Chapter 6: Equilibrium

Ans: The initial concentration of ${\text{HI}}$  is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of ${\text{HI}}$  is 0.2 – 0.04 = 0.16. The given reaction is: 

At equilibrium the pressure of ${\text{HI}}$ is 0.04 atm, the pressure for ${{\text{H}}_{\text{2}}} = \dfrac{{0.16}}{2} = 0.08{\text{atm}}$ and the pressure for ${{\text{O}}_{\text{2}}} = \dfrac{{0.16}}{2} = 0.08{\text{atm}}$.

Therefore, the value of ${{\text{K}}_{\text{p}}}$ at equilibrium is 4.0.

Ans: For the given reaction: ${{\text{N}}_{\text{2}}}{\text{(g)}} + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$

The given concentration of various species is

${\left[ {{{\text{N}}_2}} \right] = \dfrac{{1.57}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}\quad \left[ {{{\text{H}}_2}} \right] = \dfrac{{1.92}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}}$ 

${\left[ {{\text{N}}{{\text{H}}_3}} \right] = \dfrac{{8.13}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}}$ 

Now, reaction quotient ${{\text{Q}}_{\text{c}}}$  is:

${{Q_C}} { = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}} $ 

${{Q_C}} { = \dfrac{{{{\left( {\dfrac{{(8.13)}}{{20}}} \right)}^2}}}{{\left( {\dfrac{{1.57}}{{20}}} \right){{\left( {\dfrac{{1.92}}{{20}}} \right)}^3}}}} $ 

${{Q_C}} { = 2.4 \times {{10}^3}} $

Since, 𝑄𝑐 ≠ 𝐾𝑐, the reaction mixture is not at equilibrium. 

Again, 𝑄𝑐 >𝐾𝑐. Hence, the reaction will proceed in the reverse direction

Ans: The balanced chemical equation corresponding to the given expression can be written as:

$4{\text{NO (g}}) + 6{{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons 4{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}}) + 5{{\text{O}}_{\text{2}}}{\text{(g)}}$ 

Ans: The given reaction is:

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

${\dfrac{{\text{1}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{\text{1}}}{{{\text{10}}}}{\text{M}}} {\text{0}} {\text{0}} $ 

${\dfrac{{{\text{1 - 0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{1 - 0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}} $ 

${{\text{ = 0}}{\text{.06M}}}{{\text{ = 0}}{\text{.06M}}}{{\text{ = 0}}{\text{.04M}}}{{\text{ = 0}}{\text{.04M}}} $

Therefore, the equilibrium constant for the reaction:

${{\text{K}}_{\text{c}}} = \dfrac{{[{{\text{H}}_{\text{2}}}][{\text{C}}{{\text{O}}_{\text{2}}}]}}{{[{{\text{H}}_{\text{2}}}{\text{O}}][{\text{CO}}]}}$

${{\text{K}}_{\text{c}}} = \dfrac{{0.04 \times 0.04}}{{0.06 \times 0.06}}$

${{\text{K}}_{\text{c}}} = 0.444$

Ans: It is given that equilibrium constant ${{\text{K}}_{\text{c}}}$  for the reaction

${{\text{H}}_{\text{2}}}{\text{(g}}) + {{\text{I}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{HI (g)}}$ is 54.8

Therefore, at equilibrium, the equilibrium constant ${{\text{K}}^{\text{'}}}_{\text{c}}$  for the reaction

$2{\text{HI (g}}) \rightleftharpoons {{\text{H}}_{\text{2}}}{\text{(g)}} + {{\text{I}}_{\text{2}}}{\text{(g}})$ 

$[{\text{HI}}] = 0.5{\text{mol}}{{\text{L}}^{ - 1}}$ will be: $\dfrac{1}{{54.8}}$ 

Let the concentrations of hydrogen and iodine at equilibrium be $x\;{\text{mol}}{{\text{L}}^{ - 1}}$ 

$[{{\text{H}}_{\text{2}}}] = [{{\text{I}}_{\text{2}}}] = x\;{\text{mol}}{{\text{L}}^{ - 1}}$ 

Therefore, $\dfrac{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{] [}}{{\text{I}}_{\text{2}}}{\text{]}}}}{{{{{\text{[HI]}}}^{\text{2}}}}} = {{\text{K}}^{\text{'}}}_{\text{c}}$ 

${\text{}}\dfrac{{x \times x}}{{{{(0.5)}^2}}} = \dfrac{1}{{54.8}}$ 

$ \Rightarrow {x^2} = \dfrac{{0.25}}{{54.8}} \Rightarrow x = 0.06754$

$x = 0.06{\text{8 mol}}{{\text{L}}^{{\text{  -  1}}}}$ (approximately)

Hence, at equilibrium, $[{{\text{H}}_{\text{2}}}] = [{{\text{I}}_2}] = 0.06{\text{8 mol}}{{\text{L}}^{{\text{  -  1}}}}$

Ans: For the given reaction:

$2{\text{ICI (g}}) \rightleftharpoons {{\text{I}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

The initial concentration of ${\text{ICI}}$ is 0.78M and for ${{\text{I}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$ is zero (0).

At equilibrium, the concentration for ${\text{ICI}}$ is $(0.78 - 2x){\text{M}}$ and for ${{\text{I}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$is $x\;{\text{M}}$.

