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**Sol.**Soda-water bottle.

**Sol.**The amount of the gas dissolved is very high due to high pressure. On opening the bottle, the pressure tends to decrease to atmospheric pressure. So the solubility decreases,

*i.e.,*the dissolved gas escapes out.

$K_{c}=\frac{\left[N H_{3}\right]^{4}\left[O_{2}\right]^{5}}{[N O]^{4}\left[H_{2} O\right]^{6}}$

Write the balance chemical reaction to this expression.

**Sol.**$4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2}$

**Sol.**No, the reaction does not stop. It continues to take place in the forward as well as backward direction but at equal speed.

**Sol.**Two processes, namely, dissolution and precipitation will continue to take place at equal rates. It is called a state of equilibrium.

The rate of an exothermic reaction increases with increasing temperature.

**Sol.**False because in an exothermic reaction, heat is evolved. Increase of temperature will shift the equilibrium in the backward direction,

*i.e.,*the rate of reaction decreases.

**Sol.**The state of equilibrium is not disturbed but it attained quickly because both the rate of forward and backward reaction increases to the same extent.

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{c}$ for the reverse reaction?

**Sol.**For the reverse reaction $K_{c}^{\prime}=\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}$

$=1.59 \times 10^{-15}$

**Sol.**It is defined as product of molar concentration of reaction products each raised to power equal to the respective stoichiometric coefficient in balance chemical equation divided by the product of concentration of the reactants raised to power equal to their individual stoichiometric coefficients at any stage of reaction, e.g.,

$a A+b B \rightleftharpoons c C+d D$

$Q_{C}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

What is the relationship between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}} ?$

**Sol.**$K_{p}=K_{c}$

$\mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(\mathrm{g})$

**Sol.**The increase of pressure

*i.e.*compression will favour the formation of reactants because reactants have less number of moles than products.

**Sol.**The equilibrium constant is not affected by catalyst.

*K*are 2.34, 12.36 and 5.85

*kPa*respectively. Which of these has the lowest and the highest boiling point ? At 293

*K*, which of these evaporates least in a sealed container before equilibrium is established?

**Sol.**Acetone has highest vapour pressure, therefore has lowest boiling point.

Water has lowest vapour pressure, therefore has highest boiling point.

At $293 K$, water will evaporates least in a sealed container before equilibrium is established.

**Sol.**(i) If $Q=K,$ the reaction is in equilibrium.

(ii) If $Q>K, Q$ will tend to decrease so as to become equal to $K .$ As a result, the reaction will proceed in the backward direction.

(iii) If $Q<K, Q$ will tend to increase. As a result, the reaction will proceed in the forward direction.

**Sol.**(Pure liquid) or (Pure solid)

$=\frac{\text { No. of moles }}{\text { Volume of } \mathrm{L}}=\frac{\text { Mass/mol. mass }}{\text { Volume }}$

$=\frac{\text { Mass }}{\text { Volume }} \times \frac{1}{\text { Mol. mass }}=\frac{\text { Density }}{\text { Mol. mass }}$ As density of a pure liquid or pure solid is constant constant temperature and molecular mass is also. constant, therefore, their molar concentrations are constant and included into the equilibrium constant.

**Sol.**Partial pressure of $I$ atoms $\left(p_{1}\right)=\frac{40}{100} \times 10^{5}$ $P a=0.4 \times 10^{5} \mathrm{Pa}$

Partial pressure of $I_{2}\left(p_{I_{2}}\right)=\frac{60}{100} \times 10^{5} P a=0.60 \times 10^{5} P a$

$K_{p}=\frac{p_{I}^{2}}{P_{I_{2}}}=\frac{\left(0.4 \times 10^{5}\right)^{2}}{0.60 \times 10^{5}}=2.67 \times 10^{4}$

at $25^{\circ} \mathrm{C}$. If the equilibrium concentration of $\mathrm{O}_{2}$ in air at $25^{\circ} \mathrm{C}$ is $1.6 \times 10^{-2},$ what is the concentration of $\mathrm{O}_{3} ?$

