Additional Questions - 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

Additional Questions
I. Choose the correct answer.
Question 1.
Which of the following represents physical equilibrium?
(a) $\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$
(b) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
(c) $\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(d) $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Answer:
(c) $\mathrm{H}_2 \mathrm{O}(1) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
Solution:
Physical states arc in equilibrium i.e., liquid - vapour equilibrium.
Question 2.
Which one of the following is an example of chemical equilibrium?
(a) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2$ (g) $\rightleftharpoons 2 \mathrm{NO}_2$ (g)
(b) $\mathrm{I}_2(\mathrm{~s}) \rightleftharpoons \mathrm{I}_2(\mathrm{~g})$
(c) $\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(1)$
(d) $\mathrm{NH}_2 \mathrm{CI}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_4 \mathrm{CI}(\mathrm{g})$

Answer:
(a) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$
Solution:
All the other three are physical equilibrium. Only (a) is chemical equilibrium.
Question 3.
Which one of the following does not undergo sublimation?
(a) Iodine
(b) water
(c) Camphor
(d) Ammonium chloride
Answer:
(b) Water
Question 4.
At chemical equilibrium,
(a) rate of forward reaction $=$ rate of backward reaction
(b) rate of forward reaction > rate of backward reaction
(c) rate of forward reaction < rate of backward reaction
(d) rate of forward reaction = rate of backward reaction

Answer:
(a) rate of forward reaction $=$ rate of backward reaction
Question 5.
Which of the following is an example of homogeneous equilibrium?
(a) $\mathrm{H}_2 \mathrm{O}(1) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(b) $\mathrm{CaCO}_3$ (s) $\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$
(c) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
(d) $2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s})$
Answer:
(c) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
Solution:
Here all reactants and products are in same phase i.e., gaseous phase.
Question 6.
Which of the following is an example of heterogeneous equilibrium?
(a) Synthesis of $\mathrm{HI}$
(b) Dissociation of $\mathrm{PCI}_5$
(c) Acid hydrolysis of ester
(d) Decomposition of limestone
Answer:
(d) Decomposition of limestone
Solution:
$\mathrm{CaCO}_3(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
Here $\mathrm{CO}_2$ is in gaseous state while $\mathrm{CaCO}_3$ and $\mathrm{CaO}$ are in solid state.
Question 7.
Statement I: In dissociation of $\mathrm{PCI}_5$ to $\mathrm{PCI}_3$ and $\mathrm{CI}_2, \mathrm{~K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$
Statement II: In dissociation of $\mathrm{PCI}_5, \Delta \mathrm{n}_{\mathrm{g}}=-$ ve and so $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$.

(a) Statement I \& II are correct and statement II is the correct explanation of statement I.
(b) Statement I \& II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(c) Statement I is correct but statement II is wrong.
Solution:
$
\Delta \mathrm{n}_{\mathrm{g}}=2-1=1=+ \text { ve }
$
Question 8 .
In the reaction, $2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$
(a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$
(b) $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}}$
(c) $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$
(d) $\mathrm{K}_{\mathrm{p}}=\frac{1}{K_C}$
Answer:
(c) $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$
Solution:

$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \text { and } \Delta \mathrm{n}_{\mathrm{g}}=4-2=2 \\
& \therefore \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^2=\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Question 9.
In which of the following reaction, $\mathrm{K}_{\mathrm{p}}$ is equal to $\mathrm{K}_{\mathrm{C}}$ ?
(a) $\mathrm{N}_2$ (g) $\pm \mathrm{O}_2$ (g) $\rightleftharpoons 2 \mathrm{NO}$ (g)
(b) $2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2$ (g)
(c) $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(d) $\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCI}_3(\mathrm{~g})+\mathrm{CI}_2$
Answer:
(a) $\mathrm{N}_2(\mathrm{~g}) \pm \mathrm{O}_2$ (g) $\rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
Solution:

Question 10 .
In the equilibrium reaction $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ whose concentration remains constant at a given temperature?
(a) $\mathrm{CaO}$
(b) $\mathrm{CO}_2$
(e) $\mathrm{CaCO}_3$
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)
Solution:
Concentration of solids remains constant at a particular temperature.

Question 11.
Consider the following equilibrium reaction and relate their equilibrium constants
(i) $\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; \mathrm{K}_1$
(ii) $2 \mathrm{NO}+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; \mathrm{K}_2$
(iii) $\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; \mathrm{K}_3$
(a) $\mathrm{K}_3=\mathrm{K}_2=\mathrm{K}_1$
(b) $\mathrm{K}_1 \times \mathrm{K}_3=\mathrm{K}_2$
(c) $\mathrm{K}_1 \times \mathrm{K}_2=\mathrm{K}_3$
(d) $\frac{K_1}{K_2}=\mathrm{K}_3$
Answer:
(c) $\mathrm{K}_1 \times \mathrm{K}_2=\mathrm{K}_3$
Solution:
$
\mathrm{K}_1=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]}, \mathrm{K}_2=\frac{\left[\mathrm{NO}_2\right]^2}{[\mathrm{NO}]^2\left[\mathrm{O}_2\right]}, \mathrm{K}_3=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}=\mathrm{K}_1 \times \mathrm{K}_2
$
Question 12.
Statement I: A pure solid in an equilibrium reaction has the same concentration at a given temperature.
Statement II: The solid does not expand to fill its container and it has same number of moles of its volume.
(a) Statement I and II are correct and statement II is the correct explanation of statement of I.
(b) Statement I and II are correct but II is not the correct explanation of!.
(c) Statement I and II are not correct.
(d) Statement $\mathrm{I}$ is wrong but 11 is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement of $\mathrm{I}$.

Question 13 .
Find the $\mathrm{Q}$ value of the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ at an instant where concentration of $\mathrm{H}_2$, $\mathrm{I}_2$ and $\mathrm{HI}$ are found to be $0.2 \mathrm{moI} \mathrm{L}^{-1}, 0.2 \mathrm{~mol} \mathrm{~L}^{-1}$, and $0.6 \mathrm{moI} \mathrm{L}^{-1}$ respectively.
(a) 48
(b) 9
(c) 0.9
(d) 90
Answer:
(b) 9
Solution:
$
\mathrm{Q}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{0.6 \times 0.6}{0.2 \times 0.2}=9
$
Question 14.
For the reaction $\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \mathrm{K}_{\mathrm{C}}=0.21$ at $373 \mathrm{~K}$. The concentrations of $\mathrm{N}_2 \mathrm{O}_4$ and $\mathrm{NO}_2$ direction of the reaction.
(a) At equilibrium

(b) reverse direction
(c) forward direction
(d) Both reverse and forward direction
Answer:
(b) reverse direction
Solution:
$
\begin{aligned}
& \mathrm{Q}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]}=\frac{0.5 \times 0.5}{0.125}=2 \\
& \mathrm{~K}_{\mathrm{C}}=0.21 \\
& \mathrm{Q}=2 \\
& \mathrm{Q}>\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Hence the reaction will proced in the reverse direction
Question 15.
Which of the following does not alter the equilibrium?
(a) catalyst
(b) concentration
(c) temperature
(d) pressure
Answer:
(a) catalyst
Question 16.
Statement I. In Haber's process, $\mathrm{NH}_3$ is liquefied and removed.
Statement II. In manufacture of $\mathrm{NH}_3$, liquefied and removal of $\mathrm{NH}_3$, keeps the reaction moving in forward direction.
(a) Statement I and II are correct and II is the correct explanation of I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is wrong but statement II is correct.

(d) Statement I is correct but statement II is wrong.
Answer:
(a) Statement I and II are correct and II is the correct explanation of I.
Solution:
Removal of $\mathrm{NH}_3$ will decrease its concentration which favours the production of $\mathrm{NH}_3$ according to the $\mathrm{Le}$ - Chatelier's principle.
Question 17.
In which of the following reaction, pressure has no effect?
(a) $\mathrm{N}_2+3 \mathrm{~N}_2 \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
(b) $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$
(c) $\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$
(d) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
Answer:
(d) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
Solution:
In the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ the volumes are equal on both sides and so pressure has no effect.

Question 18.
Among the following reactions which one has $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$
(a) $\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
(b) $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3$ (g)
(c) $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
(d) $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Answer:
(c) $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
Solution:
$
\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot \mathrm{RT}^{\Delta \mathrm{n}_{\mathrm{g}}}
$
In equation $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
$
\begin{aligned}
& \Delta \mathrm{n}_{\mathrm{g}}=0 \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot \mathrm{RT}^{\circ} \\
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
Question 19.
Statement I. Addition of an inert gas at constant volume has no effect on equilibrium. Statement II. When an inert gas is added, the total number of moles of gases present in the container increases and total pressure also increases, the partial pressure of the products and reactants are unchanged.
(a) Statement I and II are correct but statement II is not the correct explanation of I.
(b) Statement I and II are correct and statement II is the correct explanation of 1.
(c) Statement I is correct but statement II is not correct.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statement I and II are correct and statement II is the correct explanation of I.

Question 20.
Which one of the following equation is not correct?
(a) $\Delta \mathrm{G}^{\circ}=-\mathrm{RTInK}$
(b) $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$
(c) $-\mathrm{RTInK}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$
(d) In $\mathrm{k}=\frac{\Delta H^{\circ}}{T}-\frac{\Delta S^{\circ}}{R}$
Answer:
(d) In $\mathrm{k}=\frac{\Delta H^{\circ}}{T}-\frac{\Delta S^{\circ}}{R}$
Question 21.
The equilibrium expression, $\mathrm{K}_{\mathrm{C}}=\left[\mathrm{CO}_2\right]$ represents the reaction.
(a) $\mathrm{C}(\mathrm{s})+\mathrm{O}_2$ (g) $\rightleftharpoons \mathrm{CO}_2$ (g)
(b) $\mathrm{CaCO}_3$ (s) $\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2$ (g)
(c) $2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_2(\mathrm{~g})$
(d) $\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2$ (g) $\rightleftharpoons \mathrm{CaCO}_3$ (s)
Answer:
(b) $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2$ (g)

Question 22.
Hydrogen molecule $\left(\mathrm{H}_2\right)$ can be dissociated into hydrogen atoms $(\mathrm{H})$. Which one of the following changes will not increase the number of atoms present at equilibrium?
(a) adding $\mathrm{H}$ atoms
(b) increasing the temperature
(c) increasing the total pressure
(d) increasing the volume of the container
Answer:
(d) increasing the total pressure container
Solution:
It favours backward reaction i.e., formation of $\mathrm{H}_2$ molecule.
Question 23.
What is the expression for $\mathrm{K}_{\mathrm{eq}}$ for the reaction, $2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})$ ?
(a) $\frac{\left[\mathrm{N}_2\right]\left(\mathrm{O}_2\right]}{[\mathrm{NO}]}$
(b) $\frac{[\mathrm{NO}]^4}{\left[\mathrm{~N}_2 \mathrm{O}\right]^2}$
(c) $\frac{[\mathrm{NO}]^4}{\left[\mathrm{~N}_2 \mathrm{O}\right]^2\left[\mathrm{O}_2\right]}$
$(d) \frac{\left[\mathrm{N}_2 \mathrm{O}\right]^2\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^4}$

