Chapter 6: Equilibrium

Ans: Let the amount of bromine and chlorine formed at equilibrium be ‘x’. The given reaction is:

$2{\text{BrCl (g}}) \rightleftharpoons {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{BrCl}}$ is $(3.3 \times {10^{ - 3}} - 2x)$ 

Now, we can write.

$ \Rightarrow \dfrac{{[{\text{B}}{{\text{r}}_{\text{2}}}][{\text{C}}{{\text{l}}_2}]}}{{[{\text{BrCl}}]}} = {{\text{K}}_{\text{c}}}$

$ \Rightarrow \dfrac{{x \times x}}{{{{(3.3 \times {{10}^{ - 3}} - 2x)}^2}}} = 32$

$ \Rightarrow \dfrac{x}{{3.3 \times {{10}^{ - 3}} - 2x}} = 5.66$

$ \Rightarrow x = 18.678 \times {10^{ - 3}} - 11.32x$

$ \Rightarrow 12.32x = 18.678 \times {10^{ - 3}}$

$ \Rightarrow x = 1.5 \times {10^{ - 3}}$

 Therefore, at equilibrium

$[{\text{BrCl}}] = 3.3 \times {10^{ - 3}} - (2 \times 1.5 \times {10^{ - 3}}){\text{}}$

$[{\text{BrCl}}] = 3.3 \times {10^{ - 3}} - 3.0 \times {10^{ - 3}}$

$[{\text{BrCl}}] = 3.0 \times {10^{ - 4}}{\text{ mol}}{{\text{L}}^{ - 1}}$

Ans: Let, the total mass of the gas mixture = 100 gm

Mass of ${\text{CO}}$ = 90.55 gm

And, mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ = (100 – 90.55) = 9.45 gm

Now, the number of moles of ${\text{CO}}$ , ${n_{CO}} = \dfrac{{90.55}}{{28}} = 3.23{\text{4 mol}}$ 

Number of moles of ${\text{C}}{{\text{O}}_{\text{2}}}$, ${{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = \dfrac{{9.45}}{{44}} = 0.21{\text{5 mol}}$ 

Partial pressure of ${\text{CO}}$ , ${p_{CO}} = \dfrac{{{n_{CO}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {p_{{\text{total}}}}$ 

${p_{CO}} = \dfrac{{3.234}}{{3.234 + 0.215}} \times 1{\text{}}$

${p_{CO}} = 0.938\;{\text{atm}}$

Partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ ,

${p_{C{O_2}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {p_{{\text{total}}}}{\text{}}$

${p_{C{O_2}}} = \dfrac{{0.215}}{{3.234 + 0.215}} \times 1$

${p_{C{O_2}}} = 0.06{\text{2 atm}}$

Therefore, 

For the given reaction:

$\Delta n = 2 - 1 = 1$ 

We know that,

${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{{\text{(RT}})^{\Delta {\text{n}}}}14.19{\text{}}$ 

$ = {{\text{K}}_{\text{c}}}{(0.082 \times 1127)^1}$

${{\text{K}}_{\text{c}}} = \dfrac{{14.19}}{{0.082 \times 1127}}$

${{\text{K}}_{\text{c}}} = 0.154$

Ans: For the given reaction: 

${\text{NO (g}}) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{N}}{{\text{O}}_{\text{2}}}{\text{(g}})$

1. The $\Delta {{\text{G}}^\circ }$ for the given reaction will be:

$\Delta {{\text{G}}^\circ } = \Delta {{\text{G}}^\circ }({\text{Products}}) - \Delta {{\text{G}}^\circ }({\text{Reactants}}){\text{}}$ 

$\Delta {{\text{G}}^\circ } = 52.0 - [87.0 + 0]$

$\Delta {{\text{G}}^\circ } = {\text{}} - 35.{\text{0 kJmo}}{{\text{l}}^{ - 1}}$

1. We know that:

$\Delta {{\text{G}}^\circ } = {\text{RT}}\log {{\text{K}}_{\text{c}}}{\text{}}$

$\Delta {{\text{G}}^\circ } = 2.303{\text{RT}}\log {{\text{K}}_{\text{c}}}$

$\log {{\text{K}}_{\text{c}}} = \dfrac{{ - 35.0 \times {{10}^{ - 3}}}}{{ - 2.303 \times 8.314 \times 298}}$

$\log {{\text{K}}_{\text{c}}} = 6.134$

${{\text{K}}_{\text{c}}} = {\text{antilog }}(6.134)$

${{\text{K}}_{\text{c}}} = 1.36 \times {10^6}$

Hence, the equilibrium constant for the given reaction ${{\text{K}}_{\text{c}}}$ is $1.36 \times {10^6}$ .

Ans: The amount of moles produced by the reaction will rise. According to Le Chatelier's principle, as pressure is reduced, the equilibrium swings in the direction of a greater number of moles of gas. The number of moles of gaseous products in the reaction is greater than the number of moles of gaseous reactants. As a result, the reaction will be propelled ahead. As a result, there will be more moles of reaction products.

1. ${\text{CaO (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s}})$ 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will decrease.

1. $3{\text{Fe (s}}) + 4{{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}{\text{(s}}) + 4{{\text{H}}_{\text{2}}}{\text{(g}})$ 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will remain the same.

Ans: By increasing the pressure, the reactions in (i), (iii), (iv), (v), and (vi) will be changed. Because the number of moles of gaseous reactants is greater than the number of moles of gaseous products, the reaction in (iv) will proceed in the forward direction. Because the number of moles of gaseous reactants is smaller than that of gaseous products, the reactions in (i), (iii), (v), and (vi) will shift backward.

