NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container suddenly increased. 

1. What is the initial effect of the change on vapour pressure?

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.

2. 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.

3. 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.

2. What is ${\text{K}}_{\text{c}}$  for the following equilibrium when the equilibrium concentration of each substance is: $[\text{S}{\text{O}}_{\text{2}}]=0.60\text{M}$ , $[{\text{O}}_{\text{2}}]=0.82\text{M}$ and $[\text{S}{\text{O}}_3]=1.90\text{M}$ ? $2\text{S}{\text{O}}_{\text{2}}\text{(g})+{\text{O}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{S}{\text{O}}_{\text{3}}\text{(g)}$ 

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

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

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

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

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

3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of iodine atoms. ${\text{I}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{I (g)}$ . Calculate ${\text{K}}_{\text{P}}$ for the equilibrium.

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

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

= ${\displaystyle \frac{40}{100}}\times {10}^5$ 

= $4\times {10}^{4}Pa$ 

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

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

 $={\displaystyle \frac{60}{100}}\times {10}^5$

 $=6\times {10}^4\text{Pa}$ 

Now for, the given reaction:

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

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

$=2.67\times {10}^4\text{Pa}$ 

4. Write the expression for the equilibrium constant, ${\text{K}}_{\text{c}}$  for each of the following reactions:

1. $2\text{NOCl (g})\rightleftharpoons 2\text{NO (g})+\text{C}{\text{l}}_{\text{2}}\text{(g)}$ 

Ans: The expression for the equilibrium constant is:

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

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

Ans: The expression is:

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

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

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

Ans: The expression will be:

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

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

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

Ans: The expression will be:

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

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

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

Ans: The expression will be:

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

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

5. Find out the value of ${\text{K}}_{\text{c}}$  for each of the following equilibria from the value of ${\text{K}}_{\text{p}}$ :

1. $2\text{NOCl (g})\rightleftharpoons 2\text{NO (g})+\text{C}{\text{l}}_{\text{2}}\text{(g})$ ; $K_p=1.8\times {10}^{-2}$ at 500K

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

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

$\mathrm{\Delta }\text{n}=3-2=1$ 

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

T = 500K

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

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

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

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

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

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

Ans: Here,

$\mathrm{\Delta }\text{n}=2-1=1$ 

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

T = 1073K

$K_p=167$

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

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

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

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

6. For the following equilibrium, ${\text{K}}_{\text{c}}=6.3\times {10}^{14}$ at 1000K $\text{NO (g)}+{\text{O}}_{\text{3}}\text{(g})\rightleftharpoons \text{N}{\text{O}}_{\text{2}}\text{(g)}+{\text{O}}_{\text{2}}\text{(g)}$ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is ${\text{K}}_{\text{c}}$  for the reverse reaction?

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:

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

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

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

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

7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

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

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

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

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

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

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.

8. Reaction between nitrogen and oxygen takes place as follows: $2{\text{N}}_{\text{2}}\text{(g)}+{\text{O}}_{\text{2}}\text{(g})\rightleftharpoons 2{\text{N}}_{\text{2}}\text{O (g)}$ If a mixture of 0.482 mol of ${\text{N}}_{\text{2}}$  and 0.933 mol of ${\text{O}}_{\text{2}}$  is placed in a 10 L reaction vessel and allowed to form ${\text{N}}_{\text{2}}\text{O}$  at a temperature for which ${\text{K}}_{\text{c}}=2.0\times {10}^{-37}$ , determine the composition of equilibrium mixture.

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:

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

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,

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

Now, according to the equation:

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

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

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

$x^2=43.35\times {10}^{-40}$ 

$x=6.6\times {10}^{-20}$ 

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

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

9. Nitric oxide reacts with $\text{B}{\text{r}}_{\text{2}}$  and gives nitrosyl bromide as per reaction given below: $2\text{NO (g})+\text{B}{\text{r}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{NOBr (g)}$ When 0.087 mol of $\text{NO}$  and 0.0437 mol of $\text{B}{\text{r}}_{\text{2}}$  are mixed in a closed container at constant temperature, 0.0518 mol of $\text{NOBr}$  is obtained at equilibrium. Calculate equilibrium amount of $\text{NO}$  and $\text{B}{\text{r}}_{\text{2}}$ .

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 ${\displaystyle \frac{0.0581}{2}}$ 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:

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

10.  At 450K, ${\text{K}}_{\text{p}}=2.0\times {10}^{10}\text{ba}{\text{r}}^{\text{ - 1}}$ for the given reaction reaction at equilibrium. $2\text{S}{\text{O}}_{\text{2}}\text{(g})+{\text{O}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{S}{\text{O}}_{\text{3}}\text{(g)}$ , What is ${\text{K}}_{\text{c}}$ at this temperature?