Answer:
(c) $\frac{[\mathrm{NO}]^4}{\left[\mathrm{~N}_2 \mathrm{O}\right]^2\left[\mathrm{O}_2\right]}$
Question 24.
What is the correct expression for the representation of the solubility product constant of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
(a) $\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right]$
(b) $\left[2 \mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{2-}\right]$
(c) $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{2-}\right]$
(d) $\left[2 \mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4{ }^{2-}\right]$
Answer:
(a) $\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4{ }^{2-}\right]$

Question 25.
$\mathrm{H}_2+\mathrm{S} \rightleftharpoons \mathrm{H}_2 \mathrm{~S}+$ energy. In this reversible reaction, select the factor which will shift the equilibrium to the right.
(a) adding heat
(b) adding $\mathrm{H}_2 \mathrm{~S}$
(c) blocking hydrogen gas reaction
(d) removing hydrogen suiphide gas
Answer:
(a) removing hydrogen suiphide gas
Question 26.
What effect does a catalyst have on the equilibrium position of a reaction?
(a) a catalyst favours the formation of products
(b) a catalyst favours the formation of reactants
(c) a catalyst does not change the equilibrium position of a reaction
(d) a catalyst may favour reactants or product formation, depending upon the direction in which the reaction is written.
Answer:
(c) a catalyst does not change the equilibrium position of a reaction
Question 27.
A chemist dissolves an excess of $\mathrm{BaSO}_4$ in pure water at $25^{\circ} \mathrm{C}$ if its $\mathrm{K}_{\mathrm{sp}}=1 \times 10^{-10}$. What is the concentration of barium in the water?
(a) $10^{-4} \mathrm{M}$
(b) $10^{-5} \mathrm{M}$
(c) $10^{-15} \mathrm{M}$
(a) $10^{-6} \mathrm{M}$
Answer:
(c) $10^{-15} \mathrm{M}$

Solution:
$
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]
$
$
\begin{aligned}
& 1 \times 10^{-10}=(\mathrm{x})(\mathrm{x}) \\
& 10^{-5}=\mathrm{x}
\end{aligned}
$
Question 28.
If in a mixture where $Q=K$, then what happens?
(a) the reaction shift towards products
(b) the reaction shift towards reactants
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate (d) nothing happens
Answer:
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate
Question 29.
If dissociation for reaction $\mathrm{PCI}_5 \rightleftharpoons \mathrm{PCI}_3+\mathrm{Cl}_2$ is $20 \%$ at $1 \mathrm{~atm}$ pressure. Calculate the value of $\mathrm{K}_{\mathrm{C}}$.
(a) 0.04

(b) 0.05
(c) 0.07
(d) 0.06
Answer:
(d) 0.05
Solution:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{\left[\frac{20}{100}\right] \times\left[\frac{20}{100}\right]}{\left[\frac{80}{100}\right]}=\frac{0.2 \times 0.2}{0.8}=\frac{0.04}{0.8}=0.05
$
Question 30.
What would be the value of $\Delta \mathrm{n}_{\mathrm{g}}$ for the reaction $\mathrm{NH}_4 \mathrm{CI}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_2(\mathrm{~g})+\mathrm{HCI}(\mathrm{g})$ ?
(a) 1
(b) 0.5
(c) 2
(d) 1.5
Answer:
(c) 2
Solution:
$
\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=2-0=2
$
Question 31.
Which of the following is not a general characteristic of equilibria involving physical processes?
(a) Equilibrium is possible only in a close system at a given temperature
(b) All measurable properties of the system remain constant
(c) All the physical processes stop at equilibrium
(d) The opposing processes occur at the same rate and there is dynamic but stable condition

Answer:
(c) All the physical processes stop at equilibrium
Question 32 .
At $500 \mathrm{~K}$, equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for the following reaction is $5, \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$ what would be the equilibrium constant $\mathrm{K}_{\mathrm{C}}$. for the reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$
(a) 0.44
(b) 0.04
(c) 25
(d) 2.5
Answer:
(b) 0.04
Solution:
$
\begin{aligned}
& 2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \\
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=? \\
& \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g}) \\
& \mathrm{K}_{\mathrm{C}}^1=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{\frac{1}{2}}\left[\mathrm{I}_2\right]^{\frac{1}{2}}}=5 \Rightarrow\left(\frac{1}{\mathrm{~K}_{\mathrm{C}}^{\prime}}\right)^2=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{\left[\mathrm{HI}^2\right.}=\left(\frac{1}{5}\right)^2=\mathrm{K}_{\mathrm{C}} \\
& \mathrm{K}_{\mathrm{C}}=\frac{1}{25}=0.04
\end{aligned}
$

Question 33.
For the reaction $2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}), \mathrm{K}_{\mathrm{C}}=1.8 \times 10^{-6}$ at $185^{\circ} \mathrm{C}$. At the same temperature the value of $\mathrm{K}_{\mathrm{C}}$ for the reaction. $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{NO}_2(\mathrm{~g})$ is
(a) $0.9 \times 10^6$
(b) $7.5 \times 10^2$
(c) $1.95 \times 10^{-3}$
(d) $1.95 \times 10^3$
Answer:
(b) $7.5 \times 10^2$
Solution:
The reaction is reversed and halved.
$
\mathrm{K}_{\mathrm{C}}=\frac{1}{\sqrt{1.8 \times 10^{-6}}}=7.5 \times 10^2
$

Question 34.
Which of the following reaction will be favoured at low pressure?
(a) $\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}$
(b) $\mathrm{H}_2 \pm \mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}$
(c) $\mathrm{PCI}_5 \rightleftharpoons \mathrm{PCI}_3+\mathrm{Cl}_2$
(d) $\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$
Answer:
(c) $\mathrm{PCI}_5 \rightleftharpoons \mathrm{PCI}_3+\mathrm{Cl}_2$
Question 35 .
Consider the reaction $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ is a closed container at equilibrium. What would be the effect of addition of $\mathrm{CaCO}_3$ on the equilibrium?
(a) increases
(b) remains unaffected
(c) decreases
(d) unpredictable
Answer:
(b) remains unaffected
Question 36.
For the reaction $\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCI}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ the forward reaction at constant temperature is favoured by
(a) introducing an inert gas at constant volume
(b) introducing $\mathrm{PCl}_3(\mathrm{~g})$ at constant volume
(c) introducing $\mathrm{PCl}_5(\mathrm{~g})$ at constant volume
(d) introducing $\mathrm{CI}_2(\mathrm{~g})$ at constant volume
Answer:
(c) introducing $\mathrm{PCI}_5$ (g) at constant volume

Question 37.
The equilibrium of the reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$ will shift to product side when....
(a) $\mathrm{K}_{\mathrm{p}}>1$
(b) $\mathrm{Q}<\mathrm{K}_{\mathrm{p}}$
(c) $\mathrm{Q}=\mathrm{K}_{\mathrm{p}}$
(d) $\mathrm{Q}=2 \mathrm{~K}_{\mathrm{p}}$
Answer:
(b) $\mathrm{Q}<\mathrm{K}_{\mathrm{p}}$
Question 38 .
$\mathrm{NO}_2$ is involved in the formation of smog and acid rain. A reaction that is important in the formation of $\mathrm{NO}_2$ is $\mathrm{O}_3(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{O}_2(\mathrm{~g})+\mathrm{NO}_2(\mathrm{~g}) \mathrm{K}_{\mathrm{C}}=6.0 \times 10^{34}$. If the air over a section of New Delhi contained $1.0 \times 10^{-6} \mathrm{M}$ of $\mathrm{O}_3, 1.0 \times 10^{-5} \mathrm{M}$ of NO, $2.5 \times 10^{-4} \mathrm{M}$ of $\mathrm{NO}_2$ and $8.2 \mathrm{x}$

$10^{-3}$ of $\mathrm{O}_2$, what can we conclude?
(a) there will be a tendency to form more $\mathrm{NO}$ and $\mathrm{O}_2$
(b) there will be a tendency to form more $\mathrm{NO}_2$ and $\mathrm{O}_2$
(c) there will be a tendency to form more $\mathrm{NO}_2$ and $\mathrm{O}_3$
(d) there will no tendency for change because the reaction is at equilibrium

Answer:
(b) there will be a tendency to form more $\mathrm{NO}_2$ and $\mathrm{O}_2$
Solution:
$
\begin{aligned}
& \mathrm{Q}=\frac{\left[\mathrm{O}_2\right]\left[\mathrm{NO}_2\right]}{\left[\mathrm{O}_3\right][\mathrm{NO}]}=\frac{\left[8.2 \times 10^{-3}\right]\left[2.5 \times 10^{-4}\right]}{\left[1.0 \times 10^{-6}\right]\left[1.0 \times 10^{-5}\right]} \\
&= 8.2 \times 2.5 \times 10^4=20.5 \times 10^4
\end{aligned}
$
As $\mathrm{Q}<\mathrm{K}_{\mathrm{C}}$ the reaction will hive a tendency to move forward.
Question 39 .
Haemoglobin $(\mathrm{Hb})$ forms bond with oxygen and given oxyhaemoglobin $\left(\mathrm{HbO}_2\right)$. This process is partially regulated by the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$and dissolved $\mathrm{CO}_2$ in blood as $\mathrm{HbO}_2+\mathrm{H}_3 \mathrm{O}^{+}+$ $\mathrm{CO}_2 \rightleftharpoons \mathrm{H}^{+}-\mathrm{Hb}-\mathrm{CO}_2+\mathrm{O}_2+\mathrm{H}_2 \mathrm{O}$. If there is production of lactic acid and $\mathrm{CO}_2$ during a muscular exercise, then
(a) more $\mathrm{HbO}_2$ is formed
(b) more $\mathrm{O}_2$ is released
(c) $\mathrm{CO}_2$ is released
(d) both (b) and (c)
Answer:
(b) more $\mathrm{O}_2$ is released

Question 40.
In the reaction $\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3+\mathrm{x} k$ Cal, one mole of $\mathrm{N}_2$ reacts with. 3 moles of $\mathrm{H}_2$ at equilibrium. Then the value of a (degree of dissociation) is approximately $P$ is the pressure at equilibrium

(a) $\frac{\mathrm{P} \sqrt{27 K_P}}{8}$
(b) $\frac{8 \mathrm{P}}{\mathrm{K}_{\mathrm{P}} \sqrt{27}}$
(c) $\frac{\mathrm{P} \sqrt{27}}{8 \mathrm{~K}_{\mathrm{P}}}$
(d) $\frac{n}{v}$
Answer:
(a) $\frac{\mathrm{P} \sqrt{27 \mathrm{~K}_{\mathrm{P}}}}{8}$
Solution:
$\begin{array}{llc} & \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \\ \text { Initial moles } & 1 & 3\end{array}$
Moles of equilibrium $\quad(1-\alpha) \quad(3-3 \alpha) 2 \alpha$
Total number of moles at equilibrium $=1-\alpha+3-3 \alpha+2 \alpha=4-2 \alpha$

