \ldots .(3)$
Considering $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w} ; \mathrm{w}=-\mathrm{P} \Delta \mathrm{V}$
$\Delta \mathrm{H}=\mathrm{q}+\mathrm{w}+\mathrm{P} \Delta \mathrm{V}$
$\Delta \mathrm{H}=\mathrm{q}_{\mathrm{p}}-\mathrm{P} \Delta \mathrm{V}+\mathrm{P} \Delta \mathrm{V}$
$\Delta \mathrm{H}=\mathrm{q}_{\mathrm{p}} \ldots \ldots \ldots \ldots .(4)$
$\mathrm{q}_{\mathrm{p}}$ is the heat absorbed at constant pressure and is considered as heat content.
3. Consider a closed system of gases which are chemically reacting to produce product gases at constant temperature and pressure with $\mathrm{V}$. and as the total volumes of the reactant and product gases respectively, and $\mathrm{n} 1$ and $\mathrm{n}_{\mathrm{f}}$ are the number of moles of gaseous reactants
and products. Then,
For reactants: $P V_i=n_i R T$
For products: $P V_f=n_f R T$
Then considering reactants as initial state and products as final state, $P\left(\mathrm{~V}_{\mathrm{i}}-\mathrm{V}_{\mathrm{i}}\right)=\left(\mathrm{n}_{\mathrm{i}}-\mathrm{n}_{\mathrm{i}}\right) R T$
$\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$
\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V} \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}
\end{aligned}
$
Question 51 .
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.

Answer:
The Born - Haber cycle is used to determine the lattice enthalpy of $\mathrm{NaCl}$ as follows:

Formation of $\mathrm{NaCl}$ can be considered in 5 steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of $\mathrm{NaCl}$ is calculated.

$\mathrm{Na}_{(\mathrm{s})}+\mathrm{Vi} \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{NaCl}_{(\mathrm{s})} \Delta \mathrm{H}_{\mathrm{f}}=-411.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Sublimation : $\mathrm{Na}_{(\mathrm{s})} \rightarrow \mathrm{Na}_{(\mathrm{g})} \Delta \mathrm{H}_1^{\circ}$
Dissociation : $1 / 2 \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{Cl}_{(\mathrm{g})} \Delta \mathrm{H}_2^{\circ}$
Ionisation : $\mathrm{Na}_{(\mathrm{s})} \rightarrow \mathrm{Na}_{(\mathrm{g})}^{+}+\mathrm{e}^{-} \Delta \mathrm{H}_3^{\circ}$
E1etron affinity : $\mathrm{Cl}_{(\mathrm{g})}+\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}$(g) $\Delta \mathrm{H}_4^{\circ}$
Lattice enthalpy : $\mathrm{Na}^{+}{ }_{(\mathrm{g})}+\mathrm{Cl}_{(\mathrm{g})}^{-} \rightarrow \mathrm{NaCl}_{(\mathrm{s})} \Delta \mathrm{H}_5{ }^{\circ}=$ ?
$\Delta \mathrm{H}=\Delta \mathrm{H}_1^{\circ}+\Delta \mathrm{H}_4^{\circ}+\Delta \mathrm{H}_4^{\circ}+\Delta \mathrm{H}_4^{\circ}+\Delta \mathrm{H}_4^{\circ}$
$\Delta \mathrm{H}=\Delta \mathrm{H}_{\mathrm{f}}^{\circ}-\left(\Delta \mathrm{H}_1^{\circ}+\Delta \mathrm{H}_2^{\circ}+\Delta \mathrm{H}_3^{\circ}+\Delta \mathrm{H}_4^{\circ}\right)$
$\Delta \mathrm{H}_5{ }^{\circ}=$ Lattice enthalpy of $\mathrm{NaCl}$.
By the above method, indirectly lattice enthalpy of $\mathrm{NaCl}$ is calculated, if the values of $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}, \Delta \mathrm{H}_1^{\circ}, \Delta \mathrm{H}_2^{\circ}-\Delta \mathrm{H}_3^{\circ}$ and $\Delta \mathrm{H}_4^{\circ}$ are given.

Question 52.
List the characteristics of Gibbs free energy.
Answer:
Characteristics of Gibbs free energy:
1. Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work. $\mathrm{G}=\mathrm{H}-\mathrm{TS}$
Where
$\mathrm{H}=$ enthalpy, $\mathrm{T}=$ temperature and $\mathrm{S}=$ entropy
2. $\mathrm{G}$ is a state function and is a single value function.
3. $\mathrm{G}$ is an extensive property, whereas $\Delta \mathrm{G}$ becomes intensive property for a closed system. Both $\mathrm{G}$ and $\Delta \mathrm{G}$ values correspond to the system only.
4. $\Delta \mathrm{G}$ gives a criteria for spontaneity at constant pressure and temperature.
- If $\Delta \mathrm{G}$ is negative ( $\Delta \mathrm{G}<\mathrm{O}$ ), the process is spontaneous.
- If $\Delta \mathrm{G}$ is positive $(\Delta \mathrm{G}>\mathrm{O})$, the process is non-spontaneous.
- If $\Delta \mathrm{G}$ is zero $(\mathrm{AG}=\mathrm{O})$, the process is equilibrium.
5. For any system at constant pressure and temperature, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ (2) We know $\mathrm{AH}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{G}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}-\mathrm{T} \Delta \mathrm{S}$
6. For the first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$ $\Delta \mathrm{G}=\mathrm{q}+\mathrm{w}+\mathrm{P} \Delta \mathrm{V}-\mathrm{T} \Delta \mathrm{S}$
For second law of thermodynamics, $\Delta \mathrm{S}=\frac{q}{T}$
$
\begin{aligned}
& \Delta \mathrm{G}=\mathrm{q}+\mathrm{w}+\mathrm{P} \Delta \mathrm{V}-\mathrm{T} \frac{q}{\mathrm{~T}} \\
& \Delta \mathrm{G}=\mathrm{w}+\mathrm{P} \Delta \mathrm{V} \ldots \ldots \ldots \ldots \\
& \Delta \mathrm{G}=-\mathrm{w}-\mathrm{P} \Delta \mathrm{V} \ldots \ldots \ldots
\end{aligned}
$
7. $-\mathrm{P} \Delta \mathrm{V}$ represent the work done due to expansion against a constant external pressure. Therefore, it is clear that the decrease in free energy $(-\Delta \mathrm{G})$ accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion.
8. Unit of Gibb's free energy is $\mathrm{J} \mathrm{mol}^{-1}$

Question 53.
Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of $500 \mathrm{ml}$ to a volume of $2 \mathrm{~L}$ at $25^{\circ} \mathrm{C}$ and normal pressure.
Answer:
Given
$
\mathrm{n}=2 \text { moles }
$
$
\begin{aligned}
& \mathrm{V}_{\mathrm{i}}=500 \mathrm{ml}=0.5 \text { lit } \\
& \mathrm{V}_{\mathrm{f}}=2 \text { lit } \\
& \mathrm{T}=25^{\circ} \mathrm{C}=298 \mathrm{~K} \\
& \mathrm{w}=-2.303 \mathrm{nRT} \log \left(\frac{V}{V_i}\right) \\
& \mathrm{w}=-2.303 \times 2 \times 8.314 \times 298 \times \log \frac{2}{0.5} \\
& \mathrm{w}=-2.303 \times 2 \times 8.314 \times 298 \times \log (4) \\
& \mathrm{w}=-2.303 \times 2 \times 8.314 \times 298 \times 0.6021 \\
& \mathrm{w}=-6871 \mathrm{~J} \\
& \mathrm{w}=-6.871 \mathrm{~kJ} .
\end{aligned}
$
Question 54.
In a constant volume calorimeter, $3.5 \mathrm{~g}$ of a gas with molecular weight 28 was burnt $;$ n excess oxygen at $298 \mathrm{~K}$. The temperature of the calorimeter was found to increase from $298 \mathrm{~K}$ to $298.45 \mathrm{~K}$ due to the combustion process. Given that the calorimeter constant is $2.5 \mathrm{~kJ} \mathrm{~K}^{-1}$. Calculate the enthalpy of combustion of the gas in $\mathrm{kJ} \mathrm{mol}^{-1}$.
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}_{\mathrm{i}}=298 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=298.45 \mathrm{~K} \\
& \mathrm{k}=2.5 \mathrm{~kJ} \mathrm{~K} \\
& \mathrm{~m}=3.5 \mathrm{~g} \\
& \mathrm{M}_{\mathrm{m}}=28 \\
& \text { heat evolved }=\mathrm{k} \Delta \mathrm{T} \\
& \Delta \mathrm{H}_{\mathrm{C}}=\mathrm{k}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)
\end{aligned}
$

