Chapter 6: Equilibrium

Ans: The vapour pressure would first fall if the container's volume was rapidly increased. This is due to the fact that the amount of vapour remains constant while the volume increases rapidly. As a result, the same amount of vapour is spread over a wider surface area.

1. How do rates of evaporation and condensation change initially? 

Ans: The rate of evaporation is also constant because the temperature is constant. The density of the vapour phase reduces as the container volume increases. As a result, the rate of vapour particle collisions drops as well. As a result, the rate of condensation initially slows.

1. What happens when equilibrium is restored finally and what will be the final vapour pressure?

Ans: The rate of evaporation equals the rate of condensation when equilibrium is eventually restored. Only the volume changes in this situation, but the temperature remains fixed. Temperature, not volume, determines vapour pressure. As a result, the final vapour pressure will be the same as the system's initial vapour pressure.

Ans: The equilibrium constant for the given reaction will be:

${{{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {{\text{S}}{{\text{O}}_{\text{3}}}} \right]}^2}}}{{{{\left[ {{\text{S}}{{\text{O}}_{\text{2}}}} \right]}^2}\left[ {{{\text{O}}_2}} \right]}}} $

${ = \dfrac{{{{(1.90)}^2}{{\text{M}}^2}}}{{{{(0.60)}^2}(0.821){{\text{M}}^3}}}{\text{}}}$

${{\text{K}}_{\text{c}}} = 12.239{{\text{M}}^{ - 1}}$ (approximately)

Therefore, the equilibrium constant for the given reaction is $12.239{{\text{M}}^{ - 1}}$.

Ans: Partial pressure for ${\text{I}}$  atoms:

${{{\text{p}}_{\text{1}}} = \dfrac{{40}}{{100}} \times {{\text{p}}_{{\text{total}}}}}$ 

= $\dfrac{40}{100} \times 10^5$ 

= $4 \times 10^4 Pa$ 

Partial pressure for ${{\text{I}}_{\text{2}}}$ molecules:

${{\text{p}}_{{{\text{l}}_{\text{2}}}}} = \dfrac{{60}}{{100}} \times {{\text{p}}_{{\text{total}}}}$

 $ = \dfrac{{60}}{{100}} \times {10^5}$

 $ = 6 \times {10^4}{\text{Pa}}$ 

Now for, the given reaction:

${{\text{K}}_{\text{p}}} = \dfrac{{{{({\text{pI}})}^2}}}{{{{\text{p}}_{{{\text{1}}_{\text{2}}}}}}}$

 $ = \dfrac{{\left( {4 \times {{10}^4}} \right){\text{P}}{{\text{a}}^2}}}{{6 \times {{10}^4}{\text{Pa}}}}$

$ = 2.67 \times {10^4}{\text{Pa}}$ 

Ans: The expression for the equilibrium constant is:

${{\text{K}}_{\text{c}}}=\dfrac{{{{[{\text{NO}}]}^2}{\text{[C}}{{\text{l}}_2}]}}{{{{[{\text{NOCl}}]}^2}}}$ 

1. $2{\text{Cu(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{(s}}) \rightleftharpoons 2{\text{CuO (s)}} + 4{\text{N}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}}$ 

Ans: The expression is:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{CuO}}]}^2}{{[{\text{N}}{{\text{O}}_2}]}^4}[{{\text{O}}_{\text{2}}}]}}{{{{[{\text{Cu(N}}{{\text{O}}_3}{)_2}]}^2}}}$ 

${{\text{K}}_{\text{c}}} = {[{\text{N}}{{\text{O}}_2}]^4}[{{\text{O}}_{\text{2}}}]$ 

1. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{(aq}}) + {{\text{H}}_{\text{2}}}{\text{O (l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH (aq)}} + {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH (aq)}}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH] [}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH]}}}}{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{] [}}{{\text{H}}_{\text{2}}}{\text{O]}}}}$ 

${K_c} = \dfrac{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH] [}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH]}}}}{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}]}}$ 