$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{NH}_3}\right)^2}{\left(\mathrm{P}_{\mathrm{N}_2}\right)\left(\mathrm{P}_{\mathrm{H}_2}\right)^3}=\frac{\left(\frac{2 x}{4-2 \alpha}\right)^2 \mathrm{P}^2}{\left[\left(\frac{1-\alpha}{4-2 \alpha}\right)^{\mathrm{P}}\right]\left[\frac{3(1-a) \mathrm{P}}{4-2 \alpha}\right]^3}=\frac{4 \alpha^2 \mathrm{P}^2}{(4-2 \alpha)^2} \times \frac{(4-2 \alpha)^4}{(1-\alpha)^4 \mathrm{P}^4 27} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{4 \alpha^2(4-2 \alpha)^2}{27(1-\alpha)^4 \mathrm{P}^2}=\frac{16 \alpha^2(2-\alpha)^2}{27 \mathrm{P}^2(1-\alpha)^4}=\frac{16 \alpha^2(2)^2}{27 \mathrm{P}^2(1)^4}\{\because \alpha<<1\} \\
& \Rightarrow \sqrt{\frac{27 \mathrm{~K}_{\mathrm{P}}^2}{64}}=\alpha \Rightarrow \alpha=\frac{\mathrm{P} \sqrt{27 \mathrm{~K}_{\mathrm{P}}}}{8}
\end{aligned}
$
II. Match the following.

Answer:
(a) 2, 1, 4, 3
Question 2.

Answer:
(b) $2,3,4,1$
Question 3.

Answer:
(a) $4,1,2,3$
Question 4.

Answer:
(a) $3,4,1,2$
Question 5 .

Answer:
(b) 2, 3, 4, 1
III. Fill in the blanks.
Question 1.
Transport of oxygen by Hemoglobin in our body is .......... a reaction.
Answer:
reversible
Question 2.
The temperature at which the solid and liquid phases of a substance are at equilibrium is called ..............
Answer:
Freezing point
Question 3. 
Thejemperature at which the liquid and vapour phases are at equilibrium is called .................
Answer:
Condensation point
Question 4.

..................  law is used to explain gas-solution equilibrium processes. 
Answer:
Henry' law
Question 5.
In the reaction $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$, the $\mathrm{K}_{\mathrm{p}}$ value is equal to ....................

Answer:
$<\mathrm{K}_{\mathrm{C}}$
Solution:

Qustion 6.
The expression of $\mathrm{K}$ for the reaction $\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_3^{-}$(aq) is equal to ................
Answer:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}^{+}\right]_{a q}\left[\mathrm{HCO}_3^{-}\right]_{a q}}{\left[\mathrm{CO}_2\right]_{a q}}
$
Question 7.
The expression of $\mathrm{K}$ for the reversible reaction $2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s})$
Answer:
$
\mathrm{K}_{\mathrm{P}}=\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CO}}^2}
$
Question 8 .
The Ang value for the reaction $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$ is
Answer:
$-1$

Solution:
$
\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=2-3=-1
$
Question 9.
The correct differential form of van't Hoff equation is ............
Answer:
$
\frac{d(\ln k)}{d \Gamma}=\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{RT}^2}
$
Question 10.
For the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$, the equilibrium constant $\mathrm{K}$ is .....................
Answer:
$
\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
$
Question 11.
$\mathrm{PCI}_5$ is kept in a closed container at a temperature of $250 \mathrm{~K}$ the equilibrium concentrations of
$\mathrm{PCI}_5, \mathrm{PCl}_3$ and $\mathrm{Cl}_2$ are 0.045 moles $\mathrm{L}^{-1}, 0.096$ moles $\mathrm{L}^{-1}, 0.096$ moles $\mathrm{L}^{-1}$ respectively. The value of equilibrium constant for the reaction $\mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ will be .........................
Answer:
$
0.205
$
Solution:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.096 \times 0.096}{0.045}=0.2048=0.205
$

Question 12 .
Equilibrium constant changes with .............................
Answer:
Both temperature and pressure
Question 13.
For the reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$ at $720 \mathrm{~K}$, the equilibrium constant value is 50 . The equilibrium constant for the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ at the same temperature will be ....................
Answer:
0.02
Solution:
Forward reaction equilibrium constant $\mathrm{K}_1=50$
Reverse reaction equilibrium constant $\mathrm{K}_2=$ ?
$\mathrm{K}_2=\frac{1}{K_1}=\frac{1}{50}=0.02$
Question 14.
If equilibrium constant for the reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$ at $298 \mathrm{~K}$ is 2.54 , the value of equilibrium constant for the reaction $\frac{1}{2} \mathrm{~N}_2+\frac{3}{2} \mathrm{H}_2 \rightleftharpoons \mathrm{NH}_3$ will be ..................
Answer:
1.59
Solution:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}^{\prime} \text { for } \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})=\frac{\left(\mathrm{P}_{\mathrm{NH}_2}\right)^2}{\mathrm{P}_{\mathrm{N}_2} \times\left(\mathrm{P}_{\mathrm{H}_2}\right)^3}=2.54 \\
& \mathrm{~K}_P^{\prime \prime} \text { for } \frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightleftharpoons \mathrm{NH}_3=\frac{\left(\mathrm{P}_{\mathrm{NH}_3}\right)}{\left(\mathrm{P}_{\mathrm{N}_2}\right)^{\frac{1}{2}} \times\left(\mathrm{P}_{\mathrm{H}_2}\right)^{\frac{3}{2}}}
\end{aligned}
$
Since the reaction is halved $\mathrm{K}_{\mathrm{P}}^{\prime \prime}=\sqrt{\mathrm{K}_{\mathrm{P}}^{\prime}}=\sqrt{(2.54)}=1.59$

Question 15 .
The chemical system at equilibrium is not affected by addition of .........................
Answer:
catalyst
Question 16.
A catalyst will increase the rate of a chemical reaction by lowering the ..................
Answer:
activation energy
Question 17.
In a closed system, $\mathrm{A}(\mathrm{s}) \rightleftharpoons 3 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g})$ If partial pressure of $\mathrm{C}$ is doubled, then partial pressure of B will be .................................. times the original value.
Answer:
$\frac{1}{2 \sqrt{2}}$
Question 18 .
Consider the following gaseous equilibria with equilibrium constants $\mathrm{K}_1$ and $\mathrm{K}_2$ respectively
$
\begin{aligned}
& \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g})-\mathrm{K}_1 \\
& 2 \mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})-\mathrm{K}_2
\end{aligned}
$
The equilibrium constants are related as ...............................
Answer:
$
\mathrm{K}_1^2=\frac{1}{\mathrm{~K}_2}
$
Solution:
$
\begin{aligned}
& \mathrm{K}_1= \frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}, \mathrm{~K}_2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2} \\
& \frac{1}{\mathrm{~K}_1^2}=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}=\mathrm{K}_2 \Rightarrow \mathrm{K}_2=\frac{1}{\mathrm{~K}_1^2} \Rightarrow \mathrm{K}_1^2=\frac{1}{\mathrm{~K}_2}
\end{aligned}
$
Question 19.
$\mathrm{K}_2$ for the following reaction at $700 \mathrm{~K}$ is $1.3 \times 10^{-3} \mathrm{~atm}^{-1}$. The $\mathrm{K}$ at same temperature for the reaction $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ will be ..........................
Answer:
$7.4 \times 10^{-2}$
Solution:
$
K_P=K_C(R T)^{\Delta n_g,} \mathrm{R}=\text { gas constant }
$

$
\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}^{\Delta \mathrm{n}_{\mathrm{g}}}}=\frac{1.3 \times 10^{-3}}{(0.0821 \times 700)^{-1}}=7.4 \times 10^{-2}
$
Question 20.
For the reaction $\mathrm{PCI}_{23}(\mathrm{~g})+\mathrm{CI}_2(\mathrm{~g}) \rightleftharpoons \mathrm{PCI}_5(\mathrm{~g})$ at $250^{\circ} \mathrm{C}$, the value of $\mathrm{K}_{\mathrm{C}}$ is 26 then the value of $\mathrm{K}_{\mathrm{p}}$ on the same temperature will be ........................
Answer:
0.61
Solution:
$
\Delta n_g=1-2=-1 \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta} \mathrm{ng}=26(0.0821 \times 523)^{-1}=0.61
$
Question 21.
In the reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$, the value of the equilibrium constant depends on ............................
Answer:
the temperature
Question 22.
$\mathrm{K}_1$ and $\mathrm{K}_2$ are velocity constant o forward and backward reactions. The equilibrium constant $\mathrm{K}_{\mathrm{C}}$ of the reaction is .......................
Answer:
$\frac{K_1}{K_2}$
Question 23.
The equilibrium constant of the reaction $3 \mathrm{C}_2 \mathrm{H}_2 \rightleftharpoons \mathrm{C}_6 \mathrm{H}_6$ is 4.0 at a temperature of $\mathrm{T} \mathrm{K}$. If the equilibrium concentration of $\mathrm{C}_2 \mathrm{H}_2$ is 0.5 molesL $\mathrm{m}^{-1}$, the concentration of $\mathrm{C}_6 \mathrm{H}_6$ is ........................
Answer:
$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{C}_6 \mathrm{H}_6\right]}{\left[\mathrm{C}_2 \mathrm{H}_2\right]^3} \\
4 & =\frac{\left[\mathrm{C}_6 \mathrm{H}_6\right]}{[0.5]^3} \\
{\left[\mathrm{C}_6 \mathrm{H}_6\right] } & =4 \times 0.5 \times 0.5 \times 0.5=0.5 \text { moles } \mathrm{L}^{-1}
\end{aligned}
$
Question 24.
In an equilibrium reaction for which $\Delta \mathrm{G}^0=0$, the equilibrium constant $\mathrm{K}$ should be $=$ ..........................
Answer:
1
Question 25.
The equilibrium constant for the reaction $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ is 5 . If the equilibrium constant mixture contains equal moles of $\mathrm{SO}_3$ and $\mathrm{SO}_2$, the equilibrium partial pressure of $\mathrm{O}_2$, gas is .............................
Answer:
$0.2 \mathrm{~atm}$
Solution:
$
\mathrm{K}_{\mathrm{P}}=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{O}_2\right]\left[\mathrm{SO}_2\right]^2}=5 \Rightarrow\left[\mathrm{O}_2\right]=\frac{1}{5}=0.2
$

Question 26.
In the reaction $\mathrm{NH}_4 \mathrm{CI}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})+\mathrm{HCI}(\mathrm{g})$ the value Offlg is $\Delta \mathrm{n}_{\mathrm{g}}$
Answer:
2
Solution:
$\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}(\mathrm{g})-\mathrm{n}_{\mathrm{r}}(\mathrm{g})=2-0=2$
IV. Choose the odd one out.
Question 1.
(a) see - saw
(b) tug - of - war
(c) sublimation of camphor
(d) Acid hydrolysis of an ester
Answer:
(d) Acid hydrolysis of an ester Acid hydrolysis of an ester is a chemical equilibrium.
$\mathrm{CH}_3 \mathrm{COOCH}_{3(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{aq})} \stackrel{\mathrm{H}^{+}}{\rightleftharpoons} \mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{CH}_3 \mathrm{OH}_{(\mathrm{aq})}$
Whereas a, b, c are examples of physical equilibrium.
Question 2.
(a) Synthesis of hydrogen iodide
(b) Decomposition of calcium carbonate
(c) Sublimation of iodine
(d) Dissociation of $\mathrm{PCl}_5$
Answer:
(c) Sublimation of iodine Sublimation of iodine $\mathrm{I}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{I}_2(\mathrm{~g})$ is an example of physical equilibrium, whereas $\mathrm{a}, \mathrm{b}$ and dare examples of chemical equilibrium.
Question 3.
(a) Synthesis of $\mathrm{HI}$
(b) Dissociation of $\mathrm{PCl}_5$
(e) Synthesis of $\mathrm{NH}_3$
(d) Decomposition of $\mathrm{CaCO}_3$
Answer:
(a) Decomposition of $\mathrm{CaCO}_3$
Decomposition of $\mathrm{CaCO}_3$ is an example of heterogeneous equilibrium.