$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{C}}=2.5 \mathrm{~kJ} \mathrm{~K}^{\prime}(298.45-298) \mathrm{K}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{C}}=1.125 \mathrm{~kJ}^{1.25} \\
& \Delta \mathrm{H}_{\mathrm{C}}=\frac{1.125}{3.5} \times 28 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{C}}=9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 55.
Calculate the entropy change in the system and surroundings, and the total entropy change in the universe during a process in which $245 \mathrm{~J}$ of heat flow out of the system at $77^{\circ} \mathrm{C}$ to the surrounding at $33^{\circ} \mathrm{C}$.
Answer:
Given:
$
\begin{aligned}
& \mathrm{T}_{\text {sys }}=77^{\circ} \mathrm{C}=(77+273)=350 \mathrm{~K} \\
& \mathrm{~T}_{\text {sys }}=33^{\circ} \mathrm{C}=(33+273)=306 \mathrm{~K} \\
& \mathrm{q}=245 \mathrm{~J}
\end{aligned}
$
$
\begin{aligned}
& \Delta \mathrm{S}_{\text {sys }}=\frac{\mathrm{q}}{\mathrm{T}_{\text {sys }}}=\frac{-245}{350}=-0.7 \mathrm{JK}^{-1} \\
& \Delta \mathrm{S}_{\text {univ }}=\frac{\mathrm{q}}{\mathrm{T}_{\text {sys }}}=\frac{+242}{350}=+0.8 \mathrm{JK}^{-1} \\
& \Delta \mathrm{S}_{\text {univ }}=\Delta \mathrm{S}_{\text {sys }}+\Delta \mathrm{S}_{\text {surr }}-245=-073 \mathrm{~K}^{\prime} \\
& \Delta \mathrm{S}_{\text {univ }}=-0.7 \mathrm{JK}^{-1}+0.8 \mathrm{JK}^{-1} \\
& \Delta \mathrm{S}_{\text {univ }}=0.1 \mathrm{JK}^{-1}
\end{aligned}
$

Question 56.
1 mole of an ideal gas, maintained at $4.1 \mathrm{~atm}$ and at a certain temperature, absorbs heat $3710 \mathrm{~J}$ and expands to 2 litres. Calculate the entropy change in expansion process.
Answer:
Given,
$
\begin{aligned}
& \mathrm{n}=1 \text { mole } \\
& \mathrm{P}=4.1 \mathrm{~atm} \\
& \mathrm{~V}=2 \mathrm{Lit} \\
& \mathrm{T}=? \\
& \mathrm{q}=3710 \mathrm{~J} \\
& \Delta \mathrm{S}=\frac{q}{T} \\
& \Delta \mathrm{S}=\frac{\mathrm{q}}{\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)} \\
& \Delta \mathrm{S}=\frac{\mathrm{nRq}}{\mathrm{PV}} \\
& \Delta \mathrm{S}=\frac{1 \times 0.082 \text { lit atm } \mathrm{K}^{-1} \times 3710 \mathrm{~J}}{4.1 \mathrm{~atm} \times 2 \text { lit }} \\
& \Delta \mathrm{S}=37.10 \mathrm{JK}^{-1}
\end{aligned}
$
Question 57.
$30.4 \mathrm{~kJ}$ is required to melt one mole of sodium chloride. The entropy change during melting is $28.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Calculate the melting point of sodium chloride.
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{NaCl})=30.4 \mathrm{~kJ}=30400 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{S}_{\mathrm{f}}(\mathrm{NaCl})=28.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \mathrm{~T}_{\mathrm{f}}=?
\end{aligned}
$

$
\begin{aligned}
\Delta \mathrm{S}_{\mathrm{f}} & =\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}} \\
\mathrm{T}_{\mathrm{f}} & =\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}} \\
\mathrm{T}_{\mathrm{f}} & =\frac{30400 \mathrm{~J} \mathrm{~mol}^{-1}}{28.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}} \\
\mathrm{~T}_{\mathrm{f}} & =1070.4 \mathrm{~K}
\end{aligned}
$
Question 58.
Calculate the standard heat of formation of propane, if its heat of combustion is $-2220.2 \mathrm{KJ}$ $\mathrm{mol}^{-1}$, the heats of formation of $\mathrm{CO}_{2(\mathrm{~g})}$ and $\mathrm{H}_2 \mathrm{O}_{(1)}$ are -393.5 and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
Answer:
Given,
$
\begin{aligned}
& \mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}_{\mathrm{C}}^0=-2220.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \\
& \Delta \mathrm{H}_{\mathrm{f}}^0=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{H}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}_{\mathrm{f}}^0=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& 3 \mathrm{C}+4 \mathrm{H}_2 \rightarrow \mathrm{C}_3 \mathrm{H}_8 \\
& \Delta \mathrm{H}_{\mathrm{C}}^0=\text { ? } \\
& \text { (2) } \times 3 \Rightarrow 3 \mathrm{C}+3 \mathrm{C}_2 \rightarrow 3 \mathrm{CO}_2 \\
& \Delta \mathrm{H}_{\mathrm{f}}^0=-1180.5 \mathrm{~kJ} \\
& \text { (3) } \times 4 \Rightarrow 4 \mathrm{H}_2+2 \mathrm{O}_2 \rightarrow 4 \mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}_{\mathrm{f}}^0=-1143.2 \mathrm{~kJ} \\
& \text { (4) + (5) - (1) } \\
& \Rightarrow 3 \mathrm{C}+3 \mathrm{O}_2+4 \mathrm{H}_2+2 \mathrm{O}_2+3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}+\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \\
& \Delta \mathrm{H}_{\mathrm{f}}^0=-1180.5-1143.2-(-2220.2) \mathrm{kJ} \\
& 3 \mathrm{C}+4 \mathrm{H}_2 \rightarrow \mathrm{C}_3 \mathrm{H}_8 \\
&
\end{aligned}
$

$
\Delta \mathrm{H}_{\mathrm{f}}^0=-103.5 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Standard heat of formation of propane is $\Delta \mathrm{H}_{\mathrm{f}}{ }^0\left(\mathrm{C}_3 \mathrm{H}_8\right)=-103.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Question 59.
You are given normal boiling points and standard enthalpies of vaporization. Calculate the entropy of vaporization of liquids listed below.

Answer:
For ethanol:
Given:
$
\begin{aligned}
& \mathrm{T}_{\mathrm{b}}=78.4^{\circ} \mathrm{C}=(78.4+273)=351.4 \mathrm{~K} \\
& \Delta \mathrm{H}_{\mathrm{V}}(\text { ethanol })=+42.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{\mathrm{T}_{\mathrm{b}}} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{+42.4 \mathrm{~kJ} \mathrm{~mol}^{-1}}{351.4 \mathrm{~K}} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{+42400 \mathrm{~J} \mathrm{~mol}^{-1}}{351.4 \mathrm{~K}} \\
& \Delta \mathrm{H}_{\mathrm{V}}=+91.76 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$
For Toluene:
Given:
$
\begin{aligned}
& \mathrm{T}_{\mathrm{b}}=110.6^{\circ} \mathrm{C}=(110.6+273)=383.6 \mathrm{~K} \\
& \Delta \mathrm{S}_{\mathrm{V}} \text { (toluene) }=+35.2 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{\mathrm{T}_{\mathrm{b}}} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{+35.2 \mathrm{~kJ} \mathrm{~mol}}{383.6 \mathrm{~K}} \\
& \Delta \mathrm{S}_{\mathrm{V}}=\frac{+35200 \mathrm{~J} \mathrm{~mol}^{-1}}{383.6 \mathrm{~K}} \\
& \Delta \mathrm{S}_{\mathrm{V}}=+91.76 \mathrm{~J} \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
Question 60.
For the reaction $\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+1 / 2 \mathrm{O}_{2(\mathrm{~g})}: \Delta \mathrm{H}=30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}=6.66 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ (at $1 \mathrm{~atm}$ ). Calculate the temperature at which $\mathrm{AG}$ is equal to zero. Also predict the direction of the reaction (i) at this temperature and (ii) below this temperature.
Answer:
Given,
$
\begin{aligned}
\Delta \mathrm{H} & =30.56 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\Delta \mathrm{H} & =30560 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$

$
\begin{aligned}
& \Delta \mathrm{S}=6.66 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& \mathrm{~T}=? \text { at which } \Delta \mathrm{G}=0 \\
& \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\
& 0=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\
& \mathrm{T}=\frac{\Delta H}{\Delta S} \\
& \mathrm{~T}=\frac{30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}}{6.66 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}} \\
& \mathrm{~T}=4589 \mathrm{~K}
\end{aligned}
$
(i) At $4589 \mathrm{~K} ; \Delta \mathrm{G}=0$, the reaction is in equilibrium.
(ii) At temperature below $4598 \mathrm{~K}, \Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}>0$, the reaction in the forward direction, is non-spontaneous. In other words the reaction occurs in the backward direction.