**Sol.**$K_{c}=\frac{\left[O_{3}\right]^{2}}{\left[O_{2}\right]^{3}} \therefore 2.0 \times 10^{-50}=\frac{\left[O_{3}\right]^{2}}{\left(1.6 \times 10^{-2}\right)^{3}}$

or $\left[\mathrm{O}_{3}\right]^{2}=\left(2.0 \times 10^{-50}\right)\left(1.6 \times 10^{-2}\right)^{3}=8.192 \times 10^{-56}$

or $\left[O_{3}\right]=2.86 \times 10^{-28} M$

**Sol.**$\Delta G=13.8 \mathrm{kJ} \mathrm{mol}^{-1}=13.8 \times 10^{3} \mathrm{J} \mathrm{mol}^{-1}$

$\Delta G^{\circ}=-2.303 R T \log K$

$\log K=-\frac{\Delta G^{\circ}}{2.303 R T}=-\frac{13.8 \times 10^{3}}{2.303 \times 8.314 \times 298}=-2.418$

$K=3.82 \times 10^{-3}$

**Sol.**$N O(g)+\frac{1}{2} O_{2}(g) \rightleftharpoons N O_{2}(g)$

$\Delta G^{\circ}=\Delta_{f} G^{\circ}\left(N O_{2}\right)-\left[\Delta_{f} G^{\circ}(N O)+\frac{1}{2} \Delta_{f} G^{\circ}\left(O_{2}\right)\right]$

$=52.0-87.0-\frac{1}{2} \times 0=-35 \mathrm{kJ} \mathrm{mol}^{-1}$

$\mathrm{Now}, \log K=-\frac{\Delta G^{\circ}}{2.303 R T}$

$=-\frac{-35 \times 10^{3} \mathrm{J} \mathrm{mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \times 298 \mathrm{K}}=6.314$

or $K=1.362 \times 10^{6}$

What is the $\mathrm{K}_{\mathrm{c}}$ at this temperature?

**Sol.**$K_{p}=K_{c}(R T)^{\Delta n}$ or $K_{c}=\frac{K_{p}}{(R T)^{\Delta n}}$

$\Delta n=2-(2+1)=-1, T=450 K, R=0.083$ bar $K^{-1} \mathrm{mol}^{-1}$

$\therefore K_{C}=\frac{2.0 \times 10^{10}}{(0.083 \times 450)^{-1}}=2.0 \times 10^{10} \times(0.083 \times 450)$

$=7.47 \times 10^{11} M^{-1}$

of remains. What is the value of $\mathrm{k}_{\mathrm{C}} ?$

**Sol.**

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

**Sol.**

is present at equilibrium at $700 K,$ what are the concentrations of $\mathrm{H}_{2}(\mathrm{g})$ and $\mathrm{I}_{2}(\mathrm{g})$ assuming that we initially started with HI (g) and allowed it to reach equilibrium at $700 K ?$

**Sol.**

$\frac{1}{54.8}=\frac{x^{2}}{(0.5)^{2}} \Rightarrow x^{2}=\frac{0.5 \times 0.5}{54.8} \Rightarrow x=0.0675 \mathrm{mol} L^{-1}$

If a mixture of $0.482 \mathrm{mol}$ of $\mathrm{N}_{2}$ and $0.933 \mathrm{mol}$ of $\mathrm{O}_{2}$ is placed in a reaction vessel of volume $10 L$ and allowed to form $\mathrm{N}_{2} \mathrm{O}$ at a temperature for which $\mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}$

determine the composition of equilibrium mixture

**Sol.**$2 N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N_{2} O(g)$

Concentration of $N_{2}=0.482 \mathrm{mol}$

Molar concentration of $N_{2}$ at equilibrium $=\frac{0.482}{10 L}=0.0482 \mathrm{mol} \mathrm{L}^{-1}$

$=\frac{0.482}{10 L}=0.0482 \mathrm{mol} L^{-1}$

Molar concentration of $O_{2}$ at equilibrium $=\frac{0.953}{10 L}=0.0933 \mathrm{mol} L^{-1}$

$K_{c}=\frac{\left[N_{2} \mathrm{O}\right]^{2}}{\left[N_{2}\right]^{2}\left[\mathrm{O}_{2}\right]^{1}} \Rightarrow 2.0 \times 10^{-37}=\frac{\left[N_{2} \mathrm{O}\right]^{2}}{[0.0482]^{2}[0.0933]}$

$\left[N_{2} \mathrm{O}\right]=6.6 \times 10^{-21} \mathrm{mol} L^{-1}$

**Sol.**

$C H_{4}(g)+2 H_{2} S(g) \rightleftharpoons C S_{2}(g)+4 H_{2}(g)$ at $1173 K$

the magnitude of the equilibrium constant, $K_{c}$ is $3.6 .$ For each of the following compositions, decide whether reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go:

(i) $\left[\mathrm{CH}_{4}\right]=1.07 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{S}\right]=1.20 \mathrm{M},\left[\mathrm{CS}_{2}\right]=0.90 \mathrm{M},\left[\mathrm{H}_{2}\right]=1.78 \mathrm{M}$

(ii) $\left[C H_{4}\right]=1.45 M,\left[H_{2} S\right]=1.29 M,\left[C S_{2}\right]=1.25 M,\left[H_{2}\right]=1.75 M$

**Sol.**$C H_{4}(g)+2 H_{2} S(g) \rightleftharpoons C S_{2}(g)+4 H_{2}(g)$

(i) $\quad Q_{c}=\frac{\left[C S_{2}\right]\left[H_{2}\right]^{4}}{\left[C H_{4}\right]\left[H_{2} S\right]^{2}}=\frac{[0.90][1.78]^{4}}{[1.07][1.20]^{2}}=\frac{10.03 \times 0.9}{1.44 \times 1.07}$

$K_{c}=3.6 \quad Q_{c}=\frac{9.27}{1.54}=6.02$

since $Q_{c}>K_{c},$ the equilibrium will shift in backward direction.

(ii) $\quad Q_{c}=\frac{[1.25][1.75]^{4}}{[1.45][1.29]^{2}}=\frac{9.38 \times 1.25}{1.45 \times 1.66}=\frac{11.725}{2.407}=4.86$

Since , the equilibrium will shift in backward direction.

$P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)$

Initially, $0.0646 \mathrm{M} \mathrm{PCl}_{5}$ is introduced into a reaction

vessel $350 \mathrm{K} .$ At equilibrium, $0.0444 \mathrm{M} \mathrm{Cl}_{2}$ is found.

Calculate $K_{c}$ and $K_{p}$

**Sol.**

K. At this temperature, the equilibrium constant, $K_{c}$ for the reaction, $N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ is $1.7 \times 10^{2}$

Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction?

**Sol.**

$N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ at $500 K$ is 0.061

At a panticular time, the analysis shows that composition

of the reaction mixture is $30 \mathrm{mol}^{-1} N_{2}, 20 \mathrm{moL}^{-1} \mathrm{H}_{2}$ and

$0.5 \mathrm{mol} L^{-1} \mathrm{NH}_{3} .$ Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

**Sol.**$Q_{c}=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}=0.0104$

As $Q_{c} \neq K_{c},$ reaction is not in equilibrium.

As $Q_{c}<K_{c},$ reaction will proceed in the forward direction.

reaction vessel at $400^{\circ} \mathrm{C}$ is charged with an equimolar mixture of $\mathrm{CO}$ and steam such that $p_{\mathrm{CO}}=p_{H_{2} O}=4.0$ bar,

what will be the partial pressure of $H_{2}$ at equilibrium? $K_{p}=0.1$ at $400^{\circ} \mathrm{C}$

**Sol.**

**Sol.**For reaction (iii), as $\mathrm{K}_{\mathrm{c}}$ neither high nor very low, reactants and products will be present in comparable amounts.

**Sol.**

having concentration $1.59 M P C l_{3}, 1.59 M C l_{2}$ and

$1.41 M P C l_{5} .$ Calculate $K_{c}$ for the reaction:

$P C l_{5} \rightleftharpoons P C l_{3}+C l_{2}$

**Sol.**The equilibrium constant for the reaction may be written as $K_{c}=\frac{\left[P C l_{3}\right]\left[C l_{2}\right]}{\left[P C l_{5}\right]}$

$\left[P C l_{5}\right]=1.41 M,\left[P C l_{3}\right]=1.59 M,\left[C l_{2}\right]=1.59 M$

$=\frac{(1.59)(1.59)}{1.41}=1.79 M$

(i) $\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$

(ii) $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(l)+2 \mathrm{H}_{2} \mathrm{S}(g)$

(iii) $\mathrm{CO}_{2}(g)+C(s) \rightleftharpoons 2 \mathrm{CO}(g)$

(iv) $2 H_{2}(g)+C O(g) \rightleftharpoons C H_{3} O H(g)$

(v) $\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$

(vi) $4 N H_{3}(g)+5 O_{2}(g) \rightleftharpoons 4 N O(g)+6 H_{2} O(g)$

**Sol.**(i) Backward direction

(ii) Forward direction

(iii) Backward direction

(iv) Forward direction

(v) Backward direction.