$
\mathrm{CaCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}
$
Where as a, b, c are examples of homogeneous equilibrium.
Question 4.
(a) $2 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})}$
(b) $\mathrm{Ag}_2 \mathrm{O}_{\text {(s) }}+2 \mathrm{NH}_{3 \text { (aq) }} \rightleftharpoons 2 \mathrm{AgNO}_{3(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{\text {(l) }}$
(c) $2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{3(\mathrm{~g})}$
(d) $\mathrm{CaCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}$
Answer:
(c) $2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{3(\mathrm{~g})}$
Synthesis of $\mathrm{SO}_3$ is an example of homogeneous equilibrium whereas the others $\mathrm{a}, \mathrm{b}$ and $\mathrm{d}$ are heterogeneous equilibrium.
Question 5.
(a) $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}$
(b) $2 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})}$
(c) $2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{3(\mathrm{~g})}$
(d) $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HCl}_{(\mathrm{g}}$
Answer:
(b) $2 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{\text {(s) }}$
(b) is an example of heterogeneous equilibrium whereas the others a, $c$ and $d$ are homogeneous equilibrium.
V. Choose the correct pair.
Question 1.
(a) $\mathrm{Q}=\mathrm{K}_{\mathrm{C}}$ : Reaction is in equilibrium state
(b) $\mathrm{Q}<\mathrm{K}_{\mathrm{C}}$ : Reaction proceed in reverse direction
(c) $\mathrm{Q}>\mathrm{K}_{\mathrm{C}}$ : Reaction proceed in ftrward direction
(d) $\mathrm{Q}=\mathrm{K}_{\mathrm{C}}:$ Reaction proceed in both directions
Answer:
(a) $\mathrm{Q}=\mathrm{K}_{\mathrm{C}}$ : Reaction is in equilibrium state

Question 2.
(a) $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \quad: \quad \Delta \mathrm{n}_{\mathrm{g}}=-\mathrm{ve}$
(b) $\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}: \Delta \mathrm{n}_{\mathrm{g}}=-\mathrm{ve}$
(c) $\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{(\mathrm{g})}: \Delta \mathrm{n}_{\mathrm{g}}=+\mathrm{ve}$
(d) $2 \mathrm{NH}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}: \Delta \mathrm{n}_{\mathrm{g}}=+\mathrm{ve}$
Answer:
(d) $2 \mathrm{NH}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}: \Delta \mathrm{n}_{\mathrm{g}}=+\mathrm{ve}$ $\Delta \mathrm{n}_{\mathrm{g}}=4-2=+2$
Question 3 .
(a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$ : Synthesis of $\mathrm{HI}$
(b) $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}:$ Dissociation of $\mathrm{PCl}_5$
(c) $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}}:$ Synthesis of $\mathrm{SO}_3$
(d) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}:$ Synthesis of $\mathrm{HI}$
Answer:
(a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$ : Synthesis of $\mathrm{HI}$
$\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}: \Delta \mathrm{n}_{\mathrm{g}}=2-2=0$
$\Delta \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$
VI. Choose the incorrect pair.
Question 1.
(a) Acid hydrolysis of an ester - Homogeneous equilibrium
(b) Synthesis of Ammonia - Homogeneous equilibrium
(d) Synthesis of HI - Homogeneous equilibrium
Ans.
(c) Decomposition of $\mathrm{CaCO}_3$ : Homogeneous equilibrium It is a heterogeneous equilibrium.
Question 2.
(a) $\Delta \mathrm{n}_{\mathrm{g}}=0 \quad: \quad \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}$
(b) $\Delta \mathrm{n}_{\mathrm{g}}=2 \quad: \quad 2 \mathrm{NH}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}$
(c) $\Delta \mathrm{n}_{\mathrm{g}}=0 \quad: \mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$
(d) $\Delta \mathrm{n}_{\mathrm{g}}=-1 \quad: \quad 2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{3(\mathrm{~g})}$
Answer:
$
\begin{aligned}
& \text { (c) } \Delta \mathrm{n}_{\mathrm{g}}=0 \quad: \mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \\
& \text { In }(c) \Delta \mathrm{n}_{\mathrm{g}}=2-1=1 \\
&
\end{aligned}
$

VII. Assertion and Reason.
Question 1.
Assertion (A): Chemical equilibrium is in a state of dynamic equilibrium. Reason (R): At equilibrium the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) $(\mathrm{A})$ is correct but $(\mathrm{R})$ is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
Question 2.
Assertion (A): In Haber's process, $\mathrm{NH}_3$ is liquefied and removed.
Reason (R): Because of the reaction keeps moving in the backward direction.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) $(\mathrm{A})$ is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
Question 3.
Assertion (A): In the dissociation of $\mathrm{PCI}_5$ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of $\mathrm{PCl}_5$.
Reason (R): Helium removes CI, from the field of action.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.
(b) Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is the not the correct explanation of $\mathrm{A}$.
(c) $A$ is true but $R$ is false
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are false
Answer:
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are false
VIII. Choose the incorrect statement.
Question 1.
(a) In equilibrium mixture of ice and water kept in perfectly insulted flask, mass of ice and water does not change with time.
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate
(c) On addition of catalyst the equilibrium constant value is not affected
(d) Equilibrium constant for a reaction with negative $\mathrm{H}$ value decreases as the temperature increases.

Answer:
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate
Solution:
Oxalate ions of oxalic acid form complex with ferric ions thus decrease its concentration thus, concentration of red complex in product decreases.
2 Mark Questionsand Answers.
I. Write brief answer to the following questions:
Question 1.
Define the state of equilibrium.
Answer:
At a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.
Question 2.
What are the different types of equilibrium? Explain with example?
Answer:
1. Physical equilibrium:
A system in which the amount of matter constituting different phases does not change with time is said to be in physical equilibrium. $\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(1)$. Solid-liquid equilibrium.
2. Chemical equilibrium:
Chemical reactions in which the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed is said to be in chemical equilibrium. $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons$ $2 \mathrm{HI}(\mathrm{g})$
Question 3.
Explain about the equilibrium involving dissolution of solid in liquid with suitable example.
Answer:

When sugar is added to water at a particular temperature. it dissolves to form sugar solution. When more sugar is added to that solution, a particular stage sugar remains as solid and results in the formation of saturated solution. Here a dynamic equilibrium is established between the solute molecules ii the solid phase and in the solution phase.
Sugar $_{(\text {solid })} \rightleftharpoons$ Sugar $_{i \text { solution) }}$
Question 4.
How is a gas - solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecuLes in the gaseous state and those dissolved in the liquid. Example - In carbonate beverages the following equilibrium exists. $\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_2$ (Solution)
Question 5.
What is meant by active mass? Give its unit.
Answer:
The active mass represents the molar concentration of the reactants (or) products.
Active mass $=\frac{n}{v}=\frac{\text { number of moles }}{\text { volume in litre }}$
Question 6.
Show that $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$ with two example
Answer:
Examples,
(i) $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$

$
\Delta \mathrm{n}_{\mathrm{g}}=2-2=0
$
(ii)
$
\begin{aligned}
\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) & \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \\
\Delta \mathrm{n}_{\mathrm{g}} & =2-2=0
\end{aligned}
$
Question 7.
Give two examples of equilibrium reactions where $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$.
Answer:
Examples,
(ii) $\begin{aligned} 2 \mathrm{NH}_3(\mathrm{~g}) & \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \\ \Delta \mathrm{n}_{\mathrm{g}} & =4-2=2, \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^2 \\ \mathrm{PCl}_5(\mathrm{~g}) & \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \\ \Delta \mathrm{n}_{\mathrm{g}} & =2-1=1, \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^1\end{aligned}$
$(i)$
So, $\mathrm{K}_{\mathrm{P}}>\mathrm{K}_{\mathrm{C}}$
Question 8.
When will be $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}}$ ? Give two example.

