Text Book Back Questions and Answers - Chapter 8 Physical and Chemical Equilibrium 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Physical and Chemical Equilibrium
Equilibrium Textual Evaluation Solved
Multiple ChoiceQuestions
Question 1.
If $\mathrm{K}_{\mathrm{b}}$ and $\mathrm{K}_{\mathrm{f}}$ for a reversible reactions are $0.8 \times 10^{-5}$ and $1.6 \times 10^{-4}$ respectively, the value of the equilibrium constant is
(a) 20
(b) $0.2 \times 10^{-1}$
(c) 0.05
(d) None of these
Answer:
(a) 20
Solution:
$
\begin{aligned}
\mathrm{K}_b & =0.8 \times 10^{-5} \\
\mathrm{~K}_f & =1.6 \times 10^{-4} \\
\mathrm{~K}_{e q} & =\frac{\mathrm{K}_f}{\mathrm{~K}_b}=\frac{1.6 \times 10^{-4}}{0.8 \times 10^{-5}}=20
\end{aligned}
$
Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria $3 \mathrm{~A}_2+\mathrm{B}_2+2 \mathrm{C} \stackrel{\mathrm{K}_1}{=} 2 \mathrm{~A}_3 \mathrm{BC}$ and $\mathrm{A}_3 \mathrm{BC} \stackrel{\mathrm{K}_2}{=} \frac{3}{2}\left[\mathrm{~A}_2\right]+\frac{1}{2} \mathrm{~B}_2+\mathrm{C}$
The relation between $\mathrm{K}_1$ and $\mathrm{K}_2$ is
(a) $\mathrm{K}_1=\frac{1}{\sqrt{\mathrm{K}_2}}$
(b) $\mathrm{K}_2=\mathrm{K}_1^{\frac{-1}{2}}$
(c) $\mathrm{K}_1^2=2 \mathrm{~K}_2$
(d) $\frac{\mathrm{K}_1}{2}=\mathrm{K}_2$

Answer:
(b) $\mathrm{K}_2=\mathrm{K}_1^{\frac{-1}{2}}$
Solution:
$
\begin{aligned}
\mathrm{K}_1 & =\frac{\left[\mathrm{A}_3 \mathrm{BC}\right]^2}{\left[\mathrm{~A}_2\right]^3\left[\mathrm{~B}_2\right][\mathrm{C}]^2} \\
\mathrm{~K}_2 & =\frac{\left[\mathrm{A}_2\right]^{\frac{3}{2}}\left[\mathrm{~B}_2\right]^{\frac{1}{2}}[\mathrm{C}]}{\left[\mathrm{A}_3 \mathrm{BC}\right]} \\
\mathrm{K}_2^2 & =\frac{\left[\mathrm{A}_2\right]^3\left[\mathrm{~B}_2\right][\mathrm{C}]^2}{\left[\mathrm{~A}_3 \mathrm{BC}\right]^2}
\end{aligned}
$
Comparing (1) \& (2)
$
\mathrm{K}_2^2=\frac{1}{\mathrm{~K}_1} \Rightarrow \mathrm{K}_2=\mathrm{K}_1^{\frac{-1}{2}}
$
Question 3.
The equilibrium constant for a reaction at room temperature is $\mathrm{K}_1$ and that at $700 \mathrm{~K}_{\text {is }} \mathrm{K}_2$. If $\mathrm{K}_1$ $>\mathrm{K}_2$ then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The reaction does not attain equilibrium
(d) The reverse reaction is exothermic
Answer:
(a) The forward reaction is exothermic
Solution:
$
\begin{aligned}
& \mathrm{T}_1=25+273=298 \mathrm{~K}, \mathrm{~T}_2=700 \mathrm{~K} \\
& \log \left[\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right]=\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right)
\end{aligned}
$
In this case, $\mathrm{T}_2>\mathrm{T}_1$ and $\mathrm{K}_1>\mathrm{K}_2$

$
\frac{2.303 \mathrm{R} \log \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)}{\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right)}=\Delta \mathrm{H}^{\circ} \Rightarrow \frac{-\mathrm{ve}}{+\mathrm{ve}}=\Delta \mathrm{H}^{\circ}
$
$\Delta \mathrm{H}^{\circ}$ is - ve i.e., forward reaction is exothermic.
Question 4.
The formation of ammonia from $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ is a reversible reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})+$ Heat
What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer:
(c) equilibrium is shifted to the left
Solution:
Increase in temperature, favours the endothermic reaction, given that formation of $\mathrm{NH}_3$ is exothermic i.e., the reverse reaction is endothermic.
$\therefore$ Increase in temperature, shift the equilibrium to left

Question 5 .
Solubility of carbon dioxide gas in cold water can be increased by
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure
Solution:

increase in pressure, favours the forward reaction.
Question 6.
Which one of the following is incorrect statement?
(a) for a system at equilibrium, $\mathrm{Q}$ is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer:
(a) for a system at equilibrium, $Q$ is always less than the equilibrium constant
Solution:
Correct statement is, for a system at equilibrium, $\mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$
Question 7.
$\mathrm{K}_1$ and $\mathrm{K}_2$ are the equilibrium constants for the reactions respectively.
$
\begin{gathered}
\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\mathrm{K}_1}{\rightleftharpoons} 2 \mathrm{NO}(\mathrm{g}) \\
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \stackrel{\mathrm{K}_2}{\rightleftharpoons} 2 \mathrm{NO}_2(\mathrm{~g})
\end{gathered}
$
What is the equilibrium constant for the reaction $\mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
(a) $\frac{1}{\sqrt{\mathrm{K}_1 \mathrm{~K}_2}}$
(b) $\left(\mathrm{K}_1=\mathrm{K}_2\right)^{\frac{1}{2}}$
(c) $\frac{1}{2 \mathrm{~K}_1 \mathrm{~K}_2}$

$(d)\left(\frac{1}{\mathrm{~K}_1 \mathrm{~K}_2}\right)^{\frac{3}{2}}$
Answer:
(a) $\frac{1}{\sqrt{\mathrm{K}_1 \mathrm{~K}_2}}$
Solution:
Let equilibrium constant for the required reaction be $\mathrm{K}$. Then,
$
\begin{aligned}
\mathrm{K}_1 & =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2} ; \mathrm{K}_2=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right.} ; \mathrm{K}=\frac{\left[\mathrm{N}_2\right]^{\frac{1}{2}}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]} \\
\sqrt{\mathrm{K}_1} & =\frac{[\mathrm{NO}]}{\left[\mathrm{N}_2\right]^{\frac{1}{2}}\left[\mathrm{O}_2\right]^{\frac{1}{2}}} ; \sqrt{\mathrm{K}_2}=\frac{\left[\mathrm{NO}_2\right]}{[\mathrm{NO}]\left[\mathrm{O}_2\right]^{\frac{1}{2}}} \\
\sqrt{\mathrm{K}_1 \cdot \mathrm{K}_2} & =\frac{\left[\mathrm{NO}_2\right]}{\left[\mathrm{N}_2\right]^{\frac{1}{2}}\left[\mathrm{O}_2\right]} \\
\mathrm{K} & =\frac{1}{\sqrt{\mathrm{K}_1 \mathrm{~K}_2}}
\end{aligned}
$
Question 8.
In the equilibrium, $2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}_2(\mathrm{~g})$ the equilibrium concentrations of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, at 400 $\mathrm{K}$ are $1 \times 10^4 \mathrm{M}, 2.0 \times 10^{-3} \mathrm{M}, 1.5 \times 10^{-4} \mathrm{M}$ respectively. The value of $\mathrm{K}_{\mathrm{c}}$. for the equilibrium at $400 \mathrm{~K}$ is
(a) 0.06
(b) 0.09

