Chemical Equilibrium - Section V

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Miscellaneous Examples

Example - 12

The partial pressure of ${{CO}_2}$ in the reaction

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

is 0.773 mm at 500ºC. Calculate ${K_p}$ at 600ºC for the above reaction, assuming that $\Delta \text{H }\!\!{}^\circ\!\!\text{ }$ for the reaction is $43.2 K cal\; {mol}^{-1}$ and does not vary much in the temperature range of interest.

Solution :

The expression for ${K_p}$ will simply be:

${{K}_{p}}={{p}_{C{{O}_{2}}}}$

$\; = 0.773\; mm\;$ at 500ºC

Let the equilibrium constant at 600ºC be represented by $k_{p}^{'}$ . Using the Van’t Hoff’s equation, we have

$\log \frac{k_{p}^{'}}{{{k}_{p}}}=\frac{\Delta H{}^\text{o}}{2.303R}\ \left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\;\;\ \ \ _{{{T}_{2}}\,=\,\,600{}^\text{o}C=873K}^{{{T}_{1}}\,=\,\,500{}^\text{o}C=773K}$

Using $\Delta H{}^\circ = 43200 cal\; {mol}^{-1}$ and $R = 1.98 \;cal\; {deg}^{-1} \; {mol}^{-1}$ (we have to make sure all the quantities have consistent units), we obtain

$\log \frac{k_{p}^{1}}{{{k}_{p}}}=\frac{43200}{2.303\times 1.98}\ \left( \frac{1}{773}-\frac{1}{873} \right)$

$\Rightarrow \;\;\;\;\; K_p^' = 19.6\;mm$

Example - 13

For the reaction $CO(g)+2{{H}_{2}}(g)\ \ \ \rightleftharpoons \ \ \ C{{H}_{3}}OH(g),\ {{H}_{2}}$ , is introduced into a five-litre flask at 327ºC, containing 0.2 mole of $CO(g)$ and a catalyst, until the pressure is 4.92 atm. At this point 0.1 mol of $C{{H}_{3}}OH$ is formed. Calculate ${K_p}$ and ${K_c}$ for this system.

Solution :

During he entire “ $H_2$ -introduction” process, assume that $x$ moles of ${H_2}$ were added to to the system. By the time “ $H_2$ - introduction” process is stopped, 0.1 mole of is formed.

Note that special emphasis has been laid on telling us that there is a catalyst present, which means that the attainment of equilibrium is rapid. Thus, as more $H_2$ is introduced, the new equilibrium position is quickly attained, so when addition of $H_2$ is stopped, the ${CH_3}OH$ moles given as 0.1 mol should be scan as the equilibrium value.

At equilibrium, therefore:

Moles $\;\;\;<tex>\underset{0.1}{{CO}}\,+\underset{x-0.2}{{2{{H}_{2}}}}\,\ \ \rightleftharpoons \ \ \ \underset{0.1}{{C{{H}_{3}}OH}}\,\,\,$ Total = x

This is because to produce 0.1 mol of ${CH_3}OH$ , 0.1 mol of $CO$ and 0.2 mol of ${H_2}$ must be utilized.

The total pressure, $P$ , at equilibrium is 4.92 atm; and the total moles at equilibrium are $x$ :

$PV=xRT$

$\Rightarrow \;\;\;\;\; x = \frac{{PV}}{{RT}} = \frac{{4.92 \times 5}}{{0.0821 \times 600}} = 0.5$

Thus, at equilibrium, the respective moles are

${x_{co}} = 0.1 & {x_{{H_2}}} = 0.3 & {x_{C{H_3}OH}} = 0.1$

Finally,

${K_p} = \frac{{\left( {\frac{{0.1}}{{0.5}}P} \right)}}{{\left( {\frac{{0.1}}{{0.5}}P} \right)\;{{\left( {\frac{{0.3}}{{0.5}}P} \right)}^2}}}\;\;\;\;$ Why?

$\; = \frac{{25}}{{9\;{P^2}}}$

$\;= 0.1147 \;{atm}^{-2}$

${K_c} = \frac{{{K_p}}}{{{{(RT)}^{\Delta n}}}}\;\;\;\;$ {or ${K_c}$ can also be calculated by evaluating the molar concentration}

$= 277.8 (mol {L}^{-1})^{-2}$

Example - 14

The equilibrium constant ${K_p}$ of the system

$2S{{O}_{2}}(g)+{{O}_{2}}(g)\ \ \rightleftharpoons \ \ 2S{{O}_{3}}(g)$

is 900 atm at 800 K. A mixture containing ${{SO}_3}$ and ${O_2}$ having initial partial pressures of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800 K.