Answer:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}} \text {. Examples, } \\
& 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
& \Delta \mathrm{n}_{\mathrm{g}}=2-3=-1 \\
& \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot \mathrm{RT}^{-1} \\
& 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) \\
& \Delta \mathrm{n}_{\mathrm{g}}=2-3=-1, \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{-1} \\
& \therefore \mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{C}} \\
&
\end{aligned}
$

Question 9.
Write the $\mathrm{K}_{\mathrm{C}}$ for the reaction $\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_{3(\mathrm{aq})}-$
Answer:
$\mathrm{H}_2 \mathrm{O}(1)$ is a pure liquid and its concentration remains constant.
Solution:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}^{+}\right]_{\mathrm{aq}} \times\left[\mathrm{HCO}_3^{-}\right]_{\mathrm{aq}}}{\left[\mathrm{CO}_2\right]_{\mathrm{g}}}
$
Question 10 .
$\mathrm{A} \stackrel{\mathrm{K}_1}{\rightleftharpoons} \mathrm{B}, \mathrm{B} \stackrel{\mathrm{K}_2}{\rightleftharpoons} \mathrm{C}, \mathrm{C} \stackrel{\mathrm{K}_3}{\rightleftharpoons} \mathrm{D}$ What is the value of $\mathrm{K}_4$ in $\mathrm{A} \rightleftharpoons \mathrm{D}$
Answer:
$
\begin{aligned}
& A \stackrel{K_1}{\rightleftharpoons} B, B \stackrel{K_2}{\rightleftharpoons} C, C \stackrel{K_3}{\rightleftharpoons} D \\
& \therefore K_4=K_1 K_2 K_3 \text { for } A \rightleftharpoons D
\end{aligned}
$
Question 11.
Write the $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$ for the following reactions
(i) $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$
(ii) $2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s})$

Answer:
(i) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}, \mathrm{K}_{\mathrm{P}}=\frac{p_{\mathrm{SO}_3}^2}{p_{\mathrm{SO}_2}^2 \cdot p_{\mathrm{O}_2}}$
(ii) $\mathrm{C}(\mathrm{s})$ has constant concentration
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{CO}_2\right]}{[\mathrm{CO}]^2}, \mathrm{~K}_{\mathrm{P}}=\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CO}}^2}
$
Question 12.
Explain how the equilibrium constant $\mathrm{K}$. predict the extent of a reaction.
Answer:
1. The value of equilibrium constant $\mathrm{K}_{\mathrm{C}}$ tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.
2. A large value of $\mathrm{K}_{\mathrm{C}}$ indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of $\mathrm{K}_{\mathrm{C}}$ indicates that the reaction reaches equilibrium with low product yield.
3. If $\mathrm{K}_{\mathrm{C}}>10^3$ the reaction proceeds nearly to completion.
4. If $\mathrm{K}_{\mathrm{C}}<10^{-3}$ the reaction rarely proceeds.
5. If the $\mathrm{K}_{\mathrm{C}}$ is in the range $10^{-3}$ to $10^3$ significant amount of both reactants and products are present at equilibrium.
Question 13.
$
\begin{gathered}
2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \mathrm{K}_{\mathrm{C}}=4.1 \times 10^{-48} \text { At } 599 \mathrm{~K} \\
\mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \mathrm{K}_{\mathrm{C}}=1 \times 10^{-30} \text { at } 1000 \mathrm{~K}
\end{gathered}
$
Predict the extent of the above two reactions.
Answer:
In the reactions, decomposition of water at $500 \mathrm{~K}$ and oxidation 01 nitrogen at $1000 \mathrm{~K}$, the value of $\mathrm{K}_{\mathrm{C}}$ is very less $\mathrm{K}_{\mathrm{C}}<10^{-3}$. So reverse reaction is favoured.
$\therefore$ Products $\ll$ reactants

Question 14.
Explain about the extent of reaction of dissociation of bromine mono chloride at $1000 \mathrm{~K}$.
Answer:
$
\begin{aligned}
& 2 \mathrm{BrCl}(\mathrm{g}) \rightleftharpoons \mathrm{Br}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \mathrm{K}_{\mathrm{C}}=5 \\
& 10^{-3}<\mathrm{K}_{\mathrm{C}}<10^3
\end{aligned}
$
both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.
Question 15.
What is the $\mathrm{K}_{\mathrm{C}}$ value for formation of $\mathrm{HCl}$ at $700 \mathrm{~K}$ ? Predict the extent of the reaction?
Answer:
$
\begin{aligned}
& \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \text { at } 700 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{C}}=57.0 \\
& 10^{-3}<\mathrm{K}_{\mathrm{C}}<10^3
\end{aligned}
$
So both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.
Question 16.
What is the $\mathrm{K}_{\mathrm{C}}$ value of formation of $\mathrm{HCI}$ at $300 \mathrm{~K}$ ? Explain it.
Answer:
$
\begin{aligned}
& \mathrm{H}_2(\mathrm{~g})-\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCI}(\mathrm{g}) \text { at } 300 \mathrm{~K} \\
& \mathrm{~K}_2=4 \times 10^{31} \\
& \mathrm{~K}_{\mathrm{C}}>10^3 \text { So[products] } \gg \text { [Reactant] }
\end{aligned}
$
Reaction nearly goes to completion. So forward reaction is favoured.
Question 17.
$2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_2(\mathrm{~g})$ at $1000 \mathrm{~K}$. What is the $\mathrm{Kc}$ this reaction? Predict the extent of this reaction.
Answer:
$
2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_2(\mathrm{~g}) \text { at } 1000 \mathrm{~K}
$

$
\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=2.2 \times 10^{22} \\
& \mathrm{~K}_{\mathrm{C}}>10^3
\end{aligned}
$
So [Products] $\gg$ [Reactants]
Reaction nearly goes to completion and forward reaction is favoured.
Question 18.
Define $Q$ value for a chemical equilibrium reaction.
Answer:
Consider a homogeneous reversible reaction $\mathrm{xA}+\mathrm{yB} \rightleftharpoons 1 \mathrm{C}+\mathrm{mD}$ For the above reaction under non-equilibrium conditions, reaction quotient $\mathrm{Q}$ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants. Under non equilibrium conditions,
$
\mathrm{Q}=\frac{d(I n k)}{d t} \Delta \mathrm{H}^{\circ}=\frac{\Delta H^0}{R T^2}
$
Question 19.
Explain the diagrammatic expression about the direction of reaction.
Answer:

reactants $\rightarrow$ products equilibrium products $\rightarrow$ reactants
In (i) $\mathrm{Q}<\mathrm{K}_{\mathrm{C}}$, the reaction will proceed in forward direction.
In (ii) $\mathrm{Q}=\mathrm{K}_{\mathrm{C}}$, the reaction is in equilibrium state.
In (iii) $\mathrm{Q}>\mathrm{K}_{\mathrm{C}}$, the reaction will proceed in the reverse direction.
Question 20.
Explain about the effect of catalyst in an equilibrium reaction?
Answer:
Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.
Question 21.
For the following equilibrium. $\mathrm{K}_{\mathrm{C}}=6.3 \times 10^{14}$ at $1000 \mathrm{~K}$
$
\mathrm{NO}(\mathrm{g})+\mathrm{O}_3(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})
$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions what is $\mathrm{K}_{\mathrm{C}}$ for the reverse reaction?
Answer:
For the reverse reaction
$
\mathrm{K}_{\mathrm{C}}=\frac{1}{\mathrm{~K}_{\mathrm{C}}}=\frac{1}{6.3 \times 10^{14}}=1.59 \times 10^{-1.6}
$
Question 22.
Explain why pure liquids and solids can be ignored while writing the value of equilibrium constants.
Answer:
This is because molar concentration of a pure solid or liquid is independent of the amount present.
$
\text { Molar concentration }=\frac{\text { No. of moles }}{\text { Volume }} \times \frac{\text { Mass }}{\text { Volume }} \times \text { Density }
$
Since density of pure liquid or solid is fixed and molar mass is also fixed, therefore molar concentration are constant.
Question 23.
A sample of $\mathrm{Hl}(\mathrm{g})$ is placed in a flask at a pressure of $0.2 \mathrm{~atm}$. At equilibrium partial pressure of $\mathrm{HI}(\mathrm{g})$ is $0.04 \mathrm{~atm}$. What is $\mathrm{K}_{\mathrm{p}}$ for the given equilibrium?
$
2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})
$
Answer:
$
\begin{aligned}
& \mathrm{pHI}=0.04 \mathrm{~atm} . \mathrm{pH}_2=0.08 \mathrm{~atm} ; \mathrm{pl}_2=0.08 \mathrm{~atm} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{\mathrm{pH}_2 \times \mathrm{pI}_2}{\mathrm{p}_{\mathrm{HI}}^2}=\frac{(0.08 \mathrm{~atm}) \times(0.08 \mathrm{~atm})}{(0.04 \mathrm{~atm}) \times(0.04 \mathrm{~atm})}=4.0
\end{aligned}
$
Question 24 .
The equilibrium constant expression for a gas reaction is
$
\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}
$
Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
$
4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g})
$
Question 25 .
Predict which of the following will have appreciable concentration of reactions and products:
(a) $\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}}=5 \times 10^{-39}$
(b) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}}=3.7 \times 108$
(c) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2 \mathrm{Cl}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}}=1.8$
Answer:
Following conclusion can be drawn from the values of
(a) Since the value of $\mathrm{K}$ is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of $\mathrm{K}$ is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of $\mathrm{K}$ is 1.8 , this means that both the products and reactants have appreciable concentrations.
Question 26.
Write the equilibrium constant $\left(\mathrm{K}_{\mathrm{C}}\right)$ expression for the following reactions.
(i) $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq})$
(ii) $4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

Answer:
(i) $\log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_2 \mathrm{~T}_1}\right]$
(ii) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Cl}_2(g)\right]^2\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2}{[\mathrm{HCl}(\mathrm{g})]^4\left[\mathrm{O}_2(g)\right]}$
Question 27.
For the equilibrium $2 \mathrm{NOCI}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{CI}_2(\mathrm{~g})$ the value of the equilibrium constant $\mathrm{K}_{\mathrm{C}}$ is $3.75 \mathrm{x}$ at $1069 \mathrm{~K}$. Calculate the $\mathrm{K}_{\mathrm{p}}$ for the reaction at this temperature?
Answer:
We know that $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$
For the above reaction, $\Delta \mathrm{n}_{\mathrm{g}}=(2+1)-2=1$
$
\mathrm{K}_{\mathrm{p}}=3.75 \times 10^{-6}(0.0831 \times 1069)=3.3 \times 10^{-4}
$
Question 28.
The value of $\mathrm{Kc}$ for the reaction $2 \mathrm{~A} \rightleftharpoons \mathrm{B}+\mathrm{C}$ is $2 \times 10^{-3}$. At a given time, the composition of reaction mixture is $[\mathrm{A}][\mathrm{B}][\mathrm{C}]=3 \times 10^{-4} \mathrm{M}$. In which direction the reaction will proceed?
Answer:
For the reaction, the reaction quotient $\mathrm{Q}$ is given by $\mathrm{Q}_{\mathrm{C}}=[\mathrm{B}][\mathrm{C}] /[\mathrm{A}]^2$
as $[\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=3 \times 10^{-4} \mathrm{M}$
$\mathrm{Q}_{\mathrm{C}}\left(3 \times 10^{-4}\right)\left(3 \times 10^{-4}\right) /\left(3 \times 10^{-4}\right)^2=1$
as $\mathrm{Q}_{\mathrm{C}}>\mathrm{K}_{\mathrm{C}}$ so, the reaction will proceed in the reverse direction.
Question 29.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
$
\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})
$
(a) Write an expression for $\mathrm{K}_{\mathrm{p}}$ for the above reaction.
(b) How will the values of $\mathrm{K}_{\mathrm{p}}$ and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst
Answer:
(a) $\mathrm{K}_{\mathrm{p}}=\frac{\left[p_{\mathrm{CO}}\right]\left[p_{\mathrm{H}_2}\right]^3}{\left[p_{\mathrm{CH}_4}\right]\left[p_{\mathrm{H}_2 \mathrm{O}}\right]}$

(b) (I) value of $\mathrm{K}_{\mathrm{p}}$ will not change, equilibrium will shift in backward direction.
(ii) value of $\mathrm{K}_{\mathrm{p}}$ will increase and reaction will proceed in forward direction.
(iii) no effect.
II. Answer the following questions:
Question 1.
Explain about the formation of solid-liquid equilibrium with suitable example.
Answer:
1. Consider melting of ice in a closed container at $273 \mathrm{~K}$. This system reach a state of physical equilibrium in which the amount of water in the solid phase and liquid phase does not change with time.
2. In this process. the total number of water molecules leaving from and returning to the solid phase at any instant are equal.
3. If some ice cubes and water are placed in a thermos flask (at $273 \mathrm{~K}$ and $\mathrm{I}$ atm) then there will be no change in the mass of ice and water.
4. At equilibrium: Rate of melting of ice Rate of freezing of water $\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
Question 2.
How is liquid - vapour equilibrium exist?
Answer:
1. Liquid water is in equilibrium with vapour at $373 \mathrm{~K}$ and I atm pressure in a closed vessel
2. Rate of evaporation $=$ Rate of condensation
Question 3.
What is meant by boiling point and condensation point of the liquid?
Answer:
The temperature at which the liquid and vapour phases are at equilibrium is called the boiling point and condensation point of the liquid.