(vi) Backward direction.

$400 \mathrm{K}, \mathrm{K}_{\mathrm{p}}=41 .$ Find the value of $\mathrm{K}_{\mathrm{p}}$ for each of the

following reactions at the same temperature:

(i) $2 N H_{3}(g) \rightleftharpoons N_{2}(g)+3 H_{2}(g)$

(ii) $\frac{1}{2} N_{2}(g)+\frac{3}{2} H_{2}(g) \rightleftharpoons N H_{3}(g)$

(iii) $2 N_{2}(g)+6 H_{2}(g) \rightleftharpoons 4 N H_{3}(g)$

**Sol.**(i) $\quad K_{p}=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}=41$

$2 N H_{3}(g) \rightleftharpoons N_{2}(g)+3 H_{2}(g)$

$K_{p}^{\prime}=\frac{\left[N_{2}\right]\left[H_{2}\right]^{3}}{\left[N H_{3}\right]^{2}}=\frac{1}{K_{p}}=\frac{1}{41}=0.024$

(ii) $\frac{1}{2} N_{2}(g)+\frac{3}{2} H_{2}(g) \rightleftharpoons N H_{3}(g)$

$K_{p}^{\prime \prime}=\frac{\left[N H_{3}\right]}{\left[N_{2}\right]^{1 / 2}\left[H_{2}\right]^{3 / 2}}=\sqrt{K_{p}}=\sqrt{41}=6.4$

(iii) $2 N_{2}(g)+6 H_{2}(g) ! 4 N H_{3}(g)$

$K_{p}^{\mathrm{n}}=\frac{\left[N H_{3}\right]^{4}}{\left[N_{2}\right]^{2}\left[H_{2}\right]^{6}}=\left(K_{p}\right)^{2}=(41)^{2}=1681$

$K_{p}^{\prime \prime}=1.68 \times 10^{3}$

$\mathrm{CO}(g)+H_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+H_{2}(g) \quad$ Calculate

equilibrium concentrations of $\mathrm{CO}_{2}, \mathrm{H}_{2}, \mathrm{CO}$ and $\mathrm{H}_{2} \mathrm{O}$ at

$800 K,$ if only $\mathrm{CO}$ and $\mathrm{H}_{2} \mathrm{O}$ are present initially at concentration of $0.10 \mathrm{Meach} ?$

**Sol.**

*M*?

**$2 I C l(g) \rightleftharpoons I_{2}(g)+C l_{2}(g), K_{c}=0.14$**

**Sol.**

What is the equilibrium concentration of $C_{2} H_{6}$ when it is placed in a flask at 4.0 atm. Pressure and allowed to come in equilibrium? $C_{2} H_{6}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$

**Sol.**

equilibrium from the value of $K_{p}$

(i) $2 N O C l(g) \rightleftharpoons 2 N O(g)+C l_{2}(g), K_{p}=1.8 \times 10^{-2}$ at 500

$$

\begin{array}{l}{K} \\ { \text { (ii) }\left.C a C O_{3}(s) \Longrightarrow C a Q_{s}\right)+C O_{2}(g), K_{p}=167 \text { at } 1073 K .}\end{array}

$$

**Sol.**(i) $\quad 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$

$K_{p}=1.8 \times 10^{-2} \Rightarrow \Delta n_{g}=3-2=1$

$K_{p}=K_{c}(R T)^{\Delta n}=K_{c}(R T)$

$K_{c}=\frac{K_{p}}{R T}=\frac{1.8 \times 10^{-2}}{0.0831 \times 500}=4.33 \times 10^{-4}$

(ii) $\quad \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

$K_{p}=167$

$\Delta n_{g}=1$

$K_{p}=K_{c}(R T)^{\Delta n}=K_{c}(R T)$

$K_{c}=\frac{K_{p}}{R T}=\frac{167}{0.0831 \times 1073}=1.88$

$\mathrm{CO}$ by mass, $\mathrm{C}(s)+\mathrm{CO}_{2} \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$

Calculate $K_{c}$ for this reaction at the above temperature.