(c) 0.62
(d) $3 \times 10^{-2}$
Answer:
(a) 0.06
Solution:
$
\begin{aligned}
& {[\mathrm{A}]=1 \times 10^{-4} \mathrm{M}} \\
& \text { [B] }=2 \times 10^{-3} \mathrm{M} \\
& {[C]=1.5 \times 10^{-4} \mathrm{M}} \\
& 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}_2(\mathrm{~g}) \\
& \mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{B}]^2\left[\mathrm{C}_2\right]}{[\mathrm{A}]^2}=\frac{\left(2 \times 10^{-3}\right)^2\left(1.5 \times 10^{-4}\right)}{\left(1 \times 10^{-4}\right)^2} \\
& =6.0 \times 10^{-2}=0.06 \\
&
\end{aligned}
$
Question 9.
An equilibrium constant of $3.2 \times 10^{-6}$ for a reaction means, the equilibrium is
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of these
Answer:
(b) largely towards reverse direction
Solution:
$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{[\text { Products }]}{[\text { Reactants }]} \\
3.2 \times 10^{-6} & =\frac{[\text { Products }]}{[\text { Reactants }]} \\
\mathrm{K}_{\mathrm{C}} & \left.<10^{-3} ; \text { indicates that [Reactant }\right] \gg[\text { Product }]
\end{aligned}
$
Question 10 .
$\frac{K_c}{K_p}$ for the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$ is .......
(a) $\frac{1}{R T}$
(b) $g \sqrt{R T}$
(c) RT
(d) $(\mathrm{RT})^2$
Answer:
(d) $(\mathrm{RT})^2$

Solution:
$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$
$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=2-4=-2 \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2} \\
& \frac{K_{\mathrm{c}}}{K_p}=(\mathrm{RT})^2
\end{aligned}
$
Question 11.
For the reaction, $\mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})$, at equilibrium, $\mathrm{AB}$ is $20 \%$ dissociated at a total pressure of $\mathrm{R}$ The equilibrium constant $\mathrm{K}$ is related to the total pressure by the expression
(a) $P=24 \mathrm{~K}_{\mathrm{p}}$
(b) $\mathrm{P}=8 \mathrm{~K}_{\mathrm{p}}$
(c) $24 \mathrm{p}=\mathrm{K}_{\mathrm{p}}$
(d) none of these
Answer:
(a) $P=24 \mathrm{~K}_{\mathrm{p}}$
Solution:

Total no. of moles at equilibrium $=80+20+20=120$
$
\begin{aligned}
\mathrm{K}_{\mathrm{P}} & =\frac{\mathrm{P}_{\mathrm{A}} \cdot \mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{AB}}}=\frac{\left(\frac{20}{120} \times \mathrm{P}\right)\left(\frac{20}{120} \times \mathrm{P}\right)}{\left(\frac{80}{120} \times \mathrm{P}\right)}=\frac{\mathrm{P}}{24} \\
24 \mathrm{~K}_{\mathrm{P}} & =\mathrm{P}
\end{aligned}
$
Question 12 .
In which of the following equilibrium, $\mathrm{K}$ and $\mathrm{K}$ are not equal?
(a) $2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
(b) $\mathrm{SO}_2(\mathrm{~g})+\mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g})+\mathrm{NO}(\mathrm{g})$
(c) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2$ (g) $\rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
(d) $\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$
Answer:
(d) $\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$
Solution:
For reactions given in options (a), (b) and (c) $\Delta \mathrm{n}_{\mathrm{g}}=0$
For option (d) $\Delta \mathrm{n}_{\mathrm{g}}=2-1=1$
$
\therefore \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})
$
Question 13.
If $\mathrm{x}$ is the fraction of $\mathrm{PCI}_5$ dissociated at equilibrium in the reaction,
$
\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2
$
then starting with $0.5 \mathrm{moIe}$ of $\mathrm{PCI}_5$ the total number of moles of reactants and products at equilibrium is
(a) $0.5-\mathrm{x}$

(b) $x+0.5$
(e) $2 \mathrm{x}+0.5$
(d) $\mathrm{x} \pm 1$
Answer:
(b) $x+0.5$
Solution:

Total no. of moles at equilibrium $=0.5-\mathrm{x}+\mathrm{x}+\mathrm{x}=0.5+\mathrm{x}$
Question 14.
The values of $\mathrm{K}_{\mathrm{p} 1}$ and $\mathrm{K}_{\mathrm{p} 2}$ for the reactions $\mathrm{X} \rightleftharpoons \mathrm{Y}+\mathrm{Z}$ and $\mathrm{A} \rightleftharpoons 2 \mathrm{~B}$ are in the ratio $9: 1$ if degree of dissociation and initial concentration of $\mathrm{X}$ and $\mathrm{A}$ be equal then total pressure at equilibrium $\mathrm{P}_1$, and $\mathrm{P}_2$ are in the ratio
(a) $36: 1$
(b) $1: 1$
(c) $3: 1$
(d) $1: 9$
Answer:
(a) $36: 1$
Solution:

$
\begin{aligned}
& \frac{\mathrm{K}_{\mathrm{P}_1}}{\mathrm{~K}_{\mathrm{P}_2}}=\frac{\mathrm{P}_y \mathrm{P}_z}{\mathrm{P}_x} \times \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}^2} \\
& \frac{\mathrm{K}_{\mathrm{P}_1}}{\mathrm{~K}_{\mathrm{P}_2}}=\left[\frac{\left(\frac{x}{a+x}\right) \mathrm{P}_1\left(\frac{x}{a+x}\right) \mathrm{P}_1}{\frac{(a-x)}{(a+x)} \times \mathrm{P}_1}\right] \times\left[\frac{\frac{(a-x)}{(a+x)} \mathrm{P}_2}{\frac{4 x^2 \mathrm{P}_2^2}{(a+x)^2}}\right] \\
& =\left(\frac{x^2 \mathrm{P}_1}{(a+x)(a-x)}\right) \times\left(\frac{(a-x)(a+x)}{4 x^2 \mathrm{P}_2}\right) \\
& \frac{K_{P_1}}{K_{\mathrm{P}_2}}=\frac{\mathrm{P}_1}{4 \mathrm{P}_2} \\
& \frac{\mathrm{K}_{\mathrm{P}_1}}{\mathrm{~K}_{\mathrm{P}_2}}=\frac{9}{1} \\
& \frac{9}{1}=\frac{P_1}{4 P_2} \Rightarrow \frac{P_1}{P_2}=\frac{36}{1} \\
&
\end{aligned}
$
Question 15.
In the reaction, $\mathrm{Fe}(\mathrm{OH})_3(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$, if the concentration of $\mathrm{OH}^{-}$ions is decreased by $1 / 4$ times, then the equilibrium concentration of $\mathrm{Fe}^{3+}$ will
(a) not changed
(b) also decreased by $1 / 4$ times
(c) increase by 4 times
(d) increase by 64 times
Answer:
(d) increase by 64 times
Solution:

$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3}{\left[\mathrm{Fe}(\mathrm{OH})_3(\mathrm{~s})\right]}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3
$
$[\because$ Concentration of solids is constant)
When concentration of $\mathrm{OH}^{-}$ions decreased by $\frac{1}{4}$ times, then
$
\mathrm{K}_{\mathrm{C}}=\left[\mathrm{Fe}^{3+}\right] \times\left(\frac{\left[\mathrm{OH}^{-}\right]}{4}\right)^3=\frac{1}{64}\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3
$
To maintain $\mathrm{K}_{\mathrm{C}}$ as constant, concentration of $\mathrm{Fe}^{3+}$ will increase by 64 times.
Question 16.
Consider the reaction where $\mathrm{K}_{\mathrm{p}}=0.5$ at a particular temperature
$\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCI}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ if the three gases are mixed in a container so that the partial pressure of each gas is initially $1 \mathrm{~atm}$, then which one of the following is true?
(a) more $\mathrm{PCI}_3$ will be produced
(b) more $\mathrm{Cl}_2$ will be produced
(c) more $\mathrm{PCI}_5$ will be produced
(d) none of these
Answer:
(c) more $\mathrm{PCI}_5$ will be produced .
Solution:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=0.5 \\
& \mathrm{Q}=\frac{\mathrm{P}_{\mathrm{PCl}_3} \cdot \mathrm{P}_{\mathrm{Cl}_2}}{\mathrm{P}_{\mathrm{PCl}_5}} \\
& \mathrm{Q}=\frac{1 x 1}{1} \\
& \mathrm{Q}>\mathrm{K}_{\mathrm{p}}
\end{aligned}
$
$\therefore$ Reverse reaction is favoured; i.e., more $\mathrm{PCI}_5$ will be produced.