Question 4.
Define melting point (or) freezing point of the substance.
Answer:
The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of substance.
Question 5.
Illustrate the formation of solid - vapour equilibrium with suitable example.
Answer:
1. Consider a system in which the solid sublimes to vapour. e.g., I, (or) camphor.
2. When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine.
3. Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained.
$\mathrm{I}_2(\mathrm{~s}) \rightleftharpoons \mathrm{I}_2(\mathrm{~g})$
Question 6.
Give three examples for solid vapour equilibrium.
Answer:
$\mathrm{I}_2(\mathrm{~s}) \rightleftharpoons \mathrm{I}_2(\mathrm{~g})$
Camphor (s) $\rightleftharpoons$ Camphor $(\mathrm{g})$
$\mathrm{NH}_4 \mathrm{CI}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_4 \mathrm{CI}(\mathrm{g})$
Question 7.
Explain the following diagrams.
Diagram - 1

Answer:
1. As the concentration of the products increases, more products collide and react in the backward direction.
2. As the rate of the reverse reaction increases, the rate of the forward reaction decreases.
3. Eventually the rate of both reactions becomes equal.
Diagram-II

1. Concentration of reactants decreases with time initially and concentration of products increases with time.
2. After sometime, equilibrium is reached i.e., concentration of reactants and products remains constant.
Question 8.
What are the types of chemical equilibrium? Explain with suitable example.
Answer:
1. Chemical equilibrium is of two types:
- Homogeneous equilibrium
- Heterogeneous equilibrium.
2. In a homogeneous equilibrium, all the reactants and products are in the same phase $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
In the above equilibrium $\mathrm{H}_2, \mathrm{I}_2$ and $\mathrm{HI}$ are in the gaseous state.
3. If the reactants and products of a reaction in equilibrium are in different phases, then it is calLed heterogeneous equilibrium.
$
\text { e.g., } \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_3(\mathrm{~g})
$
Question 9.
Write the value of $\mathrm{K}$ and $\mathrm{K}$ equation for $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
Answer:
A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container i.e., it has the same no. of moles of its volume. Therefore the concentration of pure solid is a constant. So the expression if $\mathrm{K}$ and $\mathrm{K}$ is $\mathrm{K}\left[\mathrm{CO}_2\right], \mathrm{K}=\mathrm{P}_{\mathrm{CO}_2}$
Question 10.
Consider the following equilibrium reaction and relate their equilibrium constants
1. $\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}, \mathrm{K}_1$
2. $2 \mathrm{NO}+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2, \mathrm{~K}_1$
3. $\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2, \mathrm{~K}_3$

Answer:
$
\mathrm{K}_1=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} ; \mathrm{K}_2=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right.}, \mathrm{K}_3=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]}
$
Now $\quad \mathrm{K}_1 \times \mathrm{K}_2=\frac{[\mathrm{NO}]}{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]} \times \frac{\left[\mathrm{NO}_2\right]^2}{[\mathrm{NO}]^2\left[\mathrm{O}_2\right]}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}=\mathrm{K}$
$
\therefore \quad \mathrm{K}_3=\mathrm{K}_1 \times \mathrm{K}_2
$
Question 11.
Explain the effect of concentration in an equilibrium state?
Answer:
At equilibrium, the concentration of the reactants and products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentration
According to Le - Chatelier's principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. For example, $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$

The addition of $\mathrm{H}_2$ or $\mathrm{I}_2$ to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where $\mathrm{H}_2$ and $\mathrm{I}_2$ are consumed i.e., formation of additional $\mathrm{HI}$ would balance the effect of added reactant.
Hence the equilibrium shifts to the right (forward direction). i.e., the equilibrium is re established. Similarly, removal of $\mathrm{HI}$ (Product) also favours forward reaction. If $\mathrm{HI}$ is added to the equilibrium mixture, the concentration of $\mathrm{HI}$ is increased and system proceeds in the reverse direction to nullify the effect of increase in concentration of $\mathrm{HI}$.

Question 12.
Consider the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$. Explain the effect of pressure on this equilibrium reaction.
Answer:
$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$
In the above equilibrium, if the pressure is increased, the volume wiL1 decreases. The system responds to this effect by reducing the number of gas molecules. i.e., it favours the formation of ammonia. If the pressure is reduced, the volume will increases. It favours the decomposition of ammonia.
Question 13.
Why pressure has no effect on the synthesis of $\mathrm{HI}$ ?
Answer:
(i) When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
$
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})
$
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{O}$.
Question 14 .
Explain the effect of temperature on the following equilibrium reaction.
$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \Delta \mathrm{H}=-92.2 \mathrm{~kJ}$.
Answer:
In this equilibrium, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \Delta \mathrm{H}=-92.2 \mathrm{~kJ}$.
Forward reaction is exothermic while the reverse reaction is endothermic. If the temperature of the system increased, the system responds by decomposing some of ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy.
Similarly, the system responds to a drop in the temperature by forming more ammonia molecule from nitrogen and hydrogen which release heat energy.

Question 15.
How does oxygen exchanges between maternal and fetal blood in a pregnant woman? Answer:
1. In a pregnant women, the oxygen supply for the fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus arc in close proximity. Both fetal and maternal hemoglobin binds tO oxygen reversibly as follows.
$
\begin{aligned}
& \mathrm{Hb}_{\text {(mother) }}+\mathrm{O}_2 \rightleftharpoons \mathrm{HbO}_{2 \text { (mother) }} \\
& \mathrm{Hb}_{\text {(fetus) }}+\mathrm{O}_2 \rightleftharpoons \mathrm{HbO}_{2 \text { (fetus) }}
\end{aligned}
$
2. In the above two equilibrium, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher which is due to its higher affinity for oxygen compared to adult hemoglobin. Hence in placenta, the oxygen from the mother's blood s effectively transferred to fetal hemoglobin.
Question 16.
What is $\mathrm{K}$ for the following reaction in state of equilibrium?
$
2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
$
(Given: $\left[\mathrm{SO}_2\right]=0.6 \mathrm{M} ;\left[\mathrm{O}_2\right]=0.82 \mathrm{M} ;$ and $\left[\mathrm{SO}_3\right]=1.90 \mathrm{M}$
Answer:
$
\begin{aligned}
& 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) \\
& \begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}=\frac{(1.9 \mathrm{M}) \times(1.9 \mathrm{M})}{(0.6 \mathrm{M}) \times(0.6 \mathrm{M}) \times(0.82 \mathrm{M})} \\
& =12.229 \mathrm{M}^{-1}=12.229 \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}
$
Question 17.
At a certain temperature and total pressure of $10^5 \mathrm{~Pa}$, iodine vapours contain $40 \%$ by volume of iodine atoms in the equilibrium $1_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})$. Calculate $\mathrm{K}_{\mathrm{p}}$ for the equilibrium.
Answer:
According to available data:
Total pressure of equilibrium mixture $=10^5 \mathrm{~Pa}$
Partial pressure of iodine atoms $(1)=\frac{40}{100} \times\left(10^5 \mathrm{~Pa}\right)=0.4 \times 10^5 \mathrm{~Pa}$
Partial pressure of iodine molecules $\left(\mathrm{I}_2\right)=\frac{60}{100} \mathrm{x}\left(10^5 \mathrm{~Pa}\right)=0.6 \times 10^5 \mathrm{~Pa}$

Question 18.
A mixture of $1.57 \mathrm{~mol}$ of $\mathrm{N}_2, 1.92 \mathrm{~mol}$ of $\mathrm{H}_2$ and $8.13 \mathrm{~mol}$ of $\mathrm{NH}_3$ is introduced into a $20 \mathrm{~L}$ reaction vessel at $500 \mathrm{~K}$. At this temperature, the equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for the reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$ is $1.7 \times 10^{-2}$
Is this reaction at equilibrium? if not. what is the direction of net rection?
Answer:
The reaction is: $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Concentration quotient $(\mathrm{Q})=$
$
\begin{aligned}
\left(\mathrm{Q}_{\mathrm{C}}\right) & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\
& =\frac{\left(\frac{8.13}{20} \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}{\left(\frac{1.57}{20} \mathrm{~mol} \mathrm{~L}^{-1}\right) \times\left(\frac{1.92}{20} \mathrm{~mol} \mathrm{~L}^{-1}\right)^3}=2.38 \times 10^3
\end{aligned}
$
The equilibrium constant $(\mathrm{Kr})$ for the reaction $=1.7 \times 10^{-2}$
As $\mathrm{Q}_{\mathrm{C}} \neq \mathrm{K}_{\mathrm{C}}$, this means that the reaction is not in a state of equilibrium.
Question 19.
What is the effect of:
1. addition of $\mathrm{H}_2$
2. addition of $\mathrm{CH}_3 \mathrm{OH}$
3. removal of $\mathrm{CO}$
4. removal of $\mathrm{CH}_3 \mathrm{OH}$
On the equilibrium $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
Answer:
1. Equilibrium will be shifted in the forward direction.
2. Equilibrium will be shifted in the backward direction.
3. Equilibrium will be shifted in the backward direction.
4. Equilibrium will be shifted in the forward direction.