Solution :

We will directly work in terms of partial pressures here (rather than moles):

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ... (i)$

Now, an alert reader might have noticed that the units specified for ${K_p}$ are “atm”, but for the reaction system written in the way it is, $\Delta n = - 1$ , which means the ${K_p}$ mentioned is actual for the system

$2S{{O}_{3}}\ \ \rightleftharpoons \ \ 2S{{O}_{2}}+{{O}_{2}}\;\;\;\;\;\;\;\;\; ... (ii)$

However, this is not a problem; the ${K_p}$ for (1) will be simply the reciprocal of the ${K_p}$ for (2), Thus, for (1), we get:

${{K}_{p}}=\frac{1}{900}=\frac{{{(1-2x)}^{2}}}{{{(2x)}^{2}}\ (2+x)}$

$\Rightarrow \;\;\;\;\; x = 0.475$

${P_{S{O_3}}} = 1 - 2x = 0.05\;atm$

${P_{S{O_2}}} = 2x = 0.95\;atm$

${P_{{O_2}}} = 2 + x = 2.475\;atm$

Example - 15

At a certain temperature, ${K_c}$ is 16 for the system

$S{{O}_{2}}(g)+N{{O}_{2}}(g)\ \ \rightleftharpoons \ \ S{{O}_{3}}(g)+NO(g)$

If we take one mole of each of the four gases in a one - litre container, What would be the equilibrium concentration of $NO$ and ${NO_2}$ ?

Solution :

The problem is evidently quite simply:

Since $V = IL$ , the concentrations at equilibrium are the same as the moles at equilibrium. Thus,

${{K}_{c}}=\frac{{{(1+x)}^{2}}}{{{(1-x)}^{2}}}=16$

$\Rightarrow \;\;\;\;\; x = 0.6$

$\Rightarrow \;\;\;\;\; \left[ {NO} \right] = 1.6\;mol\;{L^{ - 1}}\;,\;\left[ {N{O_2}} \right] = 0.4\;mol\;{L^{ - 1}}$

Example - 16

Two solids $X$ and $Y$ dissociate into gaseous products as follows:

$X(s)\ \ \rightleftharpoons \ \ A(g)+B(g)\;\;\;\;\;\;$ (I)

$Y(s)\ \ \rightleftharpoons \ \ \ B(g)+C(g)\;\;\;\;\;\;$ (II)

At a given temperature $T$ , the pressure over $X$ is 40 mm while that over $Y$ is 60 mm. What will be the total pressure over a mixture of $X$ and $Y$ ?

Solution :

For the first system I, we have

${{p}_{A}}={{p}_{B}}=\frac{40mm}{2}=20\,mm\;\;\;\;\;$ (Why?)

$\Rightarrow \;\;\;\;\; {K_{p\,{\rm{I}}}} = 400\,m{m^2}$

Similarly,

${K_{p\,{\rm{II}}}} = 900\,m{m^2}$

Now, when $X$ and $Y$ are mixed what happens? The pressure of $B$ will be due to both the systems.

Assume ${p_A} = {x_A}$ and ${p_C} = {x_C}$ . Thus, ${p_B} = {x_A} + {x_C}$

$\Rightarrow \;\;\;\;\; {K_{p\,{\rm{I}}}} = {x_A}({x_A} + {x_C}) = 400$

${K_{p\,{\rm{II}}}} = {x_C}({x_A} + {x_C}) = 900$

$\Rightarrow \;\;\;\;\; \frac{{{x_A}}}{{{x_C}}} = \frac{4}{9} \;\;\;\;\; \Rightarrow \;\;\;{x_A} = 11.1\;mm,\;\;{x_C} = 24.97\,mm$

$\Rightarrow \;\;\;\;\; {P_{Tot}} = {x_A} + ({x_A} + {x_C}) + {x_B}$

$= 72.15 \;mm$

Example - 17

In a vessel, two equilibria are simultaneously established:

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

${{N}_{2}}(g)+2{{H}_{2}}(g)\ \ \rightleftharpoons \ \ {{N}_{2}}{{H}_{4}}(g)$

Initially, the vessel contains ${N_2}$ and ${H_2}$ in the molar ratio 9:13. The equilibrium pressure is ${(7P}_0}$ , and ${{p}_{N{{H}_{3}}}}={{P}_{0}},\ {{p}_{{{H}_{2}}}}=2{{P}_{0}}$ . Find $K_{p}^{'}\,s$ for both the system.