Question 17.
Equimolar concentrations of $\mathrm{H}_2$ and $\mathrm{I}_2$ are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of $\mathrm{H}_2$ has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) $33 \%$
(b) $66 \%$
(c) $(33)^2 0 / 0$
(d) $16.5 \%$
Answer:
(a) $33 \%$
Solution:
$\begin{aligned} \mathrm{V} & =1 \mathrm{~L} \\ \mathrm{H}_2+\mathrm{I}_2 & \rightleftharpoons 2 \mathrm{HI} \\ {\left[\mathrm{H}_2\right]_{\text {initial }} } & =\left[\mathrm{I}_2\right]_{\text {initial }}=a \\ {\left[\mathrm{H}_2\right]_{\mathrm{eq}} } & =\left[\mathrm{I}_2\right]_{\mathrm{eq}}=(a-x) \text { and }[\mathrm{HI}]_{\mathrm{eq}}=2 x \\ \mathrm{~K}_{\mathrm{C}} & =\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \\ \mathrm{K}_{\mathrm{C}} & =\frac{4 x^2}{(a-x)^2} \\ \mathrm{~K}_{\mathrm{C}} & =\frac{\mathrm{K}_f}{\mathrm{~K}_r}=1 \\ 4 x^2 & =(a-x)^2 \\ 4 x^2 & =a^2+x^2-2 a x \\ 3 x^2 & +2 a x-a^2=0\end{aligned}$

$
x=-a \text { or } x=\frac{a}{3}
$
As degree of dissociation cannot be negative. Therefore, degree of dissociation $=\frac{a}{3} \times 100=33.33 \%$
Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is $2.5 \times 1$ and the equilibrium constant is 50 . The rate constant for the reverse reaction is
(a) 11.5
(b) 5
(c) $2 \times 10^2$
(d) $2 \times 10^3$
Answer:
(b) 5
Solution:
$
\begin{gathered}
\mathrm{K}_f=2.5 \times 10^2, \mathrm{~K}_{\mathrm{C}}=50, \mathrm{~K}_r=? \\
\mathrm{~K}_{\mathrm{C}}=\frac{\mathrm{K}_f}{\mathrm{~K}_r} \\
50=\frac{2.5 \times 10^2}{\mathrm{~K}_r} \\
\mathrm{~K}_r=5
\end{gathered}
$

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process (a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Answer:
(c) All the physical processes stop at equilibrium
Solution:
Correct statement - Physical processes occurs at the same rate at equilibrium.
Question 20.
For the fórmation of two moles of $\mathrm{SO}_3(\mathrm{~g})$ from $\mathrm{SO}_2$ and $\mathrm{O}_2$, the equilibrium constant is $\mathrm{K}_1$. The
(a) $\frac{1}{\mathrm{~K}_1}$
(b) $\mathrm{K}_1^2$
(c) $\left(\frac{1}{\mathrm{~K}_1}\right)^{\frac{1}{2}}$
(d) $\frac{\mathrm{K}_1}{2}$
Answer:
(c) $\left(\frac{1}{\mathrm{~K}_1}\right)^{\frac{1}{2}}$
Solution:
$
\begin{aligned}
& 2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3 \\
& \mathrm{~K}_1=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}
\end{aligned}
$

Dissociation of 1 mole of $2 \mathrm{SO}_3$
$
\begin{aligned}
\mathrm{SO}_3 & \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \\
\mathrm{~K}_2 & =\frac{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}{\left[\mathrm{SO}_3\right]} \\
\mathrm{K}_2 & =\frac{1}{\sqrt{\mathrm{K}_1}}
\end{aligned}
$
Question 21.
Match the equilibria with the corresponding conditions ...........
(i) Liquid $\rightleftharpoons$ Vapour -1 . Melting point
(ii) Solid $\rightleftharpoons$ Liquid -2 . Saturated solution
(iii) Solid $\rightleftharpoons$ Vapour -3 . Boiling point
(iv) Solute (s) $\rightleftharpoons$ Solute (Solution) -4 . Sublimation point
5. Unsaturated solution

Answer:
(b) $3,1,4,2$
Question 22.
Consider the following reversible reaction at equilibrium, $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}$, if the concentration of the reactants $A$ and $B$ are doubled, then the equilibrium constant will
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer:
(d) remain the same
Solution:
$
\begin{aligned}
& \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C} \\
& \mathrm{K}_{\mathrm{C}}=\frac{[C]}{[A][B]}
\end{aligned}
$
If $[A]$ and $[B]$ are doubled, $[C]$ increases 4 times to maintain $\mathrm{K}_C$ as constant.
$\therefore$ Equilibrium constant will remain the same.
Question 23.
$\left[\mathrm{CO}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (aq) (pink) $+4 \mathrm{C}^{-}$(aq) $\rightleftharpoons\left[\mathrm{COCI}_4\right]^{2-}(\mathrm{aq})$ (blue) $+6 \mathrm{H}_2 \mathrm{O}$ (1) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one of the following is true?
(a) $\Delta \mathrm{H}>0$ for the forward reaction
(b) $\Delta \mathrm{H}=0$ for the reverse reaction
(c) $\Delta \mathrm{H}<0$ for the forward reaction
(d) Sign of the $\Delta \mathrm{H}$ cannot be predicted based on this information.
Answer:

(a) $\Delta \mathrm{H}>0$ for the forward reaction
Solution:
On cooling, reverse reaction predominates and the solution is pink in colour. Decrease in temperature, favours the reverse reaction i.e. reverse reaction is exothermic $(\Delta \mathrm{H}<0)$ and for the forward reaction is endothermic $(\Delta \mathrm{H}>0)$.
Question 24.
The equilibrium constants of the following reactions are:
$
\begin{aligned}
& \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 ; \mathrm{K}_1 \\
& \mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; \mathrm{K}_2 \\
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} ; \mathrm{K}_3
\end{aligned}
$
The equilibrium constant $(\mathrm{K})$ for the reaction;
$
2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{\mathrm{K}}{\rightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}
$
will be
(a) $\mathrm{K}_2^3 \frac{\mathrm{K}_3}{\mathrm{~K}_1}$
(b) $\mathrm{K}_1 \frac{\mathrm{K}_3^3}{\mathrm{~K}_2}$
(c) $\mathrm{K}_2 \frac{\mathrm{K}_3^3}{\mathrm{~K}_1}$
(d) $\mathrm{K}_2 \frac{\mathrm{K}_3}{\mathrm{~K}_1}$
Answer:
(c) $\mathrm{K}_2 \frac{\mathrm{K}_3^3}{\mathrm{~K}_1}$

$
\begin{aligned}
\mathrm{K}_1= & \frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\
\mathrm{~K}_2= & \frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \\
\mathrm{K}_3= & \frac{\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{H}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}} \\
\mathrm{~K}= & \frac{[\mathrm{NO}]^2\left[\mathrm{H}_2 \mathrm{O}\right]^3}{\left[\mathrm{NH}_3\right]^2\left[\mathrm{O}_2\right]^{\frac{5}{2}}} \\
\mathrm{~K}= & \frac{\left(\mathrm{K}_2\right)\left(\mathrm{K}_3\right)^3}{\left(\mathrm{~K}_1\right)}=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \times \frac{\left[\mathrm{H}_2 \mathrm{O}\right]^3}{\left[\mathrm{H}_2\right]^3\left[\mathrm{O}_2\right]^{\frac{3}{2}}} \frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \\
= & \frac{\left[\mathrm{NO}^2\left[\mathrm{H}_2 \mathrm{O}\right]^3\right.}{\left[\mathrm{NH}_3\right]^2\left[\mathrm{O}_2\right]^{\frac{5}{2}}}
\end{aligned}
$

Question 25.
A 20 litre container at $400 \mathrm{~K}$ contains $\mathrm{CO}_2$ (g) at pressure 0.4 atm and an excess of $\mathrm{SrO}$ (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of $\mathrm{CO}_2$ attains its maximum value will be.
Given that: $\mathrm{SrCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
(a) 2 litre
(b) 5 litre
(c) 10 Litre
(d) 4 litre
Answer:
(b) 5 litre
Solution:
Given that $\mathrm{K}_{\mathrm{p}}=1.6 \mathrm{~atm}$
$
\begin{aligned}
& \mathrm{V}_1=20 \mathrm{~L} \\
& \mathrm{~V}_2=? \\
& \mathrm{~T}_1=400 \mathrm{~K} \\
& \mathrm{~T}_2=400 \mathrm{~K}
\end{aligned}
$