Question 20.
At $473 \mathrm{~K}$, the equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for the decomposition of phosphorus pentachioride $\left(\mathrm{PCl}_5\right)$ is $8.3 \times 10^{-3}$. If decomposition proceeds as:
$
\begin{aligned}
& \mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCI}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \\
& \Delta \mathrm{H}=+124.0 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$

1. Write an expression for $\mathrm{K}$. for the reaction.
2. What is the value of $\mathrm{K}$ for the reverse reaction at the same temperature.
3. What would be the effect on $\mathrm{K}_{\mathrm{C}}$ if
- More of $\mathrm{PCI}_5$ is added
- Temperature is increased.
Answer:
1. The expression for
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3(g)\right]\left[\mathrm{Cl}_2(g)\right]}{\left[\mathrm{PCl}_5(g)\right]}
$
2. For reverse reaction
$
\left(\mathrm{K}_{\mathrm{C}}^{\prime}\right)=\frac{\mathrm{PCl}_5(g)}{\left[\mathrm{PCl}_3(g)\right]\left[\mathrm{Cl}_2(g)\right]}=\frac{1}{8.3 \times 10^{-3}}=120.48
$
3.
(I) By adding more of $\mathrm{PCI}_5$, value of $\mathrm{K}^{\mathrm{C}}$ will remain constant because there is no change in temperature.
(ii) By increasing the temperature the forward reaction will be favoured since it is endotherniic in nature. Therefore, the value of equilibrium constant will increase.
Question 21.
Dihydrogen gas used in Haher's process is produced by reacting methane from natural gas with high temperature stam. The first stage of two stage reaction involves the formation of $\mathrm{CO}$ and $\mathrm{H}_2$. In second stage, $\mathrm{CO}$ formed in first stage is reacted with more steam in water gas shift reaction.
$
\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
$
If a reaction vessel at $400^{\circ} \mathrm{C}$ is charged with an equimolar mixture of $\mathrm{CO}$ and steam so that $\mathrm{P}_{\mathrm{CO}}$ $=\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=4.0$ bar., what will be the partial pressure of 2 at equilibrium? $\mathrm{K}_{\mathrm{p}}=0.1$ at $400^{\circ} \mathrm{C}$.
Answer:

Let the partial pressure of hydrogen $\left(\mathrm{H}_2\right)$ at equilibrium point $\mathrm{p}$ bar
$
\begin{aligned}
& \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \\
& \text { Initial pressure } 4.0 \mathrm{bar} \quad 4.0 \mathrm{bar} \quad 0 \quad 0 \\
& \text { Eqm. pressure }(4-p) \text { bar }(4-p) \text { bar } \quad p \text { bar } p \text { bar } \\
& \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{CO}_2} \times \mathrm{P}_{\mathrm{H}_2}}{\mathrm{P}_{\mathrm{CO}} \times \mathrm{P}_{\mathrm{H}_2 \mathrm{O}}} \text { or } 0.1=\frac{(p \text { bar }) \times(p \text { bar })}{(4-p) \text { bar } \times(4-p) \text { bar }} \\
& \frac{p^2}{(4-p)^2}=0.1 \text { or } \frac{p}{(4-p)}=(0.1)^{\frac{1}{2}}=0.316 \\
& p=0.316(4-p) \text { or } p=1.264-0.316 p \\
& 1.316 p=1.264 \text { or } p=\frac{1.264}{1.316}=0.96 \mathrm{bar} \\
&
\end{aligned}
$
Question 22.
The value of $\mathrm{K}_{\mathrm{C}}$ for the reaction $3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_3(\mathrm{~g})$ is $2.0 \times 10^{-50}$ So at $25^{\circ} \mathrm{C}$. If equilibrium concentration of 0 , in $25^{\circ} \mathrm{C}$ is $1.6 \times 10^2$ what is the concentration of $\mathrm{O}_3$ ?
Answer:
$
3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_3(\mathrm{~g})
$
or
$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3} \text { or }\left(2.0 \times 10^{-50}\right)=\frac{\left[\mathrm{O}_3\right]^2}{\left(1.6 \times 10^{-2}\right)^3} \\
{\left[\mathrm{O}_3\right]^2 } & =\left(2.0 \times 10^{-50}\right) \times\left(1.610^{-2}\right)^3 \\
{\left[\mathrm{O}_3\right]^2 } & =8.192 \times 10^{-56} \\
{\left[\mathrm{O}_3\right] } & =\left(8.192 \times 10^{-56}\right)^{1 / 2}=2.86 \times 10^{-28} \mathrm{M}
\end{aligned}
$
Question 23.
The reaction $\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ is at equilibrium at $1300 \mathrm{~K}$ in a $1 \mathrm{~K}$ flask. It also contain $0.30 \mathrm{~mol}$ of $\mathrm{CO}, 0.10 \mathrm{~mol}$ of $\mathrm{H}_2$ and $0.02 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ and an unknown amount of $\mathrm{CH}_4$ in the flask. Determine the concentration of $\mathrm{CH}_4$ in the mixture. The equilibrium constant, $\mathrm{K}_{\mathrm{C}}$ for the reaction at the given temperature is 3.90 .

Answer:
$
\begin{aligned}
\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{CH}_4\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}] \times\left[\mathrm{H}_2\right]^3} \text { or } 3.90=\frac{\left[\mathrm{CH}_4\right] \times[0.02]}{[0.30] \times[0.1]^3} \\
{\left[\mathrm{CH}_4\right]=\frac{(3.9) \times(0.30) \times(0.001)}{(0.02)}-5.85 \times 10^{-2} \mathrm{M} }
\end{aligned}
$
Question 24.
The following concentration were obtained for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$ at equilibrium at $500 \mathrm{~K}$
- $\left[\mathrm{N}_2(\mathrm{~g})\right] 1.5 \times 10^{-2} \mathrm{M}$
- $\left[\mathrm{H}_2(\mathrm{~g})\right]=3.0 \times 10^{-2} \mathrm{M}$
- $\left[\mathrm{NH}_3\right]=1.2 \times 10^{-2} \mathrm{M}$.
Calculate equilibrium constant.
Answer:
$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$
Calculate equilibrium constant
Answer:
$
\begin{aligned}
\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{\left[1.2 \times 10^{-2} \mathrm{M}\right]^2}{\left[1.5 \times 10^{-2} \mathrm{M}\right]\left[3.0 \times 10^{-2} \mathrm{M}\right]^{-3}} \\
& =3.55 \times 10^2 \mathrm{M}
\end{aligned}
$
Question 25.
1. In the reaction $A+B \rightarrow C+D$, what will happen to the equilibrium if concentration of $A$ is increased?
2. The equilibrium constant for a reaction is $2 \times 10^{-23}$ at $25^{\circ} \mathrm{C}$ and $2 \times 10^{-2}$ at $50^{\circ} \mathrm{C}$. Is the reaction endothermic or exothermic?
3. Mention at least three ways by which the concentration of $\mathrm{SO}_3$ can be increased in the following reaction in a state of equilibrium. $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$
Answer:

1. The reaction will shift in the forward direction.
2. Endothermic
3. following reaction
- increasing concentration of $\mathrm{SO}_2$
- increasing pressure.
- increasing concentration of oxygen.
Question 26.
$\mathrm{PCI}_5, \mathrm{PCI}_3$ and $\mathrm{CI}_2$ are at equilibrium at $500 \mathrm{~K}$ and having concentration I.59M PCl $\mathrm{PC}_3, 1.59 \mathrm{M}$ $\mathrm{CI}_2$ and $1.41 \mathrm{M} \mathrm{PCI}_5$. Calculate $\mathrm{K}$. for the reaction $\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2$
Answer:
The equilibrium constant $\mathrm{K}$. for the above reaction can be written as:
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{(1.59)^2}{1.41}=1.79
$
Question 27.
Given the equilibrium
$\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2$ (g) with $\mathrm{K}_{\mathrm{p}}=0.15 \mathrm{~atm}$ at $298 \mathrm{~K}$
(a) What is $\mathrm{K}_{\mathrm{p}}$ using pressure in torr?
(b) What is $\mathrm{K}_{\mathrm{C}}$ using units of moles per litre.
Answer:
(a) $\mathrm{K}_{\mathrm{P}}=\frac{(760 \text { torr }) \times(0.15 \mathrm{~atm})}{(1 \mathrm{~atm})}=1.14 \times 10^2$ torr

(b) $\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}}$
$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta n}}=\frac{(0.15 \mathrm{~atm})}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}\right)^{2-1}} \\
& =6.13 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$
1. Derive the values of equilibrium constants $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{C}}$ for a general reaction $\mathrm{xA}+\mathrm{y} \mathrm{B} \rightleftharpoons 1 \mathrm{C}+\mathrm{mD}$
Answer:
Let us consider a reversible reaction $x \mathrm{~A}+\mathrm{y} \mathrm{B} \rightleftharpoons 1 \mathrm{C}+\mathrm{mD}$ where $\mathrm{A}, \mathrm{B}$ are the reactants $\mathrm{C}$ and $\mathrm{D}$ are the product and $\mathrm{x}, \mathrm{y} .1$ and $\mathrm{m}$ are the stoichiometric coefficients of A. B, C and D respectively. Applying the law of mass action the rate of forward reaction.
$\mathrm{r}_{\mathrm{f}} \propto[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ or $\mathrm{r}_{\mathrm{f}} \mathrm{K}_{\mathrm{f}}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$
Similarly the rate of backward reaction
$\mathrm{r}_{\mathrm{b}} \propto[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{m}}$ or $\mathrm{r}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{m}}$

where $\mathrm{K}_{\mathrm{f}}$ and $\mathrm{K}_{\mathrm{b}}$ are proportionality constants.
At equilibrium, Rate of forward reaction $\left(\mathrm{r}_{\mathrm{f}}\right)=$ Rate of backward reaction $\left(\mathrm{r}_{\mathrm{b}}\right)$
$
\begin{array}{rlr}
\therefore & \mathrm{K}_f[\mathrm{~A}]^x \cdot[\mathrm{B}]^v & =\mathrm{K}_b[\mathrm{C}]^l[\mathrm{D}]^m \\
\text { or } & \frac{\mathrm{K}_f}{\mathrm{~K}_b} & =\frac{[\mathrm{C}]^l[\mathrm{D}]^m}{[\mathrm{~A}]^l \cdot[\mathrm{B}]^y}=\mathrm{K}_{\mathrm{C}}
\end{array}
$
where $\mathrm{K}_{\mathrm{C}}$ is the equilibrium constant in terms of concentration. At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant known as equilibrium constant.
If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures
$
\mathrm{K}_{\mathrm{P}}=\frac{p_{\mathrm{C}}^l \times p_{\mathrm{D}}^m}{p_{\mathrm{A}}^x \times p_{\mathrm{B}}^y}
$
where $P_A, P_B, P_C$ and $\mathrm{P}_D$ are the partial pressure of gases $A, B, C$ and $D$ respectively.
Question 2.
Derive the values of $\mathrm{K}$ and $\mathrm{K}$ for the synthesis of $\mathrm{HI}$.
Answer:
$
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})
$
Let us consider the formation of $\mathrm{HI}$ in which ' $a$ ' moles of hydrogen, ' $b$ ' moles of iodine gas are allowed to react in an container of volume ' $\mathrm{V}$ '. Let ' $\mathrm{x}$ ' moles of each of $\mathrm{H}_2$, and $\mathrm{I}_2$, react

$\text { together to form } 2 \mathrm{x} \text { moles of } \mathrm{HI} \text {. }$

Applying law of mass action.
$
\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
$
$
\mathrm{K}_{\mathrm{C}}=\frac{\left(\frac{2 x}{\mathrm{~V}}\right)^2}{\frac{(a-x)}{\mathrm{V}} \frac{(b-x)}{\mathrm{V}}}=\frac{4 x^2}{(a-x)(b-x)}
$
Calculation of $\mathbf{K}_{\mathbf{p}}: \quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot \mathrm{RT}^{\Delta \mathrm{n}} \mathrm{g}$
Here
Hence,
$
\begin{aligned}
\Delta \mathrm{n}_{\mathrm{g}} & =n_p-n_r=2-2=0 \\
\mathrm{~K}_{\mathrm{P}} & =\mathrm{K}_{\mathrm{C}}
\end{aligned}
$
$
\mathrm{K}_{\mathrm{P}}=\frac{4 x^2}{(a-x)(b-x)}
$
Question 3.
Derive the values of $\mathrm{K}$ and $\mathrm{K}$ for dissociation of $\mathrm{PCI}_5$.
Answer:
Consider that ' $\mathrm{a}$ ' moles of $\mathrm{PCl}_5$ is taken in container of volume ' $\mathrm{V}$ ' Let $\mathrm{x}$ moles of $\mathrm{PCI}_5$ be dissociated into $\mathrm{x}$ moles of $\mathrm{PCI}_3$ and $\mathrm{x}$ moles of $\mathrm{Cl}_2$.