1. ${\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{(aq}}) + 3{\text{O}}{{\text{H}}^ - }({\text{aq)}} \rightleftharpoons {\text{Fe(OH}}{{\text{)}}_{\text{3}}}{\text{(s)}}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{[{\text{Fe(OH}}{{\text{)}}_3}]}}{{[{\text{F}}{{\text{e}}^{3 + }}]{{[{\text{O}}{{\text{H}}^ - }]}^3}}}$ 

${{\text{K}}_{\text{c}}} = \dfrac{1}{{[{\text{F}}{{\text{e}}^{3 + }}]{{[{\text{O}}{{\text{H}}^ - }]}^3}}}$ 

1. ${{\text{I}}_{\text{2}}}{\text{(s)}} + 5{{\text{F}}_2} \rightleftharpoons 2{\text{I}}{{\text{F}}_5}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{I}}{{\text{F}}_5}]}^2}}}{{[{{\text{I}}_2}]{{[{{\text{F}}_2}]}^2}}}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{I}}{{\text{F}}_5}]}^2}}}{{{{[{{\text{F}}_2}]}^2}}}$ 

Ans: The relationship between ${{\text{K}}_{\text{p}}}$ and ${{\text{K}}_{\text{c}}}$ is given as:

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

$\Delta {\text{n}} = 3 - 2 = 1$ 

${\text{R}} = 0.0831{\text{ barLmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ 

T = 500K

${K_p} = 1.8 \times {10^{ - 2}}$

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

$1.8 \times {10^{ - 2}} = {{\text{K}}_{\text{c}}}{(0.0831 \times 500)^1}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{1.8 \times {{10}^{ - 2}}}}{{0.0831 \times 500}}$

${{\text{K}}_{\text{c}}} = 4.33 \times {10^{ - 4}}$ (Approximately)

1. ${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s}}) \rightleftharpoons {\text{CaO (s)}} + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}$ ; ${K_p} = 167$ at 1073K

Ans: Here,

$\Delta {\text{n}} = 2 - 1 = 1$ 

${\text{R}} = 0.0831{\text{ barLmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$

T = 1073K

${K_p} = 167$

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

$167 = {{\text{K}}_{\text{c}}}{(0.0831 \times 1073)^1}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{167}}{{0.0831 \times 1073}}$ 

${{\text{K}}_{\text{c}}} = 1.87$ (Approximately)

Ans: According to the question:

The ${{\text{K}}_{\text{c}}}$ for the forward reaction is $6.3 \times {10^{14}}$

Then, ${{\text{K}}_{\text{c}}}$for the reverse reaction will be:

${{\text{K}}^{\text{'}}}_{\text{c}} = \dfrac{1}{{{{\text{K}}_{\text{c}}}}}$ 

${{\text{K}}^{\text{'}}}_{\text{c}} = \dfrac{1}{{6.3 \times {{10}^{14}}}}$

${{\text{K}}^{\text{'}}}_{\text{c}} = 1.59 \times {10^{ - 15}}$

The equilibrium constant for the reverse reaction will be: $1.59 \times {10^{ - 15}}$.

Ans: For a pure substance which is both a solid and a liquid:

${\text{[Pure substance}}] = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume}}}}$ 

${\text{[Pure substance}}] = \dfrac{{{\text{Mass / Molecular weight}}}}{{{\text{Volume}}}}$

${\text{[Pure substance}}] = \dfrac{{{\text{Mass}}}}{{{\text{Volume }} \times {\text{ Molecular mass}}}}$

${\text{[Pure substance}}] = \dfrac{{{\text{Density}}}}{{{\text{Molecular mass}}}}$

The molecular mass and density of a pure substance (at a given temperature) are now always fixed and accounted for in the equilibrium constant. As a result, the equilibrium constant statement does not include the values of pure substances.

Ans: Let the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ at    equilibrium be x.

The initial concentration of ${{\text{N}}_{\text{2}}}$ is 0.482mol.

The initial concentration of ${{\text{O}}_{\text{2}}}$ is 0.933mol.

At equilibrium the concentration of ${{\text{N}}_{\text{2}}}$ is (0.482-x)mol, the concentration of ${{\text{O}}_{\text{2}}}$is (0.933-x)mol and the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ is x mol.