$\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=\mathrm{P}_{\mathrm{co}_2} \\
& \mathrm{P}_{\mathrm{co}_2}=1.6 \mathrm{~atm} \\
& \mathrm{P}_1=0.4 \mathrm{~atm} \\
& \mathrm{P}_2=1.6 \mathrm{~atm} \\
& \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} \\
& \mathrm{~S}^{\mathrm{m}} ? \\
& \mathrm{~V}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{\mathrm{T}_2}{\mathrm{P}_2}=\frac{0.4 \mathrm{~atm} \times 20 \mathrm{~L}}{400 \mathrm{~K}} \times \frac{400 \mathrm{~K}}{1.6 \mathrm{~atm}}=5 \mathrm{~L}
\end{aligned}$

Short Answer Questions
Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.
Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of $Q$ also constant? Explain.
Answer:
In the chemical reaction, as the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the $\mathrm{Q}$ value until the reaction reaches the equilibrium. So even at particular temperature, $\mathrm{Q}$ is not constant. Even once the equilibrium is achieved then change in concentration of reactants or products, pressure, volume will change the value of $\mathrm{Q}$.
Question 28.
What is the relation between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$. Give one example for which $\mathrm{K}_{\mathrm{p}}$ is equal to $\mathrm{K}_{\mathrm{C}}$.
Answer:
The relation between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ is $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$
$\mathrm{K}_{\mathrm{p}}$ = equilibrium constant is terms of partial pressure.
$\mathrm{K}_{\mathrm{c}}=$ equilibrium constant is terms of concentration.
$\mathrm{R}=$ gas constant
$\mathrm{T}=$ Temperature
$\Delta \mathrm{n}_{\mathrm{g}}=$ Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When $\Delta \mathrm{n}_{\mathrm{g}}=0$
$
\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^0=\mathrm{K}_{\mathrm{C}} \text { i.e., } \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}
$

Example: $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
$\Delta \mathrm{n}_{\mathrm{g}}=2-2=0$
$\therefore \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$ for the synthesis of $\mathrm{HI}$.
Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is $\mathrm{K}_{\mathrm{C}}$ is larger or smaller than $\mathrm{K}_{\mathrm{p}}$ ?
Answer:
For a homogeneous reaction at equilibrium, number of moles of products $\left(n_p\right)$ are greater than the number of moles of reactants $\left(n_R\right)$ then $\Delta n_g=+$ ve
$
\begin{aligned}
& \mathrm{n}_{\mathrm{p}}>\mathrm{n}_{\mathrm{R}} \\
& \Delta \mathrm{n}_{\mathrm{g}}=+ \text { ve }
\end{aligned}
$
If $\Delta \mathrm{n}_{\mathrm{g}}$ is ve, $\mathrm{K}_{\mathrm{p}}$ value is greater than $\mathrm{K}_{\mathrm{C}}$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{+\mathrm{ve}} \\
& \mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}} \\
& \text { Example: } \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \\
& 2-1=1 \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^1 \\
& \mathrm{~K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Question 30.
When the numerical value of the reaction quotient $(\mathrm{Q})$ is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?
Answer:
When $\mathrm{Q}>\mathrm{K}_{\mathrm{C}}$ the reaction will proceed in the reverse direction, i.e, formation of reactants.
Question 31.
For the reaction: $\mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) ; \Delta \mathrm{H}$ is $-\mathrm{ve}$.
Answer:
The following molecular scenes represent different reaction mixture
(A - dark grey, B - light grey)

1. Calculate the equilibrium constant $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of increase in pressure for the mixture at equilibrium?
Answer:
$
\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right]\left[\mathrm{B}_2\right]} \quad \begin{aligned}
& \mathrm{A}-\text { green } \\
& \mathrm{B}-\text { blue }
\end{aligned}
$
At equilibrium
$
\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\left(\frac{4}{\mathrm{~V}}\right)^2}{\left(\frac{2}{\mathrm{~V}}\right)\left(\frac{2}{\mathrm{~V}}\right)}=\frac{16}{4}=4 \\
& \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta n_{\mathrm{g}}} \\
& \mathrm{K}_{\mathrm{P}}=4(\mathrm{RT})^0=4
\end{aligned}
$

2. At state ' $x$ ', $Q=\frac{\left(\frac{6}{\mathrm{~V}}\right)^2}{\left(\frac{2}{\mathrm{~V}}\right)\left(\frac{1}{\mathrm{~V}}\right)}=\frac{36}{2}=18$
$Q>K_C$ i.e., reverse reaction is favoured
At Stage ' $y$ ' $Q=\frac{\left(\frac{2}{V}\right)^2}{\left(\frac{3}{V}\right)\left(\frac{3}{V}\right)}=\frac{4}{3 \times 3}=\frac{4}{9}$
$\mathrm{K}_{\mathrm{C}}>\mathrm{Q}$ i.e., forward reaction is favoured.
3. Since $\Delta \mathrm{n}_{\mathrm{g}}=2-2=0$, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.
Question 32.
State $\mathrm{Le}-$ Chatelier principle.
Answer:
It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.
Question 33.
Consider the following reactions
1. $\mathrm{H}_2(\mathrm{~g})+\mathrm{l}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
2. $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
3. $\mathrm{S}(\mathrm{s})+3 \mathrm{~F}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SF}_6(\mathrm{~g})$

In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
1. $\mathrm{H}_2(\mathrm{~g})+\mathrm{1}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
In the above equilibrium reaction, volume of gaseous molecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

2.

$0(\mathrm{~V}) \quad 0(\mathrm{~V}) \quad 1(\mathrm{~V})$
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product $\mathrm{CO}_2$, the pressure should be decreased or volume should be increased.

3.

Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.
Question 34.
State law of mass action.
Answer:
The law states that "At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant."
Question 35 .
Explain how will you predict the direction of an equilibrium reaction.
Answer:
1. A large value of $\mathrm{K}_{\mathrm{C}}$ indicates that the reaction reaches equilibrium with high product yield.
2. A low value of $\mathrm{K}_{\mathrm{C}}$ indicate that the reaction reaches equilibrium with low product formed.
3. In general, if the $\mathrm{K}$ is greater than I the reaction proceeds nearly to completion. If is less than $10^{-3}$, the reaction rarely proceeds.
4. If $\mathrm{K}<10^{-3}$, reverse reaction is favoured. If $\mathrm{K}_{\mathrm{C}}>10^3$, forward reaction is favoured.

Long Answer Questions
Question 36.
Derive a general expression for the equilibrium constant $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$ for the reaction.
$
3 \mathrm{H}_2(\mathrm{~g})+\mathrm{N}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$
Answer:
In the formation of ammonia, ' $a$ ' moles of Nitrogen and 'b' moles of hydrogen gas are allowed to react in a container of volume of ' $\mathrm{V}$ '. Let ' $\mathrm{x}$ ' moles of nitrogen react with $3 \mathrm{x}$ moles of hydrogen to give $2 \mathrm{x}$ moles of ammonia.
$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$

Applying law of mass action
$
\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{\left(\frac{2 x}{\mathrm{~V}}\right)^2}{\left(\frac{a-x}{\mathrm{~V}}\right)\left(\frac{b-3 x}{\mathrm{~V}}\right)^3}=\frac{\left(\frac{4 x^2}{\mathrm{~V}^2}\right)}{\left(\frac{a-x)}{\mathrm{V}}\right)\left(\frac{b-3 x}{\mathrm{~V}}\right)^3} \\
& \mathrm{~K}_{\mathrm{C}}=\frac{4 x^2 \cdot \mathrm{V}^2}{(a-x)(b-3 x)^3}
\end{aligned}
$
Total number of moles at equilibrium $\mathrm{n}=\mathrm{a}-\mathrm{x}+\mathrm{b}-3 \mathrm{x}+2 \mathrm{x}=\mathrm{a}+\mathrm{b}-2 \mathrm{x}$
$
\begin{aligned}
\mathrm{K}_{\mathrm{p}} & =\dot{\mathrm{K}}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}} \mathrm{g} \\
\Delta \mathrm{n}_{\mathrm{g}} & =\mathrm{n}_p-\mathrm{n}_r=2-4=-2 \\
\mathrm{~K}_{\mathrm{P}} & =\frac{4 x^2 \cdot \mathrm{V}^2}{(a-x)(b-3 x)^3}\left(\mathrm{RT})^{-2}\right.
\end{aligned}
$
Total number of moles at equilibrium