**Sol.**Let total mass of the gaseous mixture be $=100 \mathrm{g}$

Mass of $C O=90.55 \mathrm{g}$

Mass of $C O_{2}=9.45 g$

Moles of $C O=\frac{90.55}{28}=3.234$

Moles of $C O_{2}=\frac{9.45}{44}=0.215$

Total number of moles of gases $=3.234+0.215=3.449$

$p_{C O}=\frac{3.234}{3.449} \times 1=0.938 \mathrm{atm}$

$p_{C O_{2}}=\frac{0.215}{3.449} \times 1=0.0623 \mathrm{atm}$

$K_{p}=\frac{p_{C O}^{2}}{p_{C O_{2}}}=\frac{(0.938)^{2}}{0.0623}=14.12$

$K_{p}=K_{c}(R T)^{\Delta n}=K_{c}(R T) \quad \therefore \Delta n=(2-1)=1$

$K_{c}=\frac{K_{p}}{R T}=\frac{14.12}{0.0821 \times 1127}=0.153$

$H_{2}(g)+B r_{2}(g) \rightleftharpoons 2 H B r(g)$

Find the equilibrium pressure of all gases if 10.0 bar $\mathrm{HBr}$ is introduced into a sealed container at $1024 K$

**Sol.**

$\left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M}$ and $\left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M}$

$2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$

**Sol.**The equilibrium constant, $K_{c}$ is

$K_{c}=\frac{\left[S O_{3}\right]^{2}}{\left[S O_{2}\right]^{2}\left[O^{2}\right]}$

$\left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M}$ and $\left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M}$

$K_{c}=\frac{(1.90 M)^{2}}{(0.60 M)^{2}(0.82 M)}$

$=12.229 M^{-1}$ or $12.229 \mathrm{mol}^{-1} L$

$2 \mathrm{NO}(g)+B r_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$

When $0.087 \mathrm{mol}$ of $\mathrm{NO}$ and $0.0437 \mathrm{mol}$ of $\mathrm{Br}_{2}$ are mixed in a closed container at constant temperature $0.518 \mathrm{mol}$ of $\mathrm{NOBr}$ is obtained at equilibrium. Calculate (i) the equilibrium amount of nitric oxide and bromine (ii) equilibrium constant.

**Sol.**$2 \mathrm{NO}(g)+B r_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$

(i) According to the above equation

0.0518 mol of $\mathrm{NOBr}$ are formed from

0.0518 mol of $N O$

$\therefore$ Amount of nitric oxide at equilibrium

$=0.087-0.0518$

$=0.0352 \mathrm{mol}$

Similarly, $0.0518 \mathrm{mol}$ of $\mathrm{NOBr}$ are formed from

$\frac{0.0518}{2} \mathrm{mol}$ of $\mathrm{Br}_{2}$

$\therefore$ Amount of bromine at equilibrium

$=0.0437-\frac{0.0518}{2}=0.0178 \mathrm{mol}$

(ii) $\quad K_{c}=\frac{[N O B r]^{2}}{\left[N O^{2}\right]\left[B r_{2}\right]}$

At equilibrium $[\mathrm{NOBr}]=0.0518 \mathrm{mol},[\mathrm{NO}]=0.0352 \mathrm{mol}$

$\left[B r_{2}\right]=0.0178 \mathrm{mol}$

$\therefore \quad K_{c}=\frac{(0.0518)^{2}}{(0.0352)^{2} \times(0.0178)}=121.66$

(i) $2 N_{2} O(g) \longrightarrow 2 N_{2}(g)+O_{2}(g)$

(ii) $2 N H_{3}(g) \longrightarrow N_{2}(g)+3 H_{2}(g)$

(iii) $2 C u\left(N O_{3}\right)_{2}(s) \longrightarrow 2 C u O(s)+4 N O_{2}(g)+O_{2}(g)$

(iv) $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)$

(v) $\mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s) .3$

**Sol.**(i) Homogeneous Equilibria

(ii) Homogeneous Equilibria

(iii) Heterogeneous Equilibria

(iv) Homogeneous Equilibria

(v) Heterogeneous Equilibria

(i) Write an expression of $K_{p}$ for the above reaction.

(ii) How will the value of $K_{p}$ and composition of equilibrium mixture be affected by

(a) increasing the pressure; $(b)$ increasing the temperature;

(c) using a catalyst?