Question 61 .
What is the equilibrium constant $\mathrm{K}_{\mathrm{eq}}$ for the following reaction at $400 \mathrm{~K}$. $2 \mathrm{NOCl}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{~g})}$ given that $\mathrm{AH}^{\circ}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}^{\circ}=122 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}=400 \mathrm{~K} ; \Delta \mathrm{H}^{\circ}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}=77200 \mathrm{~J} \mathrm{~mol}^{-1} \text {; } \\
& \Delta \mathrm{S}^{\circ}=122 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} \\
& \log \mathrm{K}_{\mathrm{eq}}=\frac{\Delta \mathrm{G}^0}{2.303 \mathrm{RT}} \\
& \log \mathrm{K}_{\mathrm{eq}}=-\frac{\left(\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0\right)}{2.303 \mathrm{RT}} \\
& \log \mathrm{K}_{\mathrm{eq}}=-\left(\frac{77200-400 \times 122}{2.303 \times 8.314 \times 400}\right) \\
& \log \mathrm{K}_{\mathrm{eq}}=-\left(\frac{28400}{7659}\right) \\
& \log \mathrm{K}_{\mathrm{eq}}=-3.7080 \\
& \mathrm{~K}_{\mathrm{eq}}=\text { anti } \log (-3.7080) \\
& \mathrm{K}_{\mathrm{eq}}=1.95 \times 10^{-4} \\
&
\end{aligned}
$
Question 62.
Cyan-amide $\left(\mathrm{NH}_2 \mathrm{CN}\right)$ is completely burnt in excess oxygen in a bomb calorimeter, $\Delta \mathrm{U}$ was found to be $-742.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ calculate the enthalpy change of the reaction at $298 \mathrm{~K}$.
$
\mathrm{NH}_2 \mathrm{CN}_{(\mathrm{s})}+3 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}=\text { ? }
$

Answer:
Given,
$
\begin{aligned}
& \mathrm{T}=298 \mathrm{~K} ; \Delta \mathrm{U}=-742.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}=? \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{(\mathrm{g})} \mathrm{RT} \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\left(\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}\right) \mathrm{RT} \\
& \Delta \mathrm{H}=-742.4+\left[2-\frac{3}{2}\right] \times 8.314 \times 10^{-3} \times 298 \\
& \Delta \mathrm{H}=-742.4+\left(0.5 \times 8.314 \times 10^{-3} \times 298\right) \\
& \Delta \mathrm{H}=-742.4+1.24 . \\
& \Delta \mathrm{H}=-741.16 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 63.
Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of $\mathrm{C} \mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ are $414,347,618$ and $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Answer:
Given,
$
\begin{aligned}
& \mathrm{E}_{\mathrm{C}-\mathrm{H}}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{H}}=347 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{C}-\mathrm{H}}=618 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{H}-\mathrm{H}}=435 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$

$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{r}}=\sum(\text { Bond energy })_{\mathrm{I}}-\sum(\text { Bond energy })_{\mathrm{P}} \\
& \Delta \mathrm{H}_{\mathrm{r}}=\left(\mathrm{E}_{\mathrm{C}=\mathrm{C}}+4 \mathrm{E}_{\mathrm{C}-\mathrm{H}}+\mathrm{E}_{\mathrm{H}-\mathrm{H}}\right)-\left(\mathrm{E}_{\mathrm{C}-\mathrm{C}}+6 \mathrm{E}_{\mathrm{C}-\mathrm{H}}\right) \\
& \Delta \mathrm{H}_{\mathrm{r}}=(618+(4 \times 414)+435)-(347+(6 \times 414)) \\
& \Delta \mathrm{H}_{\mathrm{r}}=2709-2831 \\
& \Delta \mathrm{H}_{\mathrm{r}}=-122 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
Question 64.
Calculate the lattice enegry of $\mathrm{CaCl}_2$ from the given data
$\mathrm{Ca}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{CaCl}_{2(\mathrm{~s})} \Delta \mathrm{H}_{\mathrm{f}}{ }^0=-795 \mathrm{KJ} \mathrm{mol}^{-1}$
Atomisation : $\mathrm{Ca}_{(\mathrm{s})} \rightarrow \mathrm{Ca}(\mathrm{g}) \Delta \mathrm{H}_1{ }^{\circ}=+121 \mathrm{KJ} \mathrm{mol}^{-1}$
Ionization : $\mathrm{Ca}_{(\mathrm{g})} \rightarrow \mathrm{Ca}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} \Delta \mathrm{H}_2{ }^{\circ}=+242.8 \mathrm{KJ} \mathrm{mol}^{-1}$
Dissociation : $\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{Cl}_{(\mathrm{g})} \Delta \mathrm{H}_3{ }^{\circ}=+242.8 \mathrm{KJ} \mathrm{mol}^{-1}$
Answer:

$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+\Delta \mathrm{H}_3+2 \Delta \mathrm{H}_4+\mathrm{u} \\
& -795=121+2422+242.8+(2 \mathrm{x}-355)+\mathrm{u} \\
& -795=2785.8-710+\mathrm{u} \\
& -795=2075.8+\mathrm{u} \\
& \mathrm{u}=-795-2075.8 \\
& \mathrm{u}=-2870.8 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
Question 65.
Calculate the enthalpy change for the reaction $\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_2$ from the following data.
$
\begin{aligned}
& 2 \mathrm{Fe}+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{Fe}_2 \mathrm{O}_3 ; \Delta \mathrm{H}=-741 \mathrm{~kJ} \\
& \mathrm{C}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CO} ; \Delta \mathrm{H}=-137 \mathrm{KJ} \\
& \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \Delta \mathrm{H}=-394.5 \mathrm{KJ}
\end{aligned}
$
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Fe}_2 \mathrm{O}_3\right)=-741 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CO})=-137 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CO}_2\right)=-394.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_2 \Delta \mathrm{Hr}=? \\
& \Delta \mathrm{Hr}=\sum\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\text {products }}-\sum\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\text {reactants }} \\
& \Delta \mathrm{Hr}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{Fe})+3 \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CO}_2\right)\right]-\left[\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{Fe}_2 \mathrm{O}_3\right)+3 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CO})\right] \\
& \Delta \mathrm{Hr}=[-1183.5]-[-1152] \\
& \Delta \mathrm{Hr}=-1183.5+1152 \\
& \Delta \mathrm{Hr}=-31.5 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$

Question 66 .
When 1-pentyne (A) is treated with $4 \mathrm{~N}$ alcoholic $\mathrm{KOH}$ at $175^{\circ} \mathrm{C}$, it is converted slowly into an equilibrium mixture of 13\% 1-pentyne(A), 95.2\% 2-pentyne(B) and $3.5 \%$ of 1,2 pentadiene (C) the equilibrium was maintained at $175^{\circ} \mathrm{C}$, calculate $\mathrm{AG}^{\circ}$ for the following equilibria.
$
\begin{aligned}
& \mathrm{B} \rightleftharpoons \mathrm{A} \Delta \mathrm{G}_1^{\circ}=\text { ? } \\
& \mathrm{B} \rightleftharpoons \mathrm{C} \Delta \mathrm{G}_2^{\circ}=\text { ? }
\end{aligned}
$
Answer:
$
\mathrm{T}=175^{\circ} \mathrm{C}=175+273=448 \mathrm{~K}
$
Concentration of 1 - pentyne $[\mathrm{A}]=1.3 \%$
Concentration of 2 - pentyne $[\mathrm{B}]=95.2 \%$
Concentration of $1,2-$ pentadiene $[\mathrm{C}]=3.5 \%$
At equiLibrium
$
\begin{aligned}
& \mathrm{B} \rightleftharpoons \mathrm{A} \\
& 95.2 \% 1.3 \% \Rightarrow \mathrm{K}_1=\frac{3.5}{95.2}=0.0136 \\
& \mathrm{~B} \rightleftharpoons \mathrm{C} \\
& 95.2 \% 3.5 \% \Rightarrow \mathrm{K}_1=\frac{1.3}{95.2}=0.0367 \\
& \Rightarrow \Delta \mathrm{G}_1^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_1 \\
& \Delta \mathrm{G}_1^{\circ}=-2.303 \times 8.314 \times 448 \times \log 0.0136 \\
& \Delta \mathrm{G}_1^{\circ}=+16010 \mathrm{~J} \\
& \Delta \mathrm{G}_1^{\circ}=+16 \mathrm{~kJ} \\
& \Rightarrow \mathrm{G}_2^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_2 \\
& \Delta \mathrm{G}_2^{\circ}=-2.303 \times 8.314 \times 448 \times \log 0.0367 \\
& \Delta \mathrm{G}_2^{\circ}=+12312 \mathrm{~J} \\
& \Delta \mathrm{G}_2^{\circ}=+12.312 \mathrm{~kJ} .
\end{aligned}
$
Question 67.
At $33 \mathrm{~K} . \mathrm{N}_2 \mathrm{H}_4$ is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Answer:
$
\begin{aligned}
& \mathrm{T}=33 \mathrm{~K} \\
& \mathrm{~N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2
\end{aligned}
$
Initial concentration $100 \%$
Concentration dissociated 50\%
Concentration remaining at equilibrium $50 \%-100 \%$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{eq}}=\frac{100}{50}=2 \\
& \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} \\
& \Delta \mathrm{G}^{\circ}=-2.303 \times 8.31 \times 33 \times \log 2 \\
& \Delta \mathrm{G}^{\circ}=-190.18 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$

Question 68 .
The standard enthaipies of formation, of $\mathrm{SO}_2$ and $\mathrm{SO}_3$ are $-297, \mathrm{~kJ} \mathrm{rnol}^{-1}$ and $-396 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Calculate the standard enthalpy of reaction for the reaction:

$
\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3
$
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{G}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{SO}_2\right)=-297 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \Delta \mathrm{G}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{SO}_2\right)=-297 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \Delta \mathrm{H}_{\mathrm{I}}{ }^{\circ}=? \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right)_{\text {compound }}-\sum\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right)_{\text {elements }} \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{SO}_3\right)-\left[\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{SO}_2\right)+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{O}_2\right)\right] \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=-396 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-297 \mathrm{~kJ} \mathrm{~mol}^{-1}+0\right) \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=-396 \mathrm{~kJ} \mathrm{~mol}^{-1}+297 \\
& \Delta \mathrm{H}_{\mathrm{I}}{ }^{\circ}=-99 \mathrm{kJmol}^{-1}
\end{aligned}
$
Question 69.
For the reaction at $298 \mathrm{~K}: 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$
$
\Delta \mathrm{H}=400 \mathrm{~J} \mathrm{~mol}^{-1} \Delta \mathrm{S}=0.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
$
Determine the temperature at which the reaction would be spontaneous.
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}=298 \mathrm{~K} \\
& \Delta \mathrm{H}=400 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{S}=0.2 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{ThS} \\
& \text { if } \mathrm{T}=2000 \mathrm{~K} \\
& \Delta \mathrm{G}=400-(0.2 \times 2000)=0 \\
& \text { if } \mathrm{T}>2000 \mathrm{~K} \\
& \Delta \mathrm{G} \text { will be negative. }
\end{aligned}
$
The reaction would be spontaneous only beyond $2000 \mathrm{~K}$.
Question 70.
Find out the Value ofequilibrium constant for the following reaction at $298 \mathrm{~K}$,
$
2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{NH}_2 \mathrm{CONH}_{2(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}
$
Standard Gibbs energy change, $\mathrm{AGr}^{\circ}$ at the given temperature is $-13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}=298 \mathrm{~K} \\
& \Delta \mathrm{G}_{\mathrm{r}}^{\circ}=-13.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& =-13600 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}}
\end{aligned}
$

$\begin{aligned}
&\log \mathrm{K}_{\mathrm{eq}}=\frac{-\Delta G^{\circ}}{2.303 R T}\\
&\begin{aligned}
& =\frac{13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \times 10^{-3} \cdot \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}} \\
& \log \mathrm{K}_{\mathrm{eq}} \\
& \log \mathrm{K}_{\mathrm{eq}}=2.38 \\
& \mathrm{~K}_{\text {eq }}=\text { anti } \log (2.38) \\
& \mathrm{K}_{\mathrm{eq}}=239.88
\end{aligned}
\end{aligned}$

Question 71.
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at $25^{\circ} \mathrm{C}$ and at $\mathrm{I}$ atm pressure produce 6.11 lit of carbon dioxide. Find out the amount of heat evolved in $\mathrm{kJ}$, during this combustion. $\left(\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)=-890 \mathrm{~kJ} \mathrm{~mol}^{-1}\right.$ and $\left(\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}^2 \mathrm{H}_4\right)=-1423 \mathrm{~kJ} \mathrm{~mol}^{-1}\right.$.
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)=-890 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_2 \mathrm{H}_4\right)=-1423 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Let the mixture contain $\mathrm{x}$ lit of $\mathrm{CH}_4$ and $(3.67-\mathrm{x})$ lit of ethylene.
$
\begin{array}{ll}
\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow & \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \\
x \text { lit } & \text { x lit } \\
\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow & 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \\
(3.67-\mathrm{x}) \text { lit } & 2(3.67-\mathrm{x}) \text { lit }
\end{array}
$
Volume of carbon dioxide formed $x+2(3.67-x) 6.11$ lit
$
\begin{aligned}
& x+7.34-2 x=6.11 \\
& x=1.23 \text { lit }
\end{aligned}
$
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{r}}^0=\left[\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\text {products }}-\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\text {reactants }}\right] \\
& \Delta \mathrm{H}_{\mathrm{r}}^0=\left[2\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\mathrm{CO}_2}+3\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\mathrm{H}_2 \mathrm{O}}\right]-\left[1\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}+3\left(\Delta \mathrm{H}_{\mathrm{f}}^0\right)_{\mathrm{O}_2}\right] \\
& \Delta \mathrm{H}_{\mathrm{r}}^0=\left[\begin{array}{l}
2 \operatorname{mol}(-393.5) \mathrm{kJ} \mathrm{mol}^{-1} \\
+3 \operatorname{mol}(-285.5) \mathrm{kJ} \mathrm{mol}^{-1}
\end{array}\right]-\left[\begin{array}{l}
1 \operatorname{mol}(-277) \mathrm{kJ} \mathrm{mol}^{-1} \\
+3 \operatorname{mol}(0) \mathrm{kJ} \mathrm{mol}^{-1}
\end{array}\right] \\
& \Delta \mathrm{H}_{\mathrm{C}}=\left[-48.87 \mathrm{~kJ} \mathrm{~mol}^{-1}\right]+\left[-155 \mathrm{~kJ} \mathrm{~mol}^{-1}\right] \\
& \Delta \mathrm{H}_{\mathrm{C}}=-203.87 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&
\end{aligned}
$
In-Text Questions - Evaluate Yourself
Question 1.
Calculate $\mathrm{AH}$ for the reaction $\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ given that $\Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}$ for $\mathrm{CO}_2(\mathrm{~g})$,

$\mathrm{CO}(\mathrm{g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ are $-393.5,-111.31$ and $-242 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ} \mathrm{CO}_2=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ} \mathrm{CO}=-111.31 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-242 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=? \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=\sum\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right)_{\text {products }}-\sum\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right)_{\text {products }} \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=\left[\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}(\mathrm{CO})+\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right]-\left[\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{CO}_2\right)+\Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}\left(\mathrm{H}_2\right)\right] \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=[-111.31+(-242)]-[-393.5+(0)] \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=[-353.31]+3935 \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=40.19 \\
& \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=40.19 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$

Question 2 .
Calculate the amount of heat necessary to raise $180 \mathrm{~g}$ of water from $25^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Molar heat capacity of water is $75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Answer:
Given:
Number of moles of water $\mathrm{n}=\frac{180 \mathrm{~g}}{18 \mathrm{gmol}^{-1}}=10 \mathrm{~mol}$ molar heat capacity of water
$
\begin{aligned}
& \mathrm{C}_{\mathrm{p}}=75.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& \mathrm{~T}_2=100^{\circ} \mathrm{C}=373 \mathrm{~K} \\
& \mathrm{~T}_1=25^{\circ} \mathrm{C}=298 \mathrm{~K} \\
& \Delta \mathrm{H}=? \\
& \Delta \mathrm{H}=\mathrm{nC}_{\mathrm{p}}\left(\mathrm{T}_2-\mathrm{T}_1\right) \\
& \Delta \mathrm{H}=10 \mathrm{~mol} \mathrm{x} 75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times(373-298) \mathrm{K} \\
& \Delta \mathrm{H}=56475 \mathrm{~J} \\
& \Delta \mathrm{H}=56.475 \mathrm{~kJ}
\end{aligned}
$
Question 3.
From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition.
$
\mathrm{C}_6 \mathrm{H}_{6(1)}+\frac{7}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow 6 \mathrm{CO}_{6(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}
$
$\Delta \mathrm{U}$ at $25^{\circ} \mathrm{C}=-3268.12 \mathrm{~kJ}$
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}=25^{\circ} \mathrm{C}=298 \mathrm{~K} \text {; } \\
& \Delta \mathrm{U}=-3268.12 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \Delta \mathrm{H}=\text { ? } \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
&
\end{aligned}
$

$
\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\left(\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{I}}\right) \mathrm{RT} \\
& \Delta \mathrm{H}=3268.12+\left[6-\frac{7}{2} \times 8.314 \times 10^{-3} \times 298\right. \\
& \Delta \mathrm{U}=-3268.12+\left(2.5 \times 8.314 \times 10^{-3} \times 298\right) \\
& \Delta \mathrm{U}=-3268.12+6 . \mathrm{L} 9 \\
& \Delta \mathrm{U}=-3261.93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 4.
When a mole of magnesium bromide is prepared from 1 mole of magnesium and 1 mole of liquid bromine, $524 \mathrm{~kJ}$ of energy is released. The heat of sublimation of $\mathrm{Mg}$ metal is $148 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$. The heat of dissociation of bromine gas into atoms is $193 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The heat of vapourisation of liquid bromine is $31 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The ionisation energy of magnesium is $2187 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ and the electron affinity of bromine is $-662 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the lattice energy of magnesium bromide.
Answer:
Given,
$
\mathrm{Mg}(\mathrm{s})+\mathrm{Br}_2(\mathrm{l}) \mathrm{MgBr}_2(\mathrm{~s})-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-524 \mathrm{KJ} \mathrm{mol}^{-1}
$
Sublimation:
$
\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{g})-\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=+148 \mathrm{KJ} \mathrm{mol}^{-1}
$
Ionisation:
$
\mathrm{Mg}(\mathrm{g}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} 2187-\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=2187 \mathrm{KJ} \mathrm{mol}^{-1}
$
Vapourisation:
$
\mathrm{Br}_2(\mathrm{l}) \rightarrow \mathrm{Br}_2(\mathrm{~g})-\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=+31 \mathrm{KJ} \mathrm{mol}^{-1}
$
Dissociation:
$
\mathrm{Br}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Br}(\mathrm{g})-\Delta \mathrm{H}_{\mathrm{f}}=+193 \mathrm{KJ} \mathrm{mol}^{-1}
$
Electron affinity:
$
\mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g})-\Delta \mathrm{H}_{\mathrm{f}}=-331 \mathrm{KJ} \mathrm{mol}^{-1}
$