Ans: Given,

$K_{p}$ for the reaction i.e.,

$H_{2(g)}+B r_{2(g)} \leftrightarrow 2 H B r_{(g)}$ is $1.6 \times 10^{5}$

Therefore, for the reaction $2 H B r_{(g)} \leftrightarrow H_{2(g)}+B r_{2(g)}$ the equilibrium constant will be, $K_{p}^{\prime}=\dfrac{1}{K_{p}}$

=$\dfrac{1}{1.6 \times 10^{5}}$

=$6.25 \times 10^{-6}$

Now, let $\mathrm{p}$ be the pressure of both $\mathrm{H}_{2}$ and $\mathrm{Br}_{2}$ at equilibrium.

$\dfrac{p_{H B r} \times p_{2}}{p^{2} H B r}=K_{p}^{\prime}$

$\dfrac{p \times p}{(10-2 p)}=6.25 \times 10^{-6}$

$\dfrac{p}{10-2 p}=2.5 \times 10^{-3}$

$p=2.5 \times 10^{-2}-\left(5.0 \times 10^{-}\right) p$

$p+\left(5.0 \times 10^{-3}\right) p=2.5 \times 10^{-2}$

$\left(1005 \times 10^{-3}\right) p=2.5 \times 10^{-2}$

$p=2.49 \times 10^{-2}$ bar $=2.5 \times 10^{-2}$ (approximately) 

Therefore, at equilibrium,

${\left[\mathrm{H}_{2}\right]=\left[\mathrm{Br}_{2}\right]=2.49 \times 10^{-2} \text { bar }} $

${[\mathrm{HBr}]=10-2 \times\left(2.49 \times 10^{-1}\right) \text { bar }} $

=$9.95 \mathrm{bar}=10 \text { bar (approximately) }$

Ans: For the given equation:

${K_p} = \dfrac{{{p_{CO}} \times {p^3}_{{H_2}}}}{{{p_{C{H_4}}} \times {p_{{H_2}O}}}}$ 

1. How will the values of ${{\text{K}}_{\text{p}}}$ and composition of equilibrium mixture be affected by 

1. Increasing the pressure 

Ans: The equilibrium will shift in the backward direction, according to Le Chatelier's principle.

As the reaction shifts in the backward direction as the pressure is increased, the value of ${{\text{K}}_{\text{p}}}$ decreases.

1. Increasing the temperature 

Ans: Because the reaction is endothermic, the equilibrium will shift forward, according to Le Chatelier's principle.

The value of ${{\text{K}}_{\text{p}}}$ will grow as the temperature rises because the reaction will shift forward.

1. Using a catalyst?

Ans: The existence of a catalyst has no effect on the reaction's equilibrium. A catalyst does nothing but speed up a reaction. As a result, equilibrium will be soon achieved.

Because the catalyst only changes the rate of reaction, the value of ${{\text{K}}_{\text{p}}}$ is unaffected.

Ans: According to Le Chatelier's principle, when ${{\text{H}}_{\text{2}}}$  is added to a reaction, the equilibrium shifts in the forward direction.

1. Addition of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$  

Ans: The equilibrium will shift backwards with the addition of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ .

1. Removal of ${\text{CO}}$  

Ans: When ${\text{CO}}$  is removed from the equation, the equilibrium shifts backward.

1. Removal of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ 

Ans: When ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$  is removed from the equation, the equilibrium shifts forward.

Ans: The expression for ${{\text{K}}_{\text{c}}}$ will be:

${{\text{K}}_{\text{c}}} = \dfrac{{[{\text{PC}}{{\text{l}}_{\text{3}}}{\text{] [C}}{{\text{l}}_{\text{2}}}]}}{{[{\text{PC}}{{\text{l}}_{\text{5}}}{\text{]}}}}$ 

1. What is the value of ${{\text{K}}_{\text{c}}}$  for the reverse reaction at the same temperature? 

Ans: Value of ${{\text{K}}_{\text{c}}}$  for the reverse reaction at the same temperature is:

$K_{c^{\text{prime}}}=\dfrac{1}{K_{c}}$

$=\dfrac{1}{8.3 \times 10^{-3}}=1.2048 \times 10^{2}$

$=120-48$

1. What would be the effect on ${{\text{K}}_{\text{c}}}$  if 

1. More ${\text{PC}}{{\text{l}}_{\text{5}}}$  is added?

Ans:  Because the temperature is constant in this scenario, ${{\text{K}}_{\text{c}}}$ will remain constant.

1. Pressure is increased? 

Ans: At constant temperature, ${{\text{K}}_{\text{c}}}$ remains constant. As a result, ${{\text{K}}_{\text{c}}}$ would not alter in this scenario.

1. The temperature has increased?

Ans: In an endothermic process, the value of ${{\text{K}}_{\text{c}}}$ rises as the temperature rises. Because the reaction is an endothermic reaction, the value of ${{\text{K}}_{\text{c}}}$ will rise as the temperature rises.

Ans: Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is: 

${\text{CO (g}}) + {{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{CO}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ will be $(4.0 - p)$.

It is given that, ${{\text{K}}_{\text{p}}} = 10.1$ 

Now,

${\dfrac{{{p_{c{o_2}}} \times {p_2}}}{{{p_{co}} \times {p_{{H_2}o}}}} = {K_p}} $

${ \Rightarrow \dfrac{{p \times p}}{{(4.0 - p)(4.0 - p)}}} $

${ \Rightarrow \dfrac{p}{{40 - p}} = 3.178} $

${ \Rightarrow p = 12.712 - 3.178p} $

${ \Rightarrow 4.178p = 12.712} $

${ \Rightarrow p = 3.04}$

Hence, at equilibrium, the partial pressure of ${{\text{H}}_{\text{2}}}$  will be 3.04 bar.