$
\begin{aligned}
& \mathrm{n}=\mathrm{a}-\mathrm{x}+\mathrm{b}-3 \mathrm{x}+2 \mathrm{x}=\mathrm{a}+\mathrm{b}-2 \mathrm{x} \\
& \mathrm{K}_{\mathrm{P}}=\frac{4 x^2 \cdot \mathrm{V}^2}{(a-x)(b-3 x)^3} \times\left(\frac{\mathrm{PV}}{n}\right)^{-2} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{4 x^2 \mathrm{~V}^2}{(a-x)(b-3 x)^3} \times\left(\frac{n}{\mathrm{PV}}\right)^2 \\
& \mathrm{~K}_{\mathrm{P}}=\frac{4 x^2 \mathrm{~V}^2}{(a-x)(b-3 x)^3} \times\left(\frac{a+b-2 x}{\mathrm{PV}}\right)^2 \\
& \mathrm{~K}_{\mathrm{P}}=\frac{4 x^2(a+b-2 x)^2}{\mathrm{P}^2(a-x)(b-3 x)^3}
\end{aligned}
$
Question 37.
Write a balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression?
Answer:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6}
$
Question 38 .
What is the effect of added inert gas on the reaction at equilibrium at constant volume?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.
Question 39.
Derive the relation between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$.
Answer:
Consider a general reaction in which all reactants and products are ideal gases.
$\mathrm{x} \mathrm{A}+\mathrm{y} \mathrm{B} \rightleftharpoons 1 \mathrm{C}+\mathrm{mD}$
The equilibrium constant $\mathrm{K}_{\mathrm{C}}$ is
$
\begin{gathered}
\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}]^l[\mathrm{D}]^m}{[\mathrm{~A}]^x[\mathrm{~B}]^y} \\
\mathrm{~K}_{\mathrm{P}}=\frac{p_{\mathrm{C}}^l \times p_{\mathrm{D}}^m}{p_{\mathrm{A}}^x \times p_{\mathrm{B}}^y}
\end{gathered}
$
The ideal gas equation is

$\mathrm{PV}=\mathrm{nRT}$ or $\mathrm{P}=\frac{n}{V} \mathrm{RT}$
Since,
Active mass $=$ molar concentration $=\frac{n}{V}$
$\mathrm{P}=$ Active mass $\times \mathrm{RT}$
Based on the above expression, the partial pressure of the reacants and products can be expressed as
$
\begin{aligned}
p_{\mathrm{A}}^x & =[\mathrm{A}]^x \cdot[\mathrm{RT}]^x \\
p_{\mathrm{B}}^y & =[\mathrm{B}]^y \cdot[\mathrm{RT}]^y \\
p_{\mathrm{C}}^l & =[\mathrm{C}]^l \cdot[\mathrm{RT}]^l \\
p_{\mathrm{D}}^m & =[\mathrm{D}]^m \cdot[\mathrm{RT}]^m
\end{aligned}
$
On substitution in equation (2),
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\frac{[\mathrm{C}]^l[\mathrm{RT}]^l[\mathrm{D}]^m[\mathrm{RT}]^m}{[\mathrm{~A}]^x[\mathrm{RT}]^x[\mathrm{~B}]^y[\mathrm{RT}]^y} \\
& \mathrm{~K}_{\mathrm{p}}=\frac{[\mathrm{C}]^l[\mathrm{D}]^m}{[\mathrm{~A}]^x[\mathrm{~B}]^y} \times \mathrm{RT}^{(l+m)-(x+y)}
\end{aligned}
$
By comparing equation (1) and (4), we get
$
\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})_{\Delta \mathrm{g}}
$
where $\Delta n_g$ is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If $\Delta \mathrm{n}_{\mathrm{g}}=0, \mathrm{~K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^0$
$
\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}
$
Example: $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$

ii) where
$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=+\mathrm{Ve} \\
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{+\mathrm{ve}} \\
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Example: $2 \mathrm{NH}_3(\mathrm{~g}) \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$
(iii) When
$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=-\mathrm{Ve} \\
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{-\mathrm{ve}} \\
& \mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Example: $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$

Question 40.
One mole of $\mathrm{PCl}_5$ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.
Answer:
Given that
$
\begin{aligned}
& {\left[\mathrm{PCl}_5\right]_{\text {initial }}=\frac{1 \mathrm{~mol}}{1 \mathrm{dm}^3}} \\
& {\left[\mathrm{Cl}_2\right]_{\mathrm{eq}}=0.6 \mathrm{~mol} \mathrm{dm}^{-3}} \\
& \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2 \\
& {\left[\mathrm{PCl}_3\right]_{\mathrm{eq}}=0.6 \text { mole } \mathrm{dm}^{-3}} \\
& {\left[\mathrm{PCl}_5\right]_{\mathrm{eq}}=0.4 \text { mole dm }^{-3}} \\
& \mathrm{~K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.6 \times 0.6}{0.4} \\
& \mathrm{~K}_{\mathrm{C}}=0.9 \\
&
\end{aligned}
$
Question 41.
For the reaction: $\mathrm{SrCO}_2(\mathrm{~s}) \rightleftharpoons \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$, the value of equilibrium constant $\mathrm{K}_{\mathrm{p}}=2.2 \times 10^{-}$ 4 at $1002 \mathrm{~K}$. Calculate $\mathrm{K}_{\mathrm{C}}$ for the reaction.
Answer:
For the reaction, $\mathrm{SrCO}_2(\mathrm{~s}) \rightleftharpoons \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=1-0=1 \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT}) \\
& 2.2 \times 10^{-4}=\mathrm{K}_{\mathrm{C}}(0.0821)(1002)
\end{aligned}
$

$
\mathrm{K}_{\mathrm{C}}=\frac{2.2 \times 10^{-4}}{0.0821 \times 1002}=2.674 \times 10^{-6}
$
Question 42 .
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 $\mathrm{mol}$ of $\mathrm{HI}$ gas and allows the reaction to proceed at $500^{\circ} \mathrm{C}$. At equilibrium he found the concentration of $\mathrm{HI}$ which is equal to $0.05 \mathrm{M}$. Calculate $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$ for this reaction.
Answer:
$
\begin{aligned}
\mathrm{V} & =3 \mathrm{~L} \\
{[\mathrm{HI}]_{\text {initial }} } & =\frac{0.3 \mathrm{~mol}}{3 \mathrm{~L}}=0.1 \mathrm{M} \\
{[\mathrm{HI}]_{\text {eq }} } & =0.05 \mathrm{M} \\
2 \mathrm{HI}(\mathrm{g}) & \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})
\end{aligned}
$

$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\frac{0.025 \times 0.025}{0.05 \times 0.05} \\
\mathrm{~K}_{\mathrm{C}} & =0.25 \\
\mathrm{~K}_{\mathrm{P}} & =\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\
\Delta \mathrm{n}_{\mathrm{g}} & =2-2=0 \\
\mathrm{~K}_{\mathrm{P}} & =0.25(\mathrm{RT})^0 \\
\mathrm{~K}_{\mathrm{P}} & =0.25
\end{aligned}
$
Question 43.
Oxidation of nitrogen monoxide was studied at $200^{\circ} \mathrm{C}$ with initial pressures of $1 \mathrm{~atm} \mathrm{NO}$ and 1 atm of $\mathrm{O}_2$. At equilibrium partial pressure of oxygen is found to be $0.52 \mathrm{~atm}$. Calculate $\mathrm{K}_{\mathrm{p}}$ value.
Answer:
$
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})
$

$
\begin{aligned}
& {\left[\mathrm{O}_2\right] \text { at equilibrium } \Rightarrow \quad 1-x=0.52} \\
& x=0.48 \\
& \mathrm{P}_{\mathrm{NO}}=1-2 \mathrm{x}=1-2(0.48)=0.04 \\
& \mathrm{P}_{\mathrm{O}_2}=0.52 \\
& \mathrm{P}_{\mathrm{NO}_2}=2 x=2(0.48)=0.96 \\
& \mathrm{~K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{NO}_2}\right)}{\left(\mathrm{P}_{\mathrm{N}_{\mathrm{O}}}\right)^2\left(\mathrm{P}_{\mathrm{O}_2}\right)}=\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52} \\
& \mathrm{~K}_{\mathrm{P}}=1.107 \times 10^3 \\
&
\end{aligned}
$
Question 44 .
$1 \mathrm{~mol}$ of $\mathrm{CH}_4, 1 \mathrm{~mole}$ of $\mathrm{CS}_2$ and $2 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{~S}$ are $2 \mathrm{moI}$ of $\mathrm{H}_2$ are mixed in a $500 \mathrm{~mL}$ flask. The equilibrium constant for the reaction $\mathrm{K}_{\mathrm{C}}=4 \times 10^{-2} \mathrm{~mol}^2 \mathrm{lit}^{-2}$. In which direction will the reaction proceed to reach equilibrium?