Applying law of mass action
$
\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{\left(\frac{x}{\mathrm{~V}}\right)\left(\frac{x}{\mathrm{~V}}\right)}{\frac{a-x}{\mathrm{~V}}}=\frac{x^2}{(a-x) \mathrm{V}} \\
& \mathrm{K}_{\mathrm{P}} \text { calculation: } \quad \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot \mathrm{RT}^{\Delta \mathrm{n}_{\mathrm{g}}} \\
& \Delta \mathrm{n}_{\mathrm{g}}=2-1=1 \\
& \text { We know that } \quad \mathrm{PV}=n \mathrm{RT} \\
& \mathrm{RT}=\frac{\mathrm{PV}}{n} \\
&
\end{aligned}
$
Where ' $n$ ' is the total number of moles at equilibrium
$
\begin{aligned}
n & =a-x+x+x=a+x \\
\mathrm{~K}_{\mathrm{P}} & =\frac{x^2}{(a-x) \mathrm{V}} \cdot \frac{\mathrm{PV}}{n} \\
\mathrm{~K}_{\mathrm{P}} & =\frac{x^2 \times \mathrm{PV}}{(a-x) \mathrm{V}(a+x)} \\
\mathrm{K}_{\mathrm{P}} & =\frac{x^2 \mathrm{P}}{(a-x)(a+x)}
\end{aligned}
$
Question 4.
At certain temperature and under a pressure of $4 \mathrm{~atm}, \mathrm{PCl}_5$ is $10 \%$ dissociated. Calculate the pressure at which $\mathrm{PCI}_5$ will be $20 \%$ dissociated at temperature remaining constant.
Calculation of $\mathrm{K}_{\mathrm{p}}$

Answer:

Total no. of moles in the equilibrium mixture $=1-\alpha+\alpha+\alpha=(1+\alpha) \mathrm{mol}$.
Let the total pressure of equilibrium mixture $=\rho \mathrm{atm}$
Partial pressure of $\mathrm{PCl}_5, \mathrm{P}_{\mathrm{PCl}_5}=\frac{1-\alpha}{1+\alpha} \times p \mathrm{~atm}$
Partial pressure of $\mathrm{PCl}_3=\frac{\alpha}{1+\alpha} \times p$ atm
Partial pressure of $\mathrm{Cl}_2, \mathrm{PCl}_2=\frac{\alpha}{1+\alpha} \times p$ atm
$\mathrm{K}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{PCl}_3} \times \mathrm{P}_{\mathrm{Cl}_2}}{\mathrm{P}_{\mathrm{PCl}_5}}=\frac{\left(\frac{\alpha}{1+\alpha} p \mathrm{~atm}\right) \times\left(\frac{\alpha}{1+\alpha} p \mathrm{~atm}\right)}{\frac{1-\alpha}{1+\alpha} p \mathrm{~atm}}=\frac{\alpha^2 p}{1-\alpha^2}$
$\mathrm{P}=4 \mathrm{~atm}$ and $\alpha=10 \%=\frac{10}{100}=0.1$
$
\mathrm{K}_{\mathrm{P}}=\frac{(0.1) \times(0.1) \times(4 \mathrm{~atm})}{1-(0.1)^2}=\frac{0.04}{0.99}=0.04 \mathrm{~atm}
$
calculation of $P$ under new condition
$
\begin{aligned}
\alpha & =0.2, \mathrm{~K}_{\mathrm{p}}=0.04 \mathrm{~atm} \\
\mathrm{~K}_{\mathrm{p}} & =\frac{\alpha^2 p}{1-\alpha^2} \text { or } p=\frac{\mathrm{K}_{\mathrm{p}}\left(1-\alpha^2\right)}{\alpha^2}
\end{aligned}
$

$
\begin{aligned}
& =\frac{(0.04 \mathrm{~atm})\left[\left(1-(0.2)^2\right]\right.}{(0.2)^2}=\frac{0.04 \mathrm{~atm} \times 0.96}{0.04} \\
& =0.96 \mathrm{~atm}
\end{aligned}
$
IV. Numerical Problems
Question 1.
Find the value of $\mathrm{K}$ for each of the following equlibria from the value of $\mathrm{K}$
(a) $2 \mathrm{NOCI}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g}) ; \mathrm{K}_{\mathrm{p}}=1.8 \times 10^2$ atm at $500 \mathrm{~K}$
(b) $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}): \mathrm{K}_{\mathrm{p}}=167$ atm at $1073 \mathrm{~K}$.

Answer:
(a) $2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=1.8 \times 10^{-2} \mathrm{~atm} \\
& \Delta \mathrm{n}_{\mathrm{g}}=3-2=1 ; \mathrm{R}=0.0821 \text { litre atm } \mathrm{K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=500 \mathrm{~K} \\
& \therefore \quad \mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta n_R}}=\frac{\left(1.8 \times 10^{-2} \mathrm{~atm}\right)}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 500 \mathrm{~K}\right)^1} \\
& =4.4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \\
&
\end{aligned}
$
(b) $\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=167 \mathrm{~atm}, \Delta \mathrm{n}_{\mathrm{g}}=1 \\
& \mathrm{R}=0.0821 \text { litre } \operatorname{atm} \mathrm{K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=1073 \mathrm{~K} \\
& \mathrm{~K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta n_g}}=\frac{\left(1.8 \times 10^{-2} \mathrm{~atm}\right)}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 1073 \mathrm{~K}\right)^1} \\
& =1.9 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1} \\
&
\end{aligned}
$
Question 2.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was $0.78 \mathrm{M}$ ?
$
2 \mathrm{ICI}(\mathrm{g}) \rightleftharpoons \mathrm{I}_2(\mathrm{~g})+\mathrm{CI}_2(\mathrm{~g}) ; \mathrm{K}_{\mathrm{C}}=0.14
$
Answer:

$\text { Suppose at equilibrium, the molar concentration of both } \mathrm{I}_2(\mathrm{~g}) \text { and } \mathrm{Cl}_2(\mathrm{~g}) \text { is } \times \mathrm{mol} \mathrm{L}^{-1} \text {. }$

$
\begin{aligned}
\mathrm{K}_{\mathrm{C}} & =\frac{\left[\mathrm{I}_2(g)\right]\left[\mathrm{Cl}_2(g)\right]}{[\mathrm{ICl}(g)]^2}=\frac{(x) \times(x)}{(0.78-2 x)^2} \\
\frac{x}{(0.78-2 x)} & =(0.14)^{\frac{1}{2}}=0.374 \text { or } x=0.374(0.78-2 x) \\
x & =0.292-0.748 x \text { or } x=1.748 x=0.292 ; x=\frac{0.292}{1.748}=0.167 \\
{[\mathrm{ICl}] } & =(0.78-2 \times 0.167)=(0.78-0.334)=0.446 \mathrm{M} \\
{\left[\mathrm{I}_2\right] } & =0.167 \mathrm{M} ;\left[\mathrm{Cl}_2\right]=0.167 \mathrm{M}
\end{aligned}
$
Question 3.
Equilibrium constant $\mathrm{K}$ for the reaction. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$ at $500 \mathrm{~K}$ is 0.061 . At particular time, the analysis shows that the composition of the reaction mixture is $3.0 \mathrm{~mol} \mathrm{~L}^{-1}$ of $\mathrm{N}_2 ; 2.0 \mathrm{~mol} \mathrm{~L}^{-1}$ of $\mathrm{H}_2 ; 0.50 \mathrm{moI} \mathrm{L}^{-1}$ of $\mathrm{NH}_3$. Is the reaction at equilibrium?
Answer:
The given reaction is: $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
According to available data
$
\begin{aligned}
& \mathrm{N}_2=[3.0] ; \mathrm{H}_2=[2.0] ; \mathrm{NH}_3=[0.50] \\
& \mathrm{Q}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3(g)\right]^2}{\left[\mathrm{~N}_2(g)\right]\left[\mathrm{H}_2(g)\right]^3}=\frac{[0.50]^2}{[3.0][2.0]}=\frac{0.25}{24}=0.0104
\end{aligned}
$
Common Errors
1. In writing $\mathrm{K}_{\mathrm{p}}, \mathrm{K}_{\mathrm{C}}$ values from the equations, students may confuse to write whether products or reactants in the numerator or in the denominator
2. When $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^0$ students may confuse
3. In writing An values, students will consider all reactants and products.
4. Students are confused to understand the concept of $Q$ value and $K_C$ value.
5. In writing chemical equilibrium reaction, you may miss the physical states of reactants and products.

7. Equilibrium constant value is calculated under equilibrium condition and reaction quotient value defined at sometimes at equilibrium condition wrongly by the students.
8. Chemical equilibrium condition must be known. Students may wrongly write in an open vessel, the equilibrium take place.
9. Equilibrium symbol, students may wrongly write as $=$
Rectifications
1. Always we have to write $\mathrm{K}_{\mathrm{p}} \& \mathrm{~K}_{\mathrm{C}}$ as $\frac{\text { [Products] }}{\text { [Reactants] }}$
2. $(\text { Anything })^{\circ}=1$. So $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}$.
3. Only they have to consider gaseous reactants and gaseous products. Since solid and liquid are constant in their concentration.
4. $\mathrm{Q}$ value is calculated under non equilibrium conditions. $\mathrm{K}_{\mathrm{C}}$ value is calculated under equilibrium condition.
5. Physical states of reactants and products must be written as a subscript by the words $\mathrm{s}, 1, \mathrm{~g}$. (solid, liquid, gas)
6. $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{g}_{\mathrm{p}}}-\mathrm{n}_{\mathrm{g}_{\mathrm{r}}}$ You should consider only gaseous products and gaseous reactants.
7. Reaction quotient $\mathrm{Q}$ value is calculated only under non-equilibrium conditions.
8. Chemical equilibrium reactions are always take place at closed vessel only.
9. Equilibrium reaction symbol is $\Leftrightarrow$.