Solution :

Since the initial molar ratio ${{m}_{{{N}_{2}}}}:{{m}_{{{H}_{2}}}}=9:13$ , the initial partial pressures will be in the same ratio, so we let

${p_{{N_2},\,{\rm{init}}}} = 9x, \;\; {p_{{H_2},\;{\rm{init}}}} = 13x$

Now, consider the compositions at equilibrium (we have to consider both equilibria simultaneously):

$\underset{9x-u-v}{p{{{N}_{2}}}}\,+\underset{13x-3u-2v}{{3{{H}_{2}}}}\,\ \ \ \ \ \rightleftharpoons \ \ \ \underset{2U}{{\ 2N{{H}_{3}}}}\,$

$\underset{9x-u-v}{{{{N}_{2}}}}\,+\underset{13x-3u-2v}{{2{{H}_{2}}}}\,\ \ \ \ \ \ \rightleftharpoons \ \ \ \underset{v}{{{{N}_{2}}{{H}_{4}}}}\,$

where $U$ , $V$ correspond to ${N_2}$ utilized in the first and second systems. Since the total pressure and equilibrium is ${{7P}_0}$ , and we know ${{p}_{N{{H}_{3}}}}\ and\ {{p}_{{{H}_{2}}}}$ , we have

${{p}_{N{{H}_{3}}}}\ =2u={{P}_{0}}\Rightarrow u=\frac{{{P}_{0}}}{2}$

${{p}_{{{H}_{2}}}}\ =13x-3u-2v=2{{P}_{0}}\Rightarrow 13x-2v=\frac{7{{P}_{0}}}{2}$

${{p}_{Tot}}=\left( 9x-u-v \right)+\left( 13x-3u-2v \right)+\left( 2u \right)+\left( v \right)$

$\;\;=9x-2u+\left( 13x-2v \right)$

$\;\;=9x-{{P}_{0}}+\frac{7{{P}_{0}}}{2}=7{{P}_{0}}\;\;\;\;\;\;... (i)$

$\Rightarrow \;\;\;\;\; x = \frac{{{P_0}}}{2}$

$\Rightarrow \;\;\;\;\; v = \frac{{3{P_0}}}{2}$

Finally, to evaluate ${K_{{p_1}}}$ and ${K_{{p_2}}}$ , we note that

${p_{{N_2}}} = 9x - u - v = \frac{5}{2}{P_0}$

${p_{{H_2}}} = 13x - 3u - 2v = 2{P_0}$

${p_{N{H_3}}} = 2u = {P_0}$

${p_{{N_2}{n_4}}} = v = \frac{{3{P_0}}}{2}$

$\Rightarrow \;\;\;\;\; {K_{{p_1}}} = \frac{{{{({p_{N{H_3}}})}^2}}}{{{p_{{N_2}}} \cdot {{({p_{{H_2}}})}^3}}} = \frac{{P_0^2}}{{\frac{5}{2}{P_0} \cdot {{(2{P_0})}^3}}} = \frac{1}{{20P_0^2}}$

and $\;\; \Rightarrow \;\;\;\;\; {K_{{p_2}}} = \frac{{{p_{{N_2}{H_4}}}}}{{{p_{{N_2}}} \cdot {{({p_{{H_2}}})}^2}}} = \frac{{\frac{3}{2}{P_0}}}{{\frac{5}{2}{P_0} \cdot {{(2{P_0})}^2}}} = \frac{3}{{20P_0^2}}$

Can you infer from the initial data that the fact that ${K_{{p_2}}}$ is greater than ${K_{{p_1}}}$ should have been expected!

Example - 18

Consider the gaseous system in a closed vessel:

$X(g)\ \ \ \rightleftharpoons \ \ \ nY(g)$

Let the degree of dissociation for this be $\alpha$ . Let the vapour densities of the system be $D_0$ and ${D_{eq}}$ initially and at equilibrium respectively. Show that