$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+\Delta \mathrm{H}_3+\Delta \mathrm{H}_4+2 \mathrm{AH}_5+\mathrm{u} \\
& -524=148+2187+31+193+(2 \mathrm{x}-331)+\mathrm{u} \\
& -524=1897+\mathrm{u} \\
& \mathrm{u}=-524-1897 \\
& \mathrm{u}=-2421 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 5.
An engine operating between $127^{\circ} \mathrm{C}$ and $47^{\circ} \mathrm{C}$ takes some specified amount of heat from a high temperature reservoir. Assuming that there are no frictional losses, calculate the percentage efficiency of an engine.
Answer:
Given,
$
\begin{aligned}
& \mathrm{T}_1=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K} \\
& \mathrm{~T}_2=47^{\circ} \mathrm{C}=47+273=320 \mathrm{~K}
\end{aligned}
$
$\%$ efficiency $\eta=$ ?
$
\begin{aligned}
& \eta=\left[\frac{T_1-T_2}{T_1}\right] \times 100 \\
& \eta=\left[\frac{400-320}{400}\right] \times 100 \\
& \eta=\left[\frac{80}{400}\right] \times 100 \\
& \eta=20 \%
\end{aligned}
$
Question 6.
Urea on hydrolysis produces ammonia and carbon dioxide. The standard entropies of urea, $\mathrm{H}_2 \mathrm{O} . \mathrm{CO}_2, \mathrm{NH}_3$ are $173.8,70,213.5$ and $192.5 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}$ respectively. Calculate the entropy

change for this reaction.
Answer:
Given,
$
\begin{aligned}
& \mathrm{S}^{\circ}(\text { urea })=173.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
& \mathrm{~S}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=70 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
& \mathrm{~S}^{\circ}\left(\mathrm{CO}_2\right)=213.5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
& \mathrm{~S}^{\circ}\left(\mathrm{NH}_3\right)=192.5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
& \mathrm{NH}_2-\mathrm{CO}-\mathrm{NH}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NH}_3+\mathrm{CO}_2 \\
& \Delta \mathrm{S}_{\mathrm{r}}^{\circ}=\sum\left(\mathrm{S}^{\circ}\right)_{\text {product }}-\sum\left(\mathrm{S}^{\circ}\right)_{\text {reactants }} \\
& \Delta \mathrm{S}_{\mathrm{r}}^{\circ}=\left[2 \mathrm{~S}^{\circ}\left(\mathrm{NH}_3\right)+\mathrm{S}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\mathrm{S}^{\circ}(\text { urea })+\mathrm{S}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right] \\
& \Delta \mathrm{S}_{\mathrm{r}}^{\circ}=[2 \times 192.5+213.5]-[173.8+70] \\
& \Delta \mathrm{S}_{\mathrm{r}}^{\circ}=[598.5]-[243.8] \\
& \Delta \mathrm{S}_{\mathrm{r}}^{\circ}=354.7 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{aligned}
$

$
\begin{aligned}
& \Delta \mathrm{G}=-10 \mathrm{~kJ} \mathrm{~mol}-1-600 \mathrm{~K} \times\left(-20 \times 10^{-3}\right) \mathrm{kJ} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}=(-10+12) \mathrm{kJ} \mathrm{mol}^{-1} \\
& \Delta \mathrm{G}=+2 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
The value of $\Delta \mathrm{G}$ is negative at $300 \mathrm{~K}$ and the reaction is spontaneous, but at $600 \mathrm{~K}$ the value $\Delta \mathrm{G}$ becomes positive and the reaction is non-spontaneous.

In-Text Example Problems
Question 1.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of $1 \mathrm{~atm}$ from a volume of 5 litres to a volume of 10 litres. In doing solt absorbs $400 \mathrm{~J}$ of thermal energy from its surroundings. Determine the change in internal energy of system.
Answer:
Given data $\mathrm{q}=400 \mathrm{~J} ; \mathrm{V}_1=5 \mathrm{~L} ; \mathrm{V}_2=10 \mathrm{~L}$
$\mathrm{Au}=\mathrm{q}-\mathrm{w}$ (heat is given to the system $( \pm \mathrm{q})$; work is done by the system(-w)
$\mathrm{Au}=\mathrm{q}-\mathrm{PdV}$
$=400 \mathrm{~J}-1 \mathrm{~atm}(10-5) \mathrm{L}$
$=400 \mathrm{~J}-5 \mathrm{~atm} \mathrm{~L}[\therefore 1 \mathrm{~L} \mathrm{~atm}=101.33 \mathrm{~J}]$
$=400 \mathrm{~J}-5 \times 101.33 \mathrm{~J}$
$=4003-506.65 \mathrm{~J}$
$=-106,65 \mathrm{~J}$
Question 2.
The standard enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(1)}, \mathrm{CO}_{2(\mathrm{~g})}$ and $\mathrm{H}_2 \mathrm{O}_{(1)}$ are $-277,-393.5$ and $-285.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Calculate the standard enthalpy change for the reaction.
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\mathrm{l})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$. The enthalpy of formation of $\mathrm{O}_{2(\mathrm{~g})}$ in the standard state is zero by definition.
Answer:
The standard enthalpy change for the combustion of ethanol can be calculated from the strndard enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(1)}, \mathrm{CO}_{2(\mathrm{~g})}$ and $\mathrm{H}_2 \mathrm{O}_{(1)}$. The enthalpies of formation are $277,-393.5$ and $-285.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
$
\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\mathrm{l})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} . \\
& \Delta \mathrm{H}_{\mathrm{C}}=\left[\frac{\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{CH}_4\right)}{22.4 \text { lit }} \times(\mathrm{x}) \text { lit }\right]+\left[\frac{\Delta \mathrm{H}_{\mathrm{C}}\left(\mathrm{C}_2 \mathrm{H}_4\right)}{22.4 \text { lit }} \times(3.67-\mathrm{x}) \text { lit }\right] \\
& \Delta \mathrm{H}_{\mathrm{C}}=\left[\frac{-890 \mathrm{~kJ} \mathrm{~mol}}{22.4 \text { lit }} \times 1.23 \text { lit }\right]+\left[\frac{-1423}{22.4 \text { lit }} \times(3.67-1.23) \text { lit }\right] \\
& =[-787-856.5]-[-277] \\
& =-1643.5+277 \\
& \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=-1366.5 \mathrm{KJ} .
\end{aligned}
$

Question 3.
Calculate the value of $\mathrm{AU}$ and $\mathrm{AH}$ on heating $128 \mathrm{~g}$ of oxygen from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. $\mathrm{C}$. and $\mathrm{C}$, on an average are 21 and $29 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. (The difference is $8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ which is approximately equal to R)
Answer:
We know
$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_2-\mathrm{T}_1\right) \\
& \Delta \mathrm{H}=\mathrm{n} \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)
\end{aligned}
$
Here $\mathrm{n}=\frac{128}{32} 4$ moles;
$
\begin{aligned}
& \mathrm{T}_2=100^{\circ} \mathrm{C}=373 \mathrm{~K} \\
& \mathrm{~T}_1=\mathrm{O}^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_2-\mathrm{T}_1\right) \\
& \Delta \mathrm{U}=4 \times 21 \times(373-273) \\
& \Delta \mathrm{U}=8400 \mathrm{~J} \\
& \Delta \mathrm{U}=8.4 \mathrm{~kJ} \\
& \Delta \mathrm{H}=\mathrm{nC}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right) \\
& \Delta \mathrm{H}=4 \times 29 \times(373-273) \\
& \Delta \mathrm{H}=11600 \mathrm{~J} \\
& \Delta \mathrm{H}=11.6 \mathrm{KJ} .
\end{aligned}
$

Question 4.
Calculate the enthalpy of combustion of ethylene at $300 \mathrm{~K}$ at constant pressure, If Its heat of combustion at constant volume $(\Delta \mathrm{U})$ is $-1406 \mathrm{KJ}$.
Answer:
The complete ethylene combustion reaction can be written as,
$
\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\
& \Delta \mathrm{U}=-1406 \mathrm{~kJ} \\
& \Delta \mathrm{n}=\mathrm{n}_{\mathrm{p}(\mathrm{g})}-\mathrm{n}_{\mathrm{r}(\mathrm{g})} \\
& \Delta \mathrm{n}=2-4=-2 \\
& \Delta \mathrm{H}=\mathrm{U}+\mathrm{RT} \Delta \mathrm{n}_{(\mathrm{g})} \\
& \Delta \mathrm{H}=-1406+\left(8.314 \times 10^{-3} \times 300 \times(-2)\right) \\
& \Delta \mathrm{H}=-1410.9 \mathrm{~kJ}
\end{aligned}
$