Ans: For the given reaction:

$\mathrm{\Delta }n=2-3=-1$ 

$T=450\text{K}$ 

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

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

We know that,

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

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

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

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

11. A sample of $\text{HI (g)}$  is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of $\text{HI (g)}$  is 0.04 atm. What is ${\text{K}}_{\text{p}}$  for the given equilibrium? $2\text{HI (g})\rightleftharpoons {\text{H}}_{\text{2}}\text{(g)}+{\text{I}}_{\text{2}}\text{(g)}$ 

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 ${\text{H}}_{\text{2}}={\displaystyle \frac{0.16}{2}}=0.08\text{atm}$ and the pressure for ${\text{O}}_{\text{2}}={\displaystyle \frac{0.16}{2}}=0.08\text{atm}$.

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

12.  A mixture of 1.57 mol of ${\text{N}}_{\text{2}}$ , 1.92 mol of ${\text{H}}_{\text{2}}$  and 8.13 mol of $\text{N}{\text{H}}_{\text{3}}$  is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, ${\text{K}}_{\text{c}}$  for the reaction is $1.7\times {10}^2$ . The reaction is: ${\text{N}}_{\text{2}}\text{(g)}+3{\text{H}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{N}{\text{H}}_{\text{3}}\text{(g)}$ . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

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

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

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

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

$Q_C={\displaystyle \frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}}$ 

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

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

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

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

13. The equilibrium constant expression for a gas reaction is: 

${\text{K}}_{\text{c}}={\displaystyle \frac{{\text{[N}{\text{H}}_{\text{3}}\text{]}}^{\text{4}}\text{ [}{\text{O}}_{\text{2}}{\text{]}}^{\text{5}}}{{\text{[NO]}}^{\text{4}}\text{ [}{\text{H}}_{\text{2}}\text{O}{\text{]}}^{\text{6}}}}$ 

Write the balanced chemical equation corresponding to this expression.

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

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

14. One mole of ${\text{H}}_{\text{2}}\text{O}$  and one mole of $\text{CO}$  are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with $\text{CO}$  according to the equation: ${\text{H}}_{\text{2}}\text{(g})+\text{CO(g})\rightleftharpoons {\text{H}}_{\text{2}}\text{(g})+\text{C}{\text{O}}_{\text{2}}\text{(g)}$ Calculate the equilibrium constant for the reaction.

Ans: The given reaction is:

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

${\displaystyle \frac{\text{1}}{\text{10}}}\text{M}{\displaystyle \frac{\text{1}}{\text{10}}}\text{M}\text{0}\text{0}$ 

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

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

Therefore, the equilibrium constant for the reaction:

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

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

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

15.  At 700 K, equilibrium constant for the reaction ${\text{H}}_{\text{2}}\text{(g})+{\text{I}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{HI (g)}$ is 54.8. If $0.5\text{mol}{\text{L}}^{-1}$ of $\text{HI (g)}$  is present at equilibrium at 700 K, what are the concentration of ${\text{H}}_{\text{2}}\text{(g)}$  and ${\text{I}}_{\text{2}}\text{(g)}$  assuming that we initially started with $\text{HI (g)}$  and allowed it to reach equilibrium at 700 K?

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: ${\displaystyle \frac{1}{54.8}}$ 

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

${\displaystyle \frac{x\times x}{{(0.5)}^2}}={\displaystyle \frac{1}{54.8}}$ 

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

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

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

16.  What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $\text{ICI}$  was 0.78 M?

$2\text{ICI (g})\rightleftharpoons {\text{I}}_{\text{2}}\text{(g)}+\text{C}{\text{l}}_{\text{2}}\text{(g)}$ ; ${\text{K}}_{\text{c}}=0.14$ 

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:

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

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

${\displaystyle \frac{x}{0.78-2x}}=0.374$

$x=0.292-0.748x$

$1.748x=0.292$

$x=0.167$

Therefore, at equilibrium

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

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

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

17. ${\text{K}}_{\text{p}}=0.04\text{atm}$  at 899 K for the equilibrium shown below. What is the equilibrium concentration of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$  when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g})\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g})+{\text{H}}_{\text{2}}\text{(g)}$ 

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,

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

$p^2=0.16-0.04p$

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

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

$p={\displaystyle \frac{-0.04\pm 0.08}{2}}$

If we take positive value:

$p={\displaystyle \frac{0.76}{2}}$

$p=0.38$

So, at equilibrium:

$[C_2{H}_6]-4-p=4-0.38=3.62\text{ atm}$ 

18.  A sample of pure $\text{PC}{\text{l}}_{\text{5}}$  was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of $\text{PC}{\text{l}}_{\text{5}}$  was found to be $0.5\times {10}^{-1}\text{mol}{\text{L}}^{\text{ - 1}}$  . If value of ${\text{K}}_{\text{c}}$  is $8.3\times {10}^{-3}$  , what are the concentrations of $\text{PC}{\text{l}}_{\text{3}}$  and $\text{C}{\text{l}}_{\text{2}}$  at equilibrium? $\text{PC}{\text{l}}_{\text{5}}\text{(g})\rightleftharpoons \text{PC}{\text{l}}_{\text{3}}\text{(g)}+\text{C}{\text{l}}_{\text{2}}\text{(g)}$ 

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

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

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

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

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

$\Rightarrow x=0.0204$

Therefore at equilibrium:

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

19. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and $\text{C}{\text{O}}_{\text{2}}$ .$\text{FeO (s})+\text{CO (g})\rightleftharpoons \text{Fe (s})+\text{C}{\text{O}}_{\text{2}}\text{(g})$ ; ${\text{K}}_{\text{p}}=0.0265$ at 1050 K.