**Sol.**$C H_{4}(g)+H_{2} O(g) \rightleftharpoons C O(g)+3 H_{2}$

(i) $\quad K_{p}=\frac{\left(P_{\mathrm{CO}}\right)\left(P_{\mathrm{H}_{2}}\right)^{3}}{\left(P_{\mathrm{CH}_{4}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)}$

(ii) $\quad K_{p}$ is not affected by increasing pressure

(b) will increase with increase in temperature because it is endothermic process.

(c) is not affected by catalyst.

$2 N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N_{2} O(g)$

If mixture of $0.482 \mathrm{mol}$ of $N_{2}$ and $0.933 \mathrm{mol}$ of $\mathrm{O}_{2}$ is placed in a reaction vessel of volume $10 L$ and allowed to from $N_{2} O$ at a temperature for which $K_{c}=2.0 \times 10^{-37}$ Determine the composition of the equilibrium mixture.

**Sol.**

$F e O(s)+C O(g) \rightleftharpoons F e(s)+C O_{2}(g) ; K_{p}=0.265 \quad$ at

$1050 K$

What are the equilibrium partial pressures of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ at $1050 \mathrm{K}$ if the initial pressure are : $p_{\mathrm{CO}}=1.4$ atm and $p_{C O_{2}}=0.80$ atm ?

**Sol.**

$N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ at $500 K$ is $0.061 .$ At a

particular time, the analysis shows that composition, of the reaction mixture is 3.00 $m o l$

$L^{-1} N_{2}, 2.00$ mol $L^{-1} H_{2}, 0.500 \mathrm{mol} \mathrm{L}^{-1} \mathrm{NH}_{3} .$ Is the

reaction at equilibrium ? If not, in which direction does the reaction tend to proceed to reach equilibrium?

**Sol.**

since $Q_{c}$ is not equal to $K_{c},$ the reaction is not at equilibrium. since is less than, the reaction would tend to proceed in forward direction to attain equilibrium

(i) addition of $H_{2}$

(ii) addition of $C H_{3} O H$

(iii) removal of $\mathrm{CO}$

(iv) removal of $\mathrm{CH}_{3} \mathrm{OH}$

on the equilibrium of the reaction: $2 H_{2}(g)+C O(g) \rightleftharpoons C_{H_{3}} O H(g)$

**Sol.**(i) Equilibrium will shift in the

*forward direction.*

(ii) Equilibrium will shift in the *backward direction.*

(iii) Equilibrium will shift in the *backward direction.*

(iv) Equilibrium will shift in the *forward direction.*

(i) $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(g)$

(ii) $\mathrm{CO}_{2}(g)+C(g) \rightleftharpoons 2 \mathrm{CO}(g)$

(iii) $4 N H_{3}(g)+5 O_{2}(g) \rightleftharpoons 4 N O(g)+6 H_{2} O(g)$

(iv) $C_{2} H_{4}(g)+H_{2}(g) \rightleftharpoons C_{2} H_{6}(g)$

**Sol.**$\Delta n=$ number of moles of gaseous products – number of moles of gaseous reactants

(i) $\quad \Delta n=3-3=0 \quad:$ Therefore, no effect of pressure because number of moles are equal on both sides.

(ii) $\quad \Delta n=2-2=0:$ Therefore, no effect of pressure because number of moles are equal on both sides.

(iii) $\quad \Delta n=10-9=1:$ since there is increase in number of moles on product side. Therefore, increase in pressure will favour backward reaction $i . e .,$ the reaction will go into left direction.

(iv) $C_{2} H_{4}(g)+H_{2}(g) \rightleftharpoons C_{2} H_{6}(g)$ $\Delta n=1-2=-1 .$ since there is decrease in number of moles on product side. Increase in pressure will favour forward reaction, i.e., the reaction will go towards right direction.

(i) $P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)$

(ii) $\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})$

(iii) 3 Fe $(s)+4 H_{2} O(g) \rightleftharpoons F e_{3} O_{4}(s)+4 H_{2}(g)$

**Sol.**Applying LeChatelier’s principle, on decreasing the pressure, moles of reaction products will,

(i) Increase

(ii) Decrease

(iii) Remain same $\left(\because n_{p}=n_{r} \text { gaseous }\right)$

$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) ; \Delta_{r} H^{\circ}=-92.0 \mathrm{kJ} \mathrm{mol}^{-1}$

Concentration of hydrogen, carbon monoxide and methanol become constant at equilibrium. What will happen if:

(i) volume of the reaction vessel in which reactants and products are contained is suddenly reduced by half?