Question 5.
Calculate the standard enthalpy of formation $\Delta \mathrm{H}^{\circ} \mathrm{f}$ of $\mathrm{CH}_4$ from the values of enthalpy of combustion for $\mathrm{H}_2, \mathrm{C}$ (graphite) and $\mathrm{CH}_4$ which are $-285.8,-393.5$, and $-890.4 \mathrm{KJ} \mathrm{mol}^{-1}$ respectively.
Answer:
Let us interpret the information about enthalpy of formation by writing out the equations. It is important to note that the standard enthalpy of formation of pure elemental gases and elements is assumed to be zero under standard conditions. Thermochemical equation for the formation of methane from its constituent elements is,
$
\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{4(\mathrm{~g})}
$

$
\Delta \mathrm{H}_{\mathrm{f}}^0=\mathrm{X} \mathrm{kJ} \mathrm{mol}{ }^{-1}
$
Thermo chemical equations for the combustion of given substances are,
$
\begin{aligned}
& 2 \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\
& \Delta \mathrm{H}^0=-285.8 \mathrm{KJ} \mathrm{mol}^{-1} \ldots \ldots \ldots \ldots \ldots \text { (ii) } \\
& \mathrm{C}\left(\text { graphite) }+\mathrm{O}_2 \rightarrow \mathrm{CO}_2\right. \\
& \Delta \mathrm{H}^0=-393.5 \mathrm{KJ} \mathrm{mol}^{-1} \ldots \ldots \ldots \ldots \ldots . \text { (iii) } \\
& \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\
& \Delta \mathrm{H}^0=-890.4 \mathrm{KJ} \mathrm{mol}^{-1} \ldots \ldots \ldots \ldots \ldots \text { (iv) }
\end{aligned}
$
Since methane is in the product side of the required equation (i), we have to reverse the equation (iv)
$
\begin{aligned}
& \mathrm{CO}_{4(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_2 \\
& \Delta \mathrm{H}^{\circ}=+890.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots \ldots \ldots \ldots \ldots .
\end{aligned}
$
In order to get equation (i) from the remaining, (i) $=[$ (ii) $\times 2]+$ (iii) + (v)
$
\mathrm{X}=[(-285.8) \times 2]+[-393.5]+[+890.4]=-74.7 \mathrm{~kJ}
$
Hence, the amount of energy required for the formation of I mole of methane is $-74.7 \mathrm{~kJ}$ The heat of formation methane $-74.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Question 6.
Enthalpy for the oxidation of graphite to $\mathrm{CO}_2$ and $\mathrm{CO}$ to $\mathrm{CO}_2$ can easily be measured. For these conversions, the heat of combustion values are $-393.5 \mathrm{~kJ}$ and $-283.5 \mathrm{~kJ}$ respectively.
Answer:
From these data the enthalpy of combustion of graphite to $\mathrm{CO}$ can be calculated by applying Hess's law. The reactions involved in this process can be expressed as follows:

According to Hess law,
$
\begin{aligned}
& \Delta \mathrm{H}_1=\Delta \mathrm{H}_2+\Delta \mathrm{H}_3 \\
& -393.5 \mathrm{~kJ}=\mathrm{X}-283.5 \mathrm{~kJ} \\
& \mathrm{X}=-110.5 \mathrm{~kJ}
\end{aligned}
$
Question 7.
Calcu late the lattice energy of sodium chloride using Born-Haber cycle.
Answer:

$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}=\text { heat of formation of sodium chloride }=411.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_1=\text { heat of sublimation of } \mathrm{Na}(\mathrm{s})=108.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_2=\text { ionisation energy of } \mathrm{Na}(\mathrm{s})=495 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_3=\text { dissociation energy of } \mathrm{Cl}_2(\mathrm{~s})=244 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_4=\text { Electron affinity of } \mathrm{Cl}(\mathrm{s})=-349 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{U}=\text { lattice energy of } \mathrm{NaCl} \\
& \Delta \mathrm{H}_{\mathrm{f}}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+1 / 2 \Delta \mathrm{H}_3+\Delta \mathrm{H}_4+\mathrm{U} \\
& \therefore \mathrm{U}=\left(\Delta \mathrm{H}_{\mathrm{f}}\right)-\left(\Delta \mathrm{H}_1+\Delta \mathrm{H}_2+1 / 2 \Delta \mathrm{H}_3+\Delta \mathrm{H}_4\right) \\
& \Rightarrow \mathrm{U}=(-411.3)-(108.7+495+122-349) \\
& \mathrm{U}=(-411.3)-(376.7) \\
& \therefore \mathrm{U}=-788 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
This negative sign in lattice energy indicates that the energy is released when sodium is tormea from its constituent gaseous ions $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$
Question 8.
If an automobile engine burns petrol at a temperature of $816^{\circ} \mathrm{C}$ and if the surrounding temperature is $21^{\circ} \mathrm{C}$, calculate its maximum possible efficiency.
Answer:
$\%$ Efficiency $-\left[\frac{\mathrm{T}_{\mathrm{h}}-\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{h}}}\right] \times 100$
Here
$
\begin{aligned}
& \mathrm{T}_{\mathrm{h}}=816+273=1098 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{c}}=21+273=294 \mathrm{~K} \\
& \% \text { Efficiency }=73 \% .
\end{aligned}
$

Question 9.
Calculate the standard entropy change for the following reaction $\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right)$, given the standard entropies of $\mathrm{CO}_{\mathrm{f}}(\mathrm{g}), \mathrm{C}(\mathrm{s}), \mathrm{O}_2(\mathrm{~g})$ as $213.6,5.740$ and $205 \mathrm{JK}^{-1}$ respectively.

Answer:

$
\begin{aligned}
& \Delta \mathrm{S}_{\mathrm{r}}{ }^{\circ}=213.6-[5.74+205] \\
& \Delta \mathrm{S}_{\mathrm{r}}{ }^{\circ}=213.6-[210.74] \\
& \Delta \mathrm{S}_{\mathrm{r}}{ }^{\circ}=2.86 \mathrm{JK}^{-1} \text {. } \\
&
\end{aligned}
$
Question 10.
Calculate the entropy change during the melting of one mole of ice into water at $0^{\circ} \mathrm{C}$ and $\mathrm{I}$ atm pressure. Enthalpy of fusion of ice is $6008 \mathrm{~J} \mathrm{~mol}^{-1}$.
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{S}_{\text {fusion }}=6008 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \mathrm{~T}_{\mathrm{f}}=0^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \stackrel{273 \mathrm{~K}}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
& \Delta \mathrm{S}_{\text {fusion }}=\frac{\Delta \mathrm{H}_{\text {fusion }}}{\mathrm{T}_{\mathrm{f}}} \\
& \Delta \mathrm{S}_{\text {fusion }}=\frac{6008}{273} \\
& \Delta \mathrm{S}_{\text {fusion }}=22.007 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$

Question 11.
Show that the reaction $\mathrm{CO}+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2$ at $300 \mathrm{~K}$ is spontaneous. The standard Gibbs free energies of formation of $\mathrm{CO}_2$ and $\mathrm{CO}$ are -394.4 and $-137.2 \mathrm{~kJ}$ mole respectively.
Answer:
$
\begin{aligned}
& \mathrm{CO}+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \\
& \Delta \mathrm{G}_{\text {(reaction) }}{ }^{\circ}=\sum \mathrm{G}_{\mathrm{f}(\text { products) }}{ }^{\circ}-\sum \mathrm{G}_{\mathrm{f}(\text { reactants })}{ }^{\circ} \\
& \Delta \mathrm{G}_{\text {(reaction) }}^0=\left[\mathrm{G}_{\mathrm{CO}_2}^0\right]-\left[\mathrm{G}_{\mathrm{CO}}^0+\frac{1}{2} \mathrm{G}_{\mathrm{O}_2}^0\right] \\
& \Delta \mathrm{G}_{(\text {reaction })}{ }^{\circ}=-394.4+[137.2+0] \\
& \Delta \mathrm{G}_{\text {(reaction })}{ }^{\circ}=-257.2 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
$\Delta \mathrm{G}_{\text {(reaction) }}{ }^{\circ}$ of a reaction at a given temperature is negative hence the reaction is spontaneous.
Question 12.
Calculate $\mathrm{G}^{\circ}$ for conversion of oxygen to ozone $3 / 2 \mathrm{O}_2 \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})}$ at $298 \mathrm{~K}$, if $\mathrm{K}_{\mathrm{p}}$ for this conversion is $2.47 \times 10^{-29}$ in standard pressure units.
Answer:
$
\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{p}}
$
Where
$
\begin{aligned}
& \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \mathrm{~K}=2.47 \times 10^{-29} \\
& \mathrm{~T}=298 \mathrm{~K} \\
& \Delta \mathrm{G}^{\circ}=-2.303(8.314)(298) \log \left(2.47 \times 10^{-29}\right) \\
& \Delta \mathrm{G}^{\circ}=16300 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}^{\circ}=16.3 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
Additional:
Question 1.
Calculate the maximum $\%$ efficiency of thermal engine operating between $110^{\circ} \mathrm{C}$ and $25^{\circ} \mathrm{C}$.
Answer:
$
\begin{aligned}
& \% \text { Efficiency }=\left[\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}\right] \times 100 \\
& \mathrm{~T}_1=110^{\circ} \mathrm{C}+273=383 \mathrm{~K} . \\
& \mathrm{T}_2=25^{\circ} \mathrm{C}-273=298 \mathrm{~K} . \\
& \% \text { Efficiency }=\left[\frac{383-298}{383}\right] \times 100 \\
& \% \text { Efficiency }=\left[\frac{85 \times 100}{383}\right]=\left[\frac{8500}{383}\right] \% \text { Efficiency }=22.2 \%
\end{aligned}
$
Question 2.
Calculate the entropy change in the system. and in the surroundings and the total entropy change in the universe when during a process $75 \mathrm{~J}$ of heat flow out of the system at $55^{\circ} \mathrm{C}$ to the surrounding at $20^{\circ} \mathrm{C}$.
Answer:

Heat flow (q) $=75 \mathrm{~J}$
Entropy change $=\Delta \mathrm{S}=$ ?
Temperature of the system $=55^{\circ} \mathrm{C}+273=328 \mathrm{~K}$
Temperature of the surroundings $=20^{\circ} \mathrm{C}+273=293 \mathrm{~K}$
$
\Delta \mathrm{S}_{\text {system }}=\frac{4}{\mathrm{~T}_{\text {system }}}=\frac{75}{328}=0.2286 \mathrm{~J} \mathrm{~K}^{-1} .
$
$
\Delta \mathrm{S}_{\text {surroundings }}=\frac{q}{\mathrm{~T}_{\text {surroundings }}}=\frac{75}{293}=0.2559 \mathrm{~J} \mathrm{~K}^{-1} \text {. }
$
$\Delta \mathrm{S}_{\text {universe }}=$ Entropy change $=\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surroundings }}$
$
=0.2286+0.2559=\Delta \mathrm{S}_{\text {universe }}=0.4845 \mathrm{JK}^{-1}
$

Question 3.
1 mole of an ideal gas is maintained at $4.1 \mathrm{~atm}$ and at a certain temperature absorbs $3710 \mathrm{~J}$ heat and expands to 2 litres. Calculate the entropy change in expansion process.
Answer:
Pressure of an ideal gas $P_i=4.1 \mathrm{~atm}$.
Expansion in volume $=\Delta \mathrm{V}=2$ litres
Heat absorbed $=\mathrm{q}=3710 \mathrm{~J}$
Entropy change $=\Delta \mathrm{S}=$ ?
For an ideal gas PV $=\mathrm{RT}$ for one mole.
$
\begin{aligned}
& \mathrm{T}=\frac{P V}{R}=\frac{4.1 x 2}{0.0830}=100^{\circ} \mathrm{C} \\
& \mathrm{T}=100+273=373 \mathrm{~K} . \\
& \Delta \mathrm{S}=\frac{q}{\mathrm{~T}_{(\mathrm{K})}}=\frac{3710}{373}
\end{aligned}
$
Entropy change $=9.946 \mathrm{JK}^{-1}$.
Question 4.
Calculate the entropy change of a process $\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $373 \mathrm{~K}$. Enthalpy of vaporization of water is $40850 \mathrm{~J} \mathrm{Mole}^{-1}$.
Answer:
$\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
Temperature $\mathrm{T}=373 \mathrm{~K}$
Enthalpy of vapourisation of water $=\Delta \mathrm{V}_{\text {vap }}=40580 \mathrm{~J} \mathrm{~mol}^{-1}$
$\Delta \mathrm{S}=$ entropy change $=\frac{-\Delta \mathrm{H}_{\mathrm{vap}}}{\mathrm{T}_{\mathrm{b}}}=\frac{40850}{373}$
$\Delta \mathrm{S}($ entropy change $)=109.51 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Question 5.
$30.4 \mathrm{KJ}$ is required to melt one mole of sodium chloride. The entropy change during melting is $28.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ Calculate the melting point of sodium chloride.
Answer:
Heat required for I mole of $\mathrm{NaCl}$ for melting (q) $=30.4 \mathrm{~K} \mathrm{~J}$ $=30.4 \times 1000 \mathrm{~J}$
$\Delta \mathrm{S}-$ entropy change $=28.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
Melting point $=\mathrm{T}_{\mathrm{m}}=$ ?
$
\begin{aligned}
& \Delta \mathrm{S}=\frac{q}{\Gamma_m} \\
& \therefore \mathrm{T}_{\mathrm{m}}=\frac{q}{\Delta S} \\
& \mathrm{~T}_{\mathrm{m}}=\frac{30.4 x 1000}{28.4}=10704 \mathrm{~K}
\end{aligned}
$
Melting point of $\mathrm{NaCl}=1070.4 \mathrm{~K}$.
Question 6.
Calculate the standard heat of formation of propane, if its heat of combustion is $0-2220.2 \mathrm{KJ}$ $\mathrm{mol}^{-1}$ [he heats of formation of $\mathrm{CO}_{2(\mathrm{~g})}$ and $\mathrm{H}_{2(\mathrm{~g})} \mathrm{O}_{(\mathrm{l})}$ are -393.5 and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
respectively.

Solution:
Standard heat of formation of propane,
$
3 \mathrm{C}_{(\mathrm{g})}+4 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})} \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=\text { ? }
$
Data given:
$
\begin{aligned}
& \mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}=-2220.2 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}=-393.5 \mathrm{KJ} \mathrm{mol}^{-1} \\
& \mathrm{H}_{2(\mathrm{~g})}(\mathrm{g})+1 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}=-285.8 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
According to Hes\&s law, equation
Equation (1) is reversed.
Equation (2) is $\times 3$
Equation (3) is $\times 4$
The add all the equations.
$
\begin{aligned}
& 3 \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \Delta \mathrm{H}_1=+2220.2 \mathrm{KJ} \mathrm{mol}^{-1} \\
& 3 \mathrm{C}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 3 \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}_2=-1180.5 \mathrm{KJ} \mathrm{mol}^{-1} \\
& 4 \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 4 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}_3=-1143.2 \mathrm{KJ} \mathrm{mol}^{-1} \\
& 3 \mathrm{C}_{(\mathrm{S})}+4 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-103.5 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$
Standard enthalpy of formation of propane $\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=-103.5 \mathrm{~K} . \mathrm{J} \mathrm{mol}^{-1}$
Question 7.
The boiling point of water at a pressure of $50 \mathrm{~atm}$ is $265^{\circ} \mathrm{C}$. Compare the theoretical efficiencies of a steam engine operating between the boiling point of water at
1. $1 \mathrm{~atm}$ pressure
2. $50 \mathrm{~atm}$ pressure, assuming the temperature of the $\operatorname{sink}$ to be $35^{\circ} \mathrm{C}$ in each case.

Answer:
Boiling point of water at 50 atm pressure $\left(\mathrm{T}_{\mathrm{b}}\right)=265^{\circ} \mathrm{C}=265+273=538 \mathrm{~K}$
1. Boiling point of water at 1 atm pressure $\left(\mathrm{T}_{\mathrm{b}}\right)=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}$
$\%$ Efficiency of steam engine $=\left[\frac{T_1-\mathrm{T}_2}{\mathrm{~T}_1}\right] \times 100=\left[\frac{538-373}{538}\right] \times 100=0.3066 \times 100=30.66 \%$
2. BoiLing point of water at $50 \mathrm{~atm}$ pressure $\left(\mathrm{T}_{\mathrm{b}}\right)=35^{\circ} \mathrm{C}=35+273=328 \mathrm{~K}$
$\%$ Efficiency of steam engine $=\left[\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}\right] \times 100\left[\frac{538-328}{538}\right] \times 100=\left[\frac{210 x 100}{538}\right]=39.03 \%$
Question 8.
The standard enthalpies of formation of $\mathrm{SO}_2$ and $\mathrm{SO}_3$ are $-297 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-396 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.Calculate the standard enthalpy of reaction for the reaction: $\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow 4 \mathrm{SO}_3$ Solution:
Data given,
(1) $\Rightarrow \mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=-297 \mathrm{KJ} \mathrm{mol}^{-1}$
(2) $\Rightarrow \mathrm{S}+1^{1 / 2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=-396 \mathrm{KJ} \mathrm{mol}^{-1}$

$
\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \Delta \mathrm{H}_{\mathrm{r}}=?
$
Equation (1) is reversed and added with equestion (2)