What are the equilibrium partial pressures of $\text{CO}$  and $\text{C}{\text{O}}_{\text{2}}$  at 1050 K if the initial partial pressures are: ${\text{p}}_{\text{CO}}=1.4\text{atm}$  and ${\text{p}}_{\text{C}{\text{O}}_{\text{2}}}=0.80\text{atm}$ ?

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:

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

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, ${\text{p}}_{\text{CO}}=1.4-0.339=1.73\text{9 atm}$ 

20. Equilibrium constant, ${\text{K}}_{\text{c}}$  for the reaction ${\text{N}}_{\text{2}}\text{(g})+3{\text{H}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{N}{\text{H}}_{\text{3}}\text{(g})$ at 500 K os 0.061. At a particular time, the analysis shows that composition of the reaction mixture is $3.\text{0 mol}{\text{L}}^{-1}$  ${\text{N}}_{\text{2}}$ , $2.\text{0 mol}{\text{L}}^{-1}$  ${\text{H}}_{\text{2}}$  and $0.\text{5 mol}{\text{L}}^{-1}$  $\text{N}{\text{H}}_{\text{3}}$ . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

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:

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

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.

21. Bromine monochloride, $\text{BrCl}$  decomposes into bromine and chlorine and reaches the equilibrium: $2\text{BrCl (g})\rightleftharpoons \text{B}{\text{r}}_{\text{2}}\text{(g)}+\text{C}{\text{l}}_{\text{2}}\text{(g)}$ for which ${\text{K}}_{\text{c}}=32$ at 500 K. If initially pure $\text{BrCl}$  is present at a concentration of $3.3\times {10}^{-3}\text{ mol}{\text{L}}^{-1}$  , what is its molar concentration in the mixture at 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.

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

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

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

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

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

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

 Therefore, at equilibrium

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

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

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

22. At 1127 K and 1 atm pressure, a gaseous mixture of $\text{CO}$  and $\text{C}{\text{O}}_{\text{2}}$  in equilibrium with solid carbon has 90.55% $\text{CO}$ by mass: $\text{C (s})+\text{C}{\text{O}}_{\text{2}}\text{(g})\rightleftharpoons 2\text{CO (g)}$. Calculate ${\text{K}}_{\text{c}}$  for this reaction at the above temperature.

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 $\text{CO}$ , $n_{CO}={\displaystyle \frac{90.55}{28}}=3.23\text{4 mol}$ 

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

Partial pressure of $\text{CO}$ , $p_{CO}={\displaystyle \frac{n_{CO}}{n_{CO}+n_{C{O}_2}}}\times p_{\text{total}}$ 

$p_{CO}={\displaystyle \frac{3.234}{3.234+0.215}}\times 1$

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

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

$p_{C{O}_2}={\displaystyle \frac{n_{C{O}_2}}{n_{CO}+n_{C{O}_2}}}\times p_{\text{total}}$

$p_{C{O}_2}={\displaystyle \frac{0.215}{3.234+0.215}}\times 1$

$p_{C{O}_2}=0.06\text{2 atm}$

Therefore, 

For the given reaction:

$\mathrm{\Delta }n=2-1=1$ 

We know that,

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

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

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

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

23. Calculate:

1. $\mathrm{\Delta }{\text{G}}^{\circ }$ and

2. The equilibrium constant for the formation of $\text{N}{\text{O}}_{\text{2}}$  from $\text{NO}$  and ${\text{O}}_{\text{2}}$  at 298 K $\text{NO (g})+{\displaystyle \frac{1}{2}}{\text{O}}_{\text{2}}\text{(g})\rightleftharpoons \text{N}{\text{O}}_{\text{2}}\text{(g})$ 

Where:

${\mathrm{\Delta }}_{\text{f}}{\text{G}}^{\circ }(\text{N}{\text{O}}_{\text{2}})=52\text{kJ/mol}$

${\mathrm{\Delta }}_{\text{f}}{\text{G}}^{\circ }(\text{NO)}=87.0\text{kJ/mol}$

${\mathrm{\Delta }}_{\text{f}}{\text{G}}^{\circ }({\text{O}}_{\text{2}})=0\text{kJ/mol}$

Ans: For the given reaction: 

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

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

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

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

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

2. We know that:

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

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

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

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

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

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

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

24. Does the number of moles of reaction products increase, decrease or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

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

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.

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

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

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

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

25. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

1. $\text{COC}{\text{l}}_{\text{2}}\text{(g})\rightleftharpoons \text{CO (g})+\text{C}{\text{l}}_{\text{2}}\text{(g)}$ 

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

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

4. $\text{2}{\text{H}}_{\text{2}}\text{(g})+\text{CO (g})\rightleftharpoons \text{C}{\text{H}}_{\text{3}}\text{OH (g})$ 

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

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