(ii) the partial pressure of hydrogen is suddenly doubled?

(iii) an inert gas is added to the system?

**Sol.**$C O(g)+2 H_{2}(g) \rightleftharpoons C H_{3} O H(g)$

(i) The equilibrium will shift towards reactants because pressure will become double when volume is reduced to half. High pressure will favour forward reaction because decrease in number of moles are taking place on product side.

(ii) When partial pressure of hydrogen is doubled, the equilibrium will shift in forward direction.

(iii) When inert gas is added to the system, the equilibrium will shift in backward direction.

**Sol.**Solubility of iodine $=1.1 \times 10^{-3} \mathrm{mol} L^{-1}$

This means that $1.1 \times 10^{-3} \mathrm{mol}$ of iodine is present in $1 \mathrm{L}$ of water.

Molar mass of iodine $=254$

$\therefore$ Amount of iodine dissolved per litre of water

$=1.1 \times 10^{-3} \times 254=0.279 g$

The amount of iodine dissolved per $100 g$ of water

$=1.1 \times 10^{-3} \times 254=0.279 g$

The amount of iodine dissolved per $100 \mathrm{g}$ of water

$=\frac{0.279}{1000} \times 100=0.0279 g$

Amount of iodine added $=0.200 \mathrm{g}$

Amount of undissolved iodine in of water

$=0.2-0.0279=0.172 \mathrm{g}$

When $150 \mathrm{cm}^{3}$ of water is added after the equilibrium, the total volume becomes $100+150=250 \mathrm{cm}^{3}$

Now iodine dissolved in $250 \mathrm{g}$ of water $=\frac{0.279 \times 250}{1000}=$ $0.07 \mathrm{g}$

Amount of undissolved iodine $=0.2-0.07=0.130 \mathrm{g}$

Molar concentration of iodine

$=\frac{0.070(\mathrm{g})}{254 \mathrm{g} \mathrm{mol}^{-1} \times 250}\left(\frac{1000}{1 \mathrm{L}}\right)=0.0011 \mathrm{mol} \mathrm{L}^{-1}$

(ii) The $K_{p}$ for the reaction $N_{2} O_{4} \rightleftharpoons 2 N O_{2}$ is 640 $m m$ at $775 K .$ Calculate the $\%$ dissociation of $N_{2} O_{4}$ at equilibrium pressure of $160 \mathrm{mm} .$ At which pressure the dissociation will be $50 \%$

**Sol.**

**Sol.**

(i) Write the concentration ratio (reaction quotient), $Q_{c},$ for this reaction (Note: water is not in excess and is not a solvent in this reaction).

(ii) At $293 K,$ if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is $0.171 \mathrm{mol}$ of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at $293 \mathrm{K}, 0.214 \mathrm{mol}$ of ethyl acetate is found after some time. Has equilibrium been reached ?

**Sol.**

$P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)$

$\Delta H_{f}^{\circ}=+124.8 k_{\mathrm{d}} \mathrm{mol}^{-1}$

(i) Write expression $K_{c}$ for the reaction.

(ii) What would be the effect on $K_{c}$ if :

(a) more $P C l_{5}$ is added

(b) pressure is increased

(c) temperature is increased

(e) catalyst is added.

(iii) What is the value of Kc for the reverse reaction at the same temperature

**Sol.**

(i) $\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$

(ii) $C H_{4}(g)+2 S_{2}(g) \rightleftharpoons C S_{2}(g)+2 H_{2} S(g)$

(iii) $C O_{2}(g)+C(s) \rightleftharpoons 2 C O(g)$

(iv) $2 H_{2}(g)+C O(g) \rightleftharpoons C H_{3} O H(g)$

(v) $\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$

(vi) $4 N H_{3}(g)+5 O_{2}(g) \rightleftharpoons 4 N O(g)+6 H_{2} O(g)$

**Sol.**$P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g) ; \Delta_{f} H^{\circ}=+124.8 k J m o l^{-1}$

(i) $\quad K_{c}=\frac{\left[P C l_{3}\right]\left[C l_{2}\right]}{\left[P C l_{5}\right]}$

(ii) $(a) K_{c}$ will not be affected if more $P C l_{5}$ is added.

(b) does not change with increase in pressure.

(c) will increase, when temperature is increased.

(d) is not affected by catalyst.