Question 9.
For the reaction at $298 \mathrm{~K}: 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$
$\Delta \mathrm{H}=400 \mathrm{KJ} \mathrm{mol}^{-1}: \Delta \mathrm{S}=0.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Determine the temperature at which the reaction would be spontaneous.
Solution:
Data given,
$2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$ at $298 \mathrm{~K}$
$\Delta \mathrm{H}=400 \mathrm{KJ} \mathrm{mol}^{-1}$
$\Delta \mathrm{S}=0.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{T}=298 \mathrm{~K}$
$[\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}]$
if $\Delta \mathrm{G}=0$
$\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0$
$\Delta \mathrm{S}=\frac{\Delta H}{T}$
For $\Delta \mathrm{G}=0, \Delta \mathrm{S}=\frac{\Delta H}{T}=\frac{400}{0.2}=\frac{4000}{2}=2000 \mathrm{~K}$
$\therefore \mathrm{T}=2000 \mathrm{~K}$
At $2000 \mathrm{~K}$. the rcaction is in equilibrium. So. above $2000 \mathrm{~K}$, the reaction will be spontaneous.
Question 10.
Calculate the heat of glucose and its calorific value
1. $\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}=-395 \mathrm{KJ}$
2. $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{hO}_2 \rightarrow \mathrm{H}_2 \mathrm{O}_{(1)} ; \Delta \mathrm{H}=-269.4 \mathrm{KJ}$
3. $\mathrm{C}+6 \mathrm{H}_{2(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~S})} \Delta \mathrm{H}=-1169.8 \mathrm{KJ}$

Solution:
Calorific value of glucose $\Delta \mathrm{H}_{\mathrm{C}}=$ ?
$
\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(1)} \quad \Delta \mathrm{H}_{\mathrm{C}}=?
$
Glucose
Data given are.
1. $\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}=-395 \mathrm{KJ}$
2. $\mathrm{H}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}_{(1)} \Delta \mathrm{H}_{\mathrm{C}} \Delta \mathrm{H}=-269.4 \mathrm{KJ}$
3. $\mathrm{C}+6 \mathrm{H}_{2(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~S})} \Delta \mathrm{H}=-1169.8 \mathrm{KJ}$
Equation (1) $666 \mathrm{C}+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 \Delta \mathrm{H}=-2370 \mathrm{KJ}$
Equation (2) $\times 66 \mathrm{H}_2+3 \mathrm{O}_2 \rightarrow 6 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-1616.4 \mathrm{KJ}$
Equation (3) is reversed, $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \rightarrow 6 \mathrm{C}+6 \mathrm{H}_2+3 \mathrm{O}_2 \Delta \mathrm{H}=+1169.8 \mathrm{KJ}$

Add all equations,
$
\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}=-2370-1616.4+1169.8 \\
& \Delta \mathrm{H}=-3986.4+1169.8 \\
& \Delta \mathrm{H}=-2816.6 \mathrm{KJ}
\end{aligned}
$
Calorific value of $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6=-2816.6 \mathrm{KJ} \mathrm{mol}^{-1}$
Question 11.
Calculate the entropy change when 1 mole of ethanol is evaporated at $351 \mathrm{~K}$. The molar heat of vapourisation of ethanol is $39.84 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Answer:
Given,
$
\begin{aligned}
& \Delta \mathrm{H}_{\text {vap }}=39840 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{V}}=\frac{\Delta \mathrm{H}_{\text {vap }}}{\mathrm{T}_b} \\
& \mathrm{~T}_{\mathrm{b}}=351 \mathrm{~K} \\
& \Delta \mathrm{H}_{\mathrm{V}}=\frac{39840}{351} \\
& \Delta \mathrm{H}_{\mathrm{V}}=113.5 \mathrm{~J} \mathrm{k}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 12 .
Calculate the entropy change of a process possessing $\Delta \mathrm{H}_{\mathrm{t}}=2090 \mathrm{~J} \mathrm{~mole}^{-1}$
Answer:
Given,
$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}\left(\alpha 13^{\circ} \mathrm{C}\right)}-\mathrm{S} \mathrm{n}\left(\beta 13^{\circ} \mathrm{C}\right) \\
& \Delta \mathrm{H}_{\mathrm{t}}=2090 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{S}_{\mathrm{t}}=\frac{\Delta \mathrm{H}_t}{\mathrm{~T}_t} \\
& \Delta \mathrm{S}_{\mathrm{t}}=\frac{2090}{286}
\end{aligned}
$

$
\Delta \mathrm{S}_{\mathrm{t}}=7.307 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
$
Question 13.
Calculate the standard enthalpy of formation of $\mathrm{CH}_3 \mathrm{OH}_{(1)}$ from the following data:
(i) $\mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})}+3 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_3 \mathrm{O}_{(\mathrm{l})} \Delta_{\mathrm{r}} H^{\ominus}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) $\mathrm{C}_{(\mathrm{S})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})} ; \Delta_{\mathrm{C}} H^{\ominus}=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iii) $\mathrm{H}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta_{\mathrm{f}} H^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Answer:
The equation we aim at;
$
\mathrm{C}_{(\mathrm{S})}+2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})} \Delta_{\mathrm{f}} H^{\ominus}= \pm \text { ? }
$
Multiply equation (iii) by 2 and add to equation (ii)
$
\mathrm{C}_{(\mathrm{S})}+2 \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}=-(393+522)=-965 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Subtract equation (iv) from equation (i)
$
\mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})}+3 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}
$

Subtract:
$
\mathrm{C}_{(\mathrm{S})}+2 \mathrm{H}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})} \Delta_{\mathrm{f}} H^{\ominus}=-239 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Question 14.
The equilibrium constant for the reaction is 10. Calculate the value of $\Delta \mathrm{G}$-;
Given $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=300 \mathrm{~K}$.
Answer:
$
\begin{aligned}
& \Delta \mathrm{Ge}=-\mathrm{RT} \text { in } \mathrm{K}=-2.303 \mathrm{RT} \log \mathrm{K} \\
& \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=300 \mathrm{~K} ; \mathrm{K}=10 \\
& \Delta \mathrm{Ge}=-2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \mathrm{x}(300 \mathrm{~K}) \times \log 10 \\
& =-5527 \mathrm{~J} \mathrm{~mol}^{-1}=-5.527 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 15 .
Calculate the entropy change in surroundings when $1 \mathrm{~mol}^{-1}$ of $\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$ is formed under standard conditions. Given $\Delta \mathrm{H}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Answer:
$
\begin{aligned}
& \mathrm{q}_{\mathrm{rev}}=\left(-\Delta_{\mathrm{f}} H^{\ominus}\right)=-286 \mathrm{KJ} \mathrm{mol}=286000 \mathrm{~J} \mathrm{~mol} \\
& \Delta \mathrm{S}_{\text {(Surroundings) }}=\frac{q_{\mathrm{rev}}}{\mathrm{T}}=\frac{\left(286000 \mathrm{~J} \mathrm{~mol}^{-1}\right)}{298 \mathrm{~K}}=959 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$

Question 16.
The enthalpy of formation of methane at constant pressure and $300 \mathrm{~K}$ is $-78.84 \mathrm{~kJ}$. What will be the enthalpy of formation at constant volume?
Answer:
The equation representing the enthalpy of formation of methane is:
$
\begin{aligned}
& \mathrm{C}_{(\mathrm{S})}+2 \mathrm{H}_{2(\mathrm{~g})}-\mathrm{CH}_{2(\mathrm{~g})} ; \\
& \Delta \mathrm{H}=-78.84 \mathrm{~kJ} \\
& \Delta \mathrm{U}=78.84 \mathrm{~kJ} \\
& \Delta^{\mathrm{ng}}=1-2=-1 \mathrm{~mol} \\
& \mathrm{R}=8.314 \times 10^{-13} \mathrm{KJ} \mathrm{K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=300 \mathrm{~K}
\end{aligned}
$
According to the relation,
$
\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta^{\mathrm{ng}} \mathrm{RT}-78.84 \mathrm{~kJ} \\
& \Delta \mathrm{U}=\Delta \mathrm{H}-\Delta^{\mathrm{ng}} \mathrm{RT} \\
& \Delta \mathrm{U}=(-78.84 \mathrm{~kJ})-(1 \mathrm{~mol}) \times\left(8.314 \times 10^{-3} \mathrm{KJ} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \times 300 \mathrm{~K} . \\
& =-78.84-2.49=-8.314 \mathrm{KJ} .
\end{aligned}
$

Question 17.
Calculate $\Delta_{\mathrm{r}} \mathrm{Ge}$ for conversion of oxygen to ozone $3 / 2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{O}_{3(\mathrm{~g})}$ at $298 \mathrm{~K}_{\mathrm{p}}$ If for this conversion is $2.47 \times 10^{-29}$.
Solution:
We know
$
\begin{aligned}
& \Delta_{\mathrm{r}} G^{\ominus}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{p}} \\
& \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{r}} G^{\ominus}=-2.303\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})\left(\log 2.47 \times 10^{-29}\right) \\
& =163000 \mathrm{~J} \mathrm{~mol}^{-1}=163 \mathrm{~mol}^{-1}
\end{aligned}
$
Question 18.
(a) Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
(b) Calculate the entropy change for the following reversible process.
$\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(1)} \Delta_{\mathrm{fus}} \mathrm{H}$ is $6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(a) $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
When the reaction is carried out at $0^{\circ} \mathrm{K}$
or $\Delta \mathrm{S}=0$
$\Delta \mathrm{G}=\Delta \mathrm{H}$

(b)
$\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$
$\Delta_{\text {fus }} \mathrm{H}=6 \mathrm{~kJ} \mathrm{~mol}^{-1}=6000 \mathrm{Jmol}^{-1}$
$\Delta_{\text {fus }} \mathrm{H}=6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$6000 \mathrm{~J} \mathrm{~mol}^{-1}$