Answer:
$
\begin{aligned}
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) & \rightleftharpoons \mathrm{CS}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) \\
\mathrm{K}_{\mathrm{C}} & =4 \times 10^{-2} \mathrm{~mol}^2 \mathrm{~L}^{-2} \\
\text { Volume } & =500 \mathrm{~mL}=\frac{1}{2} \mathrm{~L} \\
{\left[\mathrm{CH}_4\right] } & =\frac{1 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}=2 \mathrm{~mol} \mathrm{~L}^{-1} \\
{\left[\mathrm{CS}_2\right] } & =\frac{1 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}=2 \mathrm{~mol} \mathrm{~L}^{-1} \\
{\left[\mathrm{H}_2 \mathrm{~S}\right] } & =\frac{2 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1} \\
{\left[\mathrm{H}_2\right] } & =\frac{2 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$

$
\begin{aligned}
& \mathrm{Q}=\frac{\left[\mathrm{CS}_2\right]\left[\mathrm{H}_2\right]^4}{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{~S}\right]^2} \\
& \mathrm{Q}=\frac{2 \times(4)^4}{2 \times(4)^2}=16 \\
& \mathrm{Q}>\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
$\therefore$ The reaction will proceed in the reverse direction to reach the equilibrium.
Question 45.
At particular temperature $\mathrm{K}_{\mathrm{C}}=4 \times 10^{-2}$ for the reaction
$\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{~S}_2(\mathrm{~g})$. Calculate $\mathrm{K}_{\mathrm{C}}$ for each of the following reaction
(i). $\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{S}_2(\mathrm{~g})$
(ii). $3 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{~S}_2(\mathrm{~g})$

Answer:
$\mathrm{K}_{\mathrm{C}}=4 \times 10^{-2}$ for the reaction
(i) $\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{~S}_2(\mathrm{~g})$
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{S}_2\right]^{\frac{1}{2}}}{\left[\mathrm{H}_2 \mathrm{~S}\right]}=4 \times 10^{-2}
$
For the reaction, $2 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{S}_2(\mathrm{~g})$
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_2\right]^2\left[\mathrm{~S}_2\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]^2}=\left(4 \times 10^{-2}\right)^2=16 \times 10^{-4}
$
(ii) For the reaction $3 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{~S}_2(\mathrm{~g})$
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_2\right]^3\left[\mathrm{~S}_2\right]^{\frac{3}{2}}}{\left[\mathrm{H}_2 \mathrm{~S}\right]^3}=\left(4 \times 10^{-2}\right)^3=64 \times 10^{-6}
$
Question 46.
$28 \mathrm{~g}$ of nitrogen and $6 \mathrm{~g}$ of hydrogen were mixed in a 1 litre closed container. At equilibrium $17 \mathrm{~g}$ $\mathrm{NH}_3$ was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer:
Given
$
m_{\mathrm{N}_2}=28 \mathrm{~g}, \quad m_{\mathrm{H}_2}=6 \mathrm{~g}, \quad \mathrm{~V}=1 \mathrm{~L}
$

$\begin{aligned}
\left(n_{\mathrm{N}_2}\right)_{\text {initial }} & =\frac{28}{28}=1 \mathrm{~mol} \\
\left(n_{\mathrm{H}_2}\right)_{\text {initial }} & =\frac{6}{2}=3 \mathrm{~mol} \\
\mathrm{~N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) & \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
\end{aligned}$

$
\left[\mathrm{NH}_3\right]=\left(\frac{17}{17}\right)=1 \mathrm{~mol}=2 x
$
$
\left[\mathrm{NH}_3\right]=\left(\frac{17}{17}\right)=1 \mathrm{~mol}=2 \mathrm{x}
$
$\mathrm{x}=0.5 \mathrm{~mol}$
At equilibrium, $\left[\mathrm{N}_2\right]=1-\mathrm{x}=0.5 \mathrm{~mol}$ $\left[\mathrm{H}_2\right]=3-3 \mathrm{x}=3-3(0.5)=1.5 \mathrm{~mol}$
Weight of $\mathrm{N}_2$ (no. of moles of $\mathrm{N}_2$ ) molar mass of $\mathrm{N}_2=0.5 \times 28=14 \mathrm{~g}$
Weight of $\mathrm{H}_2=$ (no. of moles of $\mathrm{H}_2$ ) $\times$ molar mass of $\mathrm{H}_2=1.5 \times 2=3 \mathrm{~g}$
Question 47.
The equilibrium for the dissociation of $\mathrm{XY}_2$ is given as,
$
2 \mathrm{XY}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}(\mathrm{g})+\mathrm{Y}_2(\mathrm{~g})
$
If the degree of dissociation $\mathrm{x}$ is so small compared to one. Show that $2 \mathrm{~K}_{\mathrm{p}}=\mathrm{Px}^3$ where $\mathrm{P}$ is the total pressure and $\mathrm{K}$ is the dissociation equilibrium constant of $\mathrm{XY}_2$.
Answer:
$
2 \mathrm{XY}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}(\mathrm{g})+\mathrm{Y}_2(\mathrm{~g})
$

Total No. of moles $=1-x+x+\frac{x}{2}=1+\frac{x}{2} \cong 1$
$\left[\therefore\right.$ Given that $x \ll 1 ; 1-x \cong 1$ and $\left.1+\frac{x}{2} \cong 1\right]$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{XY}}\right)^2\left(\mathrm{P}_{\mathrm{Y}_2}\right)}{\left(\mathrm{P}_{\mathrm{XY}_2}\right)^2}=\frac{\left(\frac{x}{1} \times \mathrm{P}\right)^2\left(\frac{\frac{x}{2}}{1} \times \mathrm{P}\right)}{\left(\frac{1}{1} \times \mathrm{P}\right)^2} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{x^3 \mathrm{p}}{2} \Rightarrow 2 \mathrm{~K}_{\mathrm{P}}=x^3 \mathrm{P} \\
&
\end{aligned}
$
Question 48 .
A sealed container was filled with I $\mathrm{mol}$ of A, ( $\mathrm{g})$, I $\mathrm{mol} \mathrm{B}, \mathrm{g})$ at $800 \mathrm{~K}$ and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that $\mathrm{K}=1$ for the reaction:
$
\mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})
$
Answer:
$
\mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})
$

Total no. of moles $=1-x+1-x+2 x=2$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{AB}}\right)^2}{\left(\mathrm{P}_{\mathrm{A}_2}\right)\left(\mathrm{P}_{\mathrm{B}_2}\right)}=\frac{\left(\frac{2 x}{2} \times \mathrm{P}\right)^2}{\left(\frac{(1-x)}{2} \times \mathrm{P}\right)\left(\frac{(1-x)}{2} \times \mathrm{P}\right)} \\
& \mathrm{K}_{\mathrm{P}}=\frac{4 x^2}{(1-x)^2}
\end{aligned}
$
Given that
$
\begin{aligned}
& \Rightarrow \quad 4 x^2=(1-x)^2 \Rightarrow 4 x^2=1+x^2-2 x \\
& 3 x^2+2 x-1=0 \\
& x=\frac{-2 \pm \sqrt{4-4 \times 3 \times(-1)}}{2(3)} \\
& x=\frac{-2 \pm \sqrt{4+12}}{6}=\frac{-2 \pm \sqrt{16}}{6}=\frac{-2+4}{6} ; \frac{-2-4}{6}=\frac{2}{6} ; \frac{-6}{6} \\
& x=0.33 ;-1 \text { (not possible) } \\
& \therefore \quad\left[\mathrm{A}_2\right]_{\mathrm{eq}}=1-x=1-0.33=0.67 \\
& {\left[\mathrm{~B}_2\right]_{\mathrm{eq}}=1-\mathrm{x}=1-0.33=0.667} \\
& {[\mathrm{AB}]_2=2 \mathrm{x}=2 \times 0.33=0.66} \\
&
\end{aligned}
$

Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives the quantitative temperature dependance of equilibrium constant $\mathrm{K}$. The relation between standard free energy change $\Delta \mathrm{G}^{\circ}$ and equilibrium constant is $\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$
We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$
Substituting (1) in equation (2)
$-\mathrm{RT} \ln \mathrm{K}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$
Rearranging, $\ln \mathrm{K}=\frac{-\Delta H^*}{R T}+\frac{\Delta S^*}{R}$
Differentiating equation (3) with respect to temperature $\mathrm{d}(\operatorname{lnK}) \Delta \mathrm{H}^{\circ} \frac{d(\ln K)}{d T}=\frac{\Delta H^0}{R T^2}$
Equation (4) is known as differential form of van't Hoff equation. On integrating the equation 4 , between $T_1$ and $T_2$ with their respective equilibrium constants and $K_2$

$
\begin{aligned}
\int_{\mathrm{K}_1}^{\mathrm{K}_2} d(\ln \mathrm{K}) & =\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{R}} \int_{\mathrm{T}_1}^{\mathrm{T}_2} \frac{d \mathrm{~T}}{\mathrm{~T}^2} \\
{[\ln \mathrm{K}]_{\mathrm{K}_1}^{\mathrm{K}_2} } & =\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{R}}\left[-\frac{1}{\mathrm{~T}}\right]_{\mathrm{T}_1}^{\mathrm{T}_2} \\
\ln \mathrm{K}_2-\ln \mathrm{K}_1 & =\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{R}}\left[-\frac{1}{\mathrm{~T}_2}+\frac{1}{\mathrm{~T}_1}\right] \\
\ln \frac{\mathrm{K}_2}{\mathrm{~K}_1} & =\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_2 \mathrm{~T}_1}\right] \\
\log \frac{\mathrm{K}_2}{\mathrm{~K}_1} & =\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_2 \mathrm{~T}_1}\right]
\end{aligned}
$
Equation is known as integrated form of van't Hoff equation.
Question 50.
The equilibrium constant $\mathrm{K}$ for the reaction
$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})=2 \mathrm{NH}_3(\mathrm{~g}) \text { is } 8.19 \times 10^2
$
at $298 \mathrm{~K}$ and $4.6 \times 10^{-1}$ at $498 \mathrm{~K}$. Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction.
Answer:

$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}_1}=8.19 \times 10^2 ; \mathrm{T}_1=298 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{P}_2}=4.6 \times 10^{-1} ; \mathrm{T}_2=498 \mathrm{~K} \\
& \log \left[\frac{K_{P_2}}{\mathrm{~K}_{\mathrm{P}_1}}\right]=\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right) \\
& \log \left(\frac{4.6 \times 10^{-1}}{8.19 \times 10^2}\right)=\frac{\Delta H^{\circ}}{2.303 \times 8.314}\left(\frac{498-298}{498 \times 298}\right) \\
& \frac{-3.2505 \times 2.303 \times 8.314 \times 498 \times 298}{200}=\Delta \mathrm{H}^{\circ} \\
& \Delta \mathrm{H}^{\circ}=-46181 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}^{\circ}=-46.18 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&
\end{aligned}
$
Question 51.
The partial pressure of carbon dioxide in the reaction $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ is $1.017 \mathrm{x}$ $10^{-3} \mathrm{~atm}$ at $500^{\circ} \mathrm{C}$. Calculate $\mathrm{K}_{\mathrm{p}}$, at $600^{\circ} \mathrm{C}$ for the reaction. $\Delta \mathrm{H}$ for the reaction is $181 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and does not change in the given range of temperature.

Answer:
$
\begin{aligned}
& \mathrm{P}_{\mathrm{CO}_2}=1.017 \times 10^{-3} \mathrm{~atm} ; \mathrm{T}=500^{\circ} \mathrm{C} \\
& \mathrm{K}_{\mathrm{P}}=\mathrm{P}_{\mathrm{CO}_2} \\
& \therefore \quad K_{P_1}=1.017 \times 10^{-3}, \mathrm{~T}=500+273=773 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{P}_2}=?, \mathrm{~T}=600+273=873 \mathrm{~K} \\
& \Delta \mathrm{H}^{\circ}=181 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \log \left(\frac{K_{P_2}}{K_{P_1}}\right)=\frac{\Delta H^{\circ}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right) \\
& \log \left(\frac{\mathrm{K}_{\mathrm{P}_2}}{1.017 \times 10^{-3}}\right)=\frac{181 \times 10^3}{2.303 \times 8.314}\left(\frac{873-773}{873 \times 773}\right) \\
& \log \left(\frac{\mathrm{K}_{\mathrm{P}_2}}{1.017 \times 10^{-3}}\right)=\frac{181 \times 10^3 \times 100}{2.303 \times 8.314 \times 873 \times 773} \\
& \frac{\mathrm{K}_{\mathrm{P}_2}}{1.017 \times 10^{-3}}=\text { Antilog of }(1.40) \\
& \frac{\mathrm{K}_{\mathrm{P}_2}}{1.017 \times 10^{-3}}=25.12 \\
& \Rightarrow \quad K_{\mathrm{P}_2}=25.12 \times 1.017 \times 10^{-3} \\
&
\end{aligned}
$

$
\mathrm{K}_{\mathrm{P}_2}=25.54 \times 10^{-3}
$
Text problems Solved
Question 1.
Consider the following reaction
$
\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq})[\mathrm{Fe}(\mathrm{SCN})]^{2+}(\mathrm{aq})
$
A solution is made with initial $\mathrm{Fe}^{3+}, \mathrm{SCN}^{-}$concentration of $1 \times 10^{-3} \mathrm{M}$ and $8 \times 10^{-4} \mathrm{M}$ respectively. At equilibrium $[\mathrm{Fe}(\mathrm{SCN})]^{2+}$ concentration is $2 \times 10^{-4} \mathrm{M}$. Calculate the value of equilibrium constant.
Answer:

$
\begin{aligned}
& \mathrm{K}_{\mathrm{eq}}=\frac{[\mathrm{Fe}(\mathrm{SCN})]^{2+}}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}=\frac{2 \times 10^{-4} \mathrm{M}}{8 \times 10^{-4} \mathrm{M} \times 6 \times 10^{-4} \mathrm{M}}=0.0416 \times 10^4 \\
& \mathrm{~K}_{\mathrm{eq}}=4.16 \times 10^2 \mathrm{M}^{-1}
\end{aligned}
$
Question 2.
The atmospheric oxidation of $\mathrm{NO}$
$
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})
$
was studied with initial pressure of $1 \mathrm{~atm}$ of $\mathrm{NO}$ and $\mathrm{I}$ atm of $\mathrm{O}_2$. At equilibrium, partial pressure of oxygen is $0.52 \mathrm{~atm}$. Calculate $\mathrm{K}_{\mathrm{p}}$ of the reaction.
Answer:

As,
$
\begin{aligned}
& 1-\mathrm{x}=0.52 \\
& \mathrm{x}=0.48 \\
& =\text { At equilibrium, } \\
& \mathrm{P}_{\mathrm{NO}}=1-2 \mathrm{x}=1-2(0.48)=0.04 \\
& \mathrm{P}_{\mathrm{NO}_2}=2 \mathrm{x}=2(0.48)=0.96 \\
& \mathrm{~K}_{\mathrm{eq}}=\frac{\mathrm{P}_{\mathrm{NO}_2}^2}{\mathrm{P}_{\mathrm{NO}}^2 \cdot \mathrm{P}_{\mathrm{O}_2}}=\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52}=11.07 \times 10^2(\mathrm{~atm})^{-1}
\end{aligned}
$
Question 3.
The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
$\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2$ (g) At a given temperature $\mathrm{K}_{\mathrm{p}}=2.7$. If $0.13 \mathrm{moI}$ of $\mathrm{CO}, 0.56$ moI of water, $0.78 \mathrm{~mol}$ of $\mathrm{CO}_2$ and $0.28 \mathrm{~mol} \mathrm{of}_2$ are introduced into a $2 \mathrm{~L}$ flask, find out in which direction must the reaction proceed to reach equilibrium.
Answer:

$
\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
$
Given
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=2.7 \\
& {[\mathrm{CO}]=0.13 \mathrm{~mol},\left[\mathrm{H}_2 \mathrm{O}\right]=0.56 \mathrm{~mol}} \\
& {\left[\mathrm{CO}_2\right]=0.78 \mathrm{~mol} ;\left[\mathrm{H}_2\right]=0.28 \mathrm{~mol}} \\
& \mathrm{~V}=2 \mathrm{~L} \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\
& 2.7=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\circ} \\
& \mathrm{K}_{\mathrm{C}}=2.7
\end{aligned}
$
$
\mathrm{Q}_{\mathrm{C}}=\frac{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]}{[\mathrm{CO}]\left[\mathrm{H}_2 \mathrm{O}\right]}=\frac{\left(\frac{0.78}{2}\right)\left(\frac{0.28}{2}\right)}{\left(\frac{0.13}{2}\right)\left(\frac{0.56}{2}\right)}
$
$
\mathrm{Q}=3
$
$\mathrm{Q}>\mathrm{K}_{\mathrm{C}}$ Hence, the reaction proceed in the reverse direction.
Question 4
$1 \mathrm{moI}$ of $\mathrm{PCl}_5$, kept in a closed container of volume $1 \mathrm{dm}^3$ and was allowed to attain equilibrium at $423 \mathrm{~K}$. Calculate the equilibrium composition of reaction mixture. (The $\mathrm{K}_{\mathrm{C}}$ value for $\mathrm{PCl}_5$ dissociation at $423 \mathrm{~K}$ is 2 )
Answer:
$
\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCL}_3+\mathrm{Cl}_2
$
Given that $\left[\mathrm{PCl}_5\right]_{\text {initial }}=1 \mathrm{~mol}$
$
\begin{aligned}
& \mathrm{V}=1 \mathrm{dm}^3 \\
& \mathrm{~K}_{\mathrm{C}}=2
\end{aligned}
$

$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} \\
2 & =\frac{x \times x}{(1-x)} \\
2-2 x & =x^2 \\
x^2+2 x-2 & =0
\end{aligned}
$
Solution for a quadratic equation
$
\begin{aligned}
& a x^2+b x+c=0 \text { are, } x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& a=1, b=2, c=-2 \\
& x=\frac{-2 \pm \sqrt{4-4 \times 1 \times(-2)}}{2 \times 1}=\frac{-2 \pm \sqrt{12}}{2}=\frac{-2 \pm \sqrt{4 \times 3}}{2} \\
& x=\frac{-2 \pm 2 \sqrt{3}}{2}=\frac{-2+2 \sqrt{3}}{2}, \frac{-2-2 \sqrt{3}}{3} \\
& {\left[\mathrm{PCl}_5\right]_{\mathrm{eq}}=\frac{1-x}{1}=1-0.732=0.268 \mathrm{M}} \\
& {\left[\mathrm{PCl}_3\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \\
& {\left[\mathrm{Cl}_2\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \\
&
\end{aligned}
$
Question 5.
The equilibrium constant for the following reaction is 0.15 at $298 \mathrm{~K}$ and 1 atm pressure.
Answer:
$
\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})
$

$\begin{aligned}
&\begin{aligned}
& \mathrm{T}_1=298 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{p}_1}=0.15 \\
& \mathrm{~T}_2=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}
\end{aligned}\\
&\begin{aligned}
& \mathrm{K}_{\mathrm{p}_2}=? \\
& \log \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right] \\
& \log \left(\frac{\mathrm{K}_{\mathrm{P}_2}}{0.15}\right)=\frac{57.32 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}\left[\frac{373-298}{373 \times 298}\right] \\
& \log \left(\frac{\mathrm{K}_{\mathrm{P}_2}}{0.15}\right)=\frac{57.32 \times 10^3 \times 75}{2.303 \times 8.314 \times 373 \times 298} \\
& \log \left(\frac{\mathrm{K}_{\mathrm{P}_2}}{0.15}\right)=2.02
\end{aligned}
\end{aligned}$

$
\begin{aligned}
\frac{\mathrm{K}_{\mathrm{P}_2}}{0.15} & =104.7 \\
\mathrm{~K}_{\mathrm{P}_2} & =104.7 \times 0.15 \\
\mathrm{~K}_{\mathrm{P}_2} & =15.705
\end{aligned}
$
Text problems Solved
Question 1.
One mole of $\mathrm{H}_2$ and one mole of $\mathrm{I}_2$ are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data: $\left[\mathrm{H}_2\right]=\mathrm{I}$ mole, $\left[\mathrm{I}_2\right]=1$ mole
At equilibrium, $[\mathrm{HI}]=0.4$ mole, $\mathrm{K}_{\mathrm{C}}=$ ?
$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$

$\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{0.4 \times 0.4}{0.8 \times 0.8}=0.25$

Question 2.
The equilibrium concentrations of $\mathrm{NH}_3, \mathrm{~N}_2$ and $\mathrm{H}_2$ are $1.8 \times 10^2 \mathrm{M}, 1.2 \times 10^{-2} \mathrm{M}$ and $3 \times 10^{-2} \mathrm{M}$ respectively. Calculate the equilibrium constant for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$. [Hint: $\left.\mathrm{M}=\mathrm{mol} \mathrm{lit}^{-1}\right]$
Answer:
Given data:
$
\begin{aligned}
& {\left[\mathrm{NH}_3\right] 1.8 \times 10^{-2} \mathrm{M},\left[\mathrm{N}_2\right]=1.2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right] 3 \times 10^{-2} \mathrm{M}, \mathrm{K}_{\mathrm{C}}=?} \\
& \mathrm{~N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \\
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{1.8 \times 10^{-2} \times 1.8 \times 10^{-2}}{1.2 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2}}=1 \times 10^3 \mathrm{~L}^2 \mathrm{~mol}^{-2}
\end{aligned}
$
Question 3.
The equilibrium constant at $298 \mathrm{~K}$ for a reaction is 100 .
$\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$
If the initial concentration of all the four species is $1 \mathrm{M}$, the equilibrium concentration of $\mathrm{D}$ (in $\mathrm{mol}$ lit $^{-1}$ ) will be
Answer:
Given data: $[\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=[\mathrm{D}]=1 \mathrm{M}, \mathrm{K}_{\mathrm{C}}=100,[\mathrm{D}]_{\mathrm{eq}}=$ ?

$
\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \\
& 100=\frac{(1+x)(1+x)}{(1-x)(1-x)} \\
& \sqrt{100}=\sqrt{\frac{(1+x)(1+x)}{(1-x)(1-x)}} \\
& 10=\frac{1+x}{1-x} \\
& 10(1-x)=1+x \\
& 10-10 x-1-x=0 \\
& 9-11 x=0 \\
& 11 x=9 \\
& x=\frac{9}{11}=0.818 \\
& {[\mathrm{D}]_{\mathrm{eq}}=1+x=1+0.818=1.818 \mathrm{M}} \\
&
\end{aligned}
$
Question 4
For an equilibrium reaction $\mathrm{K}_{\mathrm{p}}=0.0260$ at $25^{\circ} \mathrm{C}$ and $? \mathrm{H}=32.4 \mathrm{~kJ} \mathrm{mor}^{-1}$. Calculate $\mathrm{K}_{\mathrm{p}}$ at $37^{\circ} \mathrm{C}$.
Answer:
$
\begin{aligned}
& \mathrm{T}_1=25+273=298 \mathrm{~K} \\
& \mathrm{~T}_2=37+273=310 \mathrm{~K} \\
& \Delta \mathrm{H}=32.4 \mathrm{~kJ} \mathrm{mor}^{-1}=32400 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \mathrm{~K}_{\mathrm{p}_1}=0.0260 \\
& \mathrm{~K}_{\mathrm{p}_2}=?
\end{aligned}\\
&\begin{aligned}
\log \frac{\mathrm{K}_2}{0.0260}= & \frac{\Delta \mathrm{H}^{\circ}}{2.303 \times 8.314}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_2 \mathrm{~T}_1}\right] \\
\log \frac{\mathrm{K}_2}{0.0260} & =\frac{32400}{2.303 \times 8.314}\left(\frac{310-298}{310 \times 298}\right) \\
& =\frac{32400 \times 12}{2.303 \times 8.314 \times 310 \times 298}=0.2198
\end{aligned}
\end{aligned}$

$
\begin{aligned}
\log \frac{\mathrm{K}_2}{0.0260} & =\text { anti } \log 0.2198=1.6588 \\
\mathrm{~K}_2 & =1.6588 \times 0.026=0.0431
\end{aligned}
$