The reaction occurs in a 10L reaction vessel:

$[{{\text{N}}_{\text{2}}}] = \dfrac{{0.482 - x}}{{10}}$ , $[{O_2}] = \dfrac{{0.933 - x/2}}{{10}}$ , $[{{\text{N}}_{\text{2}}}{\text{O}}] = \dfrac{x}{{10}}$ 

The value of equilibrium constant i.e. ${{\text{K}}_{\text{c}}} = 2.0 \times {10^{ - 37}}$  is very small. Therefore, the amount of nitrogen and oxygen reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of nitrogen and oxygen. Then,

$[{{\text{N}}_{\text{2}}}] = \dfrac{{0.482}}{{10}} = 0.048{\text{2 mol}}{{\text{L}}^{ - 1}}$ and $[{{\text{O}}_{\text{2}}}] = \dfrac{{0.933}}{{10}} = 0.093{\text{3 mol}}{{\text{L}}^{{\text{ - 1}}}}$

Now, according to the equation:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{{\text{N}}_{\text{2}}}{\text{O}}]}^2}}}{{{{[{{\text{N}}_2}]}^2}[{{\text{O}}_{\text{2}}}]}}$ 

$2.0 \times {10^{ - 37}} = \dfrac{{{{(\dfrac{x}{2})}^2}}}{{{{(0.0482)}^2}(0.0933)}}$ 

$\dfrac{{{x^2}}}{{100}} = 2.0 \times {10^{ - 37}} \times {(0.0482)^2}(0.0933)$ 

${x^2} = 43.35 \times {10^{ - 40}}$ 

$x = 6.6 \times {10^{ - 20}}$ 

$[{{\text{N}}_{\text{2}}}{\text{O}}] = \dfrac{x}{{10}} = \dfrac{{6.6 \times {{10}^{ - 20}}}}{{10}}$ 

$[{{\text{N}}_{\text{2}}}{\text{O}}] = 6.6 \times {10^{ - 21}}$ 

Ans: The given reaction is:

$2{\text{NO (g}}) + {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{NOBr (g)}}$

Now, 2 mol of ${\text{NOBr}}$  are formed from 2 mol of ${\text{NO}}$ . Therefore, 0.0518 mol of ${\text{NOBr}}$  are formed from 0.0518 mol of ${\text{NO}}$. Again, 2 mol of ${\text{NOBr}}$  are formed from 1 mol of ${\text{Br}}$ .

Therefore, 0.0518 mol of ${\text{NOBr}}$ are formed from $\dfrac{{0.0581}}{2}$ mol of ${\text{Br}}$, or 0.0259 mol of ${\text{NO}}$.

The amount of ${\text{NO}}$ and ${\text{Br}}$ present initially is as follows: 

$[{\text{NO}}] = 0.087{\text{mol}}$ and $[{\text{B}}{{\text{r}}_2}] = 0.0437{\text{mol}}$ 

Therefore, the amount of ${\text{NO}}$ present at equilibrium is:

$[{\text{NO}}] = 0.087 - 0.0518 = 0.0352{\text{mol}}$ 

And, the amount of ${\text{Br}}$  present at equilibrium is:

$[{\text{B}}{{\text{r}}_2}] = 0.0437 - 0.0259 = 0.0178{\text{mol}}$ 

Ans: For the given reaction:

${{\Delta}n = 2 - 3 =  - 1} $ 

${T = 450{\text{K}}} $ 

${R = 0.0831{\text{bar L bar }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}} $ 

${{K_p} = 2.0 \times {{10}^{10}}{\text{ba}}{{\text{r}}^{ - 1}}}$ 

We know that,

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

$2.0 \times {10^{10}} = {{\text{K}}_{\text{c}}}{(0.0831 \times 450)^{ - 1}}{\text{}}$

${{\text{K}}_{\text{c}}} = \dfrac{{2.0 \times {{10}^{10}}}}{{(0.0831 \times 450)}}$

${{\text{K}}_{\text{c}}} = 74.79 \times {10^{14}}{\text{Lmol}}$