Now, we can write:

$\dfrac{{[{{\text{I}}_{\text{2}}}][{\text{C}}{{\text{l}}_{\text{2}}}]}}{{{{[{\text{ICI}}]}^2}}} = {{\text{K}}_{\text{c}}}$

$\dfrac{{x \times x}}{{{{(0.78 - 2x)}^2}}} = 0.14$

$\dfrac{x}{{0.78 - 2x}} = 0.374$

$x = 0.292 - 0.748x$

$1.748x = 0.292$

$x = 0.167$

Therefore, at equilibrium

$[{{\text{H}}_2}]$ = $[{{\text{I}}_2}]$ = $0.167{\text{}}$

${\text{M}}[{\text{HI}}] = (0.78 - 2 \times 0.167)$

${\text{M}}[{\text{HI}}] = 0.446{\text{M}}$

Ans: Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction

${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{(g}}) \rightleftharpoons {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{(g)}}$

The initial concentration of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ is 4.0 atm and that of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ and ${{\text{H}}_{\text{2}}}$ is zero (0).

At equilibrium, the concentration of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ is (4.0 – p) and that of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ and ${{\text{H}}_{\text{2}}}$is p. 

Then we can write,

$ \Rightarrow {K_p}\dfrac{{p \times p}}{{40 - p}} = 0.04$

${p^2} = 0.16 - 0.04p{\text{}}$

${p^2} + 0.04p - 0.16 = 0$

$p = \dfrac{{ - 0.04 \pm \sqrt {{{(0.04)}^2} - 4 \times 1 \times ( - 0.16)} }}{{2 \times 1}}$

$p = \dfrac{{ - 0.04 \pm 0.08}}{2}$

If we take positive value:

$ p=\dfrac{0.76}{2}$

$p=0.38 $

So, at equilibrium:

$[{C_2}{H_6}] - 4 - p = 4 - 0.38 = 3.62{\text{ atm}}$ 

Ans: Let the concentrations of both ${\text{PC}}{{\text{l}}_{\text{3}}}$  and ${\text{C}}{{\text{l}}_{\text{2}}}$  at equilibrium be $x{\kern 1pt} \;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ . The given reaction is:

${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g}}) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{PC}}{{\text{l}}_{\text{5}}}$is $0.5 \times {10^{ - 1}}\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$

It is given that the value of equilibrium constant ${{\text{K}}_{\text{c}}}$  is $8.3 \times {10^{ - 3}}$

The expression for equilibrium will be:

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

$ \Rightarrow \dfrac{x\times x}{0.5\times {{10}^{-1}}}=8.3\times {{10}^{-3}} $

$ \Rightarrow {{x}^{2}}=4.15\times {{10}^{-4}} $

$ \Rightarrow x=2.04\times {{10}^{-2}} $

$ \Rightarrow x=0.0204 $

Therefore at equilibrium:

$[{\text{PC}}{{\text{l}}_3}] = [{\text{C}}{{\text{l}}_2}] = 0.02{\kern 1pt} {\text{mol}}{{\text{L}}^{ - 1}}$ 

Ans: For the given reaction:

${\text{FeO (s}}) + {\text{CO (g}}) \rightleftharpoons {\text{Fe (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}})$

The initial concentration of ${\text{FeO}}$ is 1.4 atm and the initial concentration of ${\text{C}}{{\text{O}}_{\text{2}}}$ is 0.80 atm. 

It is given in the that, ${{\text{K}}_{\text{p}}} = 0.265$.

 Since ${{\text{Q}}_{\text{p}}} > {{\text{K}}_{\text{p}}}$ , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of ${\text{CO}}$  will increase while the pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$  will decrease. Now, let the increase in pressure of ${\text{CO}}$  be equal to the  decrease in pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$  be ‘p’. Then, we can write:

${{\text{Q}}_{\text{p}}} = \dfrac{{{{\text{p}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}}}{{{{\text{p}}_{{\text{CO}}}}}} = \dfrac{{0.80}}{{1.4}} = 0.571{\text{}}$

Therefore, equilibrium partial of ${\text{C}}{{\text{O}}_{\text{2}}}$ will be, ${{\text{p}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = 0.080 - 0.339 = 0.46{\text{1 atm}}$ 

And, equilibrium partial pressure of ${\text{CO}}$ will be, ${{\text{p}}_{{\text{CO}}}} = 1.4 - 0.339 = 1.73{\text{9 atm}}$ 

Ans: The given reaction is:

${{\text{N}}_{\text{2}}}{\text{(g}}) + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}})$

At a particular time, the concentration of ${{\text{N}}_{\text{2}}}$ is $3.{\text{0 mol}}{{\text{L}}^{ - 1}}$ , for ${{\text{H}}_{\text{2}}}$ is $2.{\text{0 mol}}{{\text{L}}^{ - 1}}$ and for ${\text{N}}{{\text{H}}_{\text{3}}}$ is $0.{\text{5 mol}}{{\text{L}}^{ - 1}}$.

Now, we know that:

${{\text{Q}}_{\text{c}}} = \dfrac{{{{[{\text{N}}{{\text{H}}_{\text{3}}}]}^{\text{2}}}}}{{[{{\text{N}}_{\text{2}}}]{{[{{\text{H}}_{\text{2}}}]}^3}}} = \dfrac{{{{(0.5)}^2}}}{{(3.0){{(2.0)}^3}}} = 0.0104{\text{}}$

It is given that ${{\text{K}}_{\text{c}}} = 0.061$ 

Since ${{\text{Q}}_{\text{c}}} \ne {{\text{K}}_{\text{c}}}$  , the reaction is not at equilibrium. 

Since ${{\text{Q}}_{\text{c}}} < {{\text{K}}_{\text{c}}}$  , the reaction will proceed in the forward direction to reach equilibrium.