Chemical Equilibrium

Addition of a solid component to a system at constant pressure has no effect on the equilibrium. Therefore, addition of Fe₂O₃ will not disturb the equilibrium.

To determine the equilibrium constant, $K_p$, for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:

$ \text{XY (s)} \rightleftharpoons \text{X (g)} + \text{Y (g)} $

For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products. Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression. Hence the expression for $K_p$ is:

$ K_p = P_X \cdot P_Y $

Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.

Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:

$ P_X = P_Y = \frac{10}{2} = 5 \, \text{bar} $

Substituting these partial pressures into the expression for $K_p$ gives:

$ K_p = P_X \cdot P_Y = 5 \, \text{bar} \cdot 5 \, \text{bar} = 25 \, \text{bar}^2 $

Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is:

Option C - 25

Given,



$\therefore$ $\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100$

$\therefore$ $\,\,\,{{1 + a} \over {1 - a}} = 10$

On solving

$a=0.81$

${\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81$

$\Delta {G^ \circ } = 2494.2J$

$2A\,\rightleftharpoons\,B + C$

$R = 8.314\,J/K/mol.$

$e = 2.718$

$\left[ A \right] = {1 \over 2},\left[ B \right] = 2,$

$\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}$

$ = {{2 \times 1/2} \over {{{\left( {{1 \over 2}} \right)}^2}}} = 4$

$\Delta {G^ \circ } = - 2.303\,\,RT\,\log \,{K_c}.$

$2494.2J = - 2.303 \times \left( {8.314J/K/mol} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {300K} \right)\log {K_c}$

$ \Rightarrow \log \,{K_c}$

$ = - {{2494.2\,J} \over {2.303 \times 8.314\,J/K/mol \times 300\,K}}$

$ \Rightarrow \log \,{K_c} = - 0.4341;\,\,{K_c} = 0.37;\,\,Q > {K_c}.$

$S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)$

${K_p} = {K_C}{\left( {RT} \right)^x}$

where $x = \Delta {n_g} = $ number of gaseous moles in product

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$-$ number of gaseous moles in reactant

$ = 1 - \left( {1 + {1 \over 2}} \right)$

$ = 1 - {3 \over 2} = - {1 \over 2}$

For the reaction

${N_2} + {O_2} \to 2NO$

$\,K = 4 \times {10^{ - 4}}$

Hence for the reaction

$NO \to {1 \over 2}{N_2} + {1 \over 2}{O_2}$

$K' = {1 \over {\sqrt K }}$

$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$

$ = 50$

To determine the equilibrium constant $ K $, consider the equilibrium reaction:

$ \text{CO}_2 + \text{C (graphite)} \rightleftharpoons 2\text{CO} $

Initially, the pressure of $\text{CO}_2$ is 0.5 atm, and there is no $\text{CO}$:

Initial Pressure: $\text{CO}_2 = 0.5 \text{ atm}, \quad \text{CO} = 0$

At equilibrium, assume that $ x $ atm of $\text{CO}_2$ is converted into $\text{CO}$:

Final Pressure: $\text{CO}_2 = 0.5 - x \text{ atm}, \quad \text{CO} = 2x \text{ atm}$

The total pressure at equilibrium is given as 0.8 atm:

$ 0.5 - x + 2x = 0.5 + x = 0.8 $

Solving for $ x $:

$ x = 0.8 - 0.5 = 0.3 \text{ atm} $

The equilibrium constant $ K_p $ is calculated as:

$ K_p = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} $

Substitute the equilibrium pressures:

$ K_p = \frac{(2x)^2}{0.5 - x} = \frac{(2 \times 0.3)^2}{0.5 - 0.3} = \frac{0.6^2}{0.2} $

$ K_p = \frac{0.36}{0.2} = 1.8 \text{ atm} $

$\mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} \, + \,{H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {HCO_3^ - \left( {aq} \right)}\limits_x \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x $

${K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}$

$ = {{x \times x} \over {0.034 - x}}$

$ \Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}$

$ \Rightarrow x = 1.195 \times {10^{ - 4}}$

As ${H_2}C{O_3}$ is a weak acid so the concentration of

${H_2}C{O_3}$ will remain $0.034\,\,$ as $0.034 > > x.$

$x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]$

$ = 1.195 \times {10^{ - 4}}$

Now, $\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y $

As $HCO_3^ - \,$ is again a weak acid (weaker than ${H_2}C{O_3})$

with $x > > y.$

${K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}$

$ = {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}$

Note : $\left[ {{H_3}{O^ + }} \right] = {H^ + }\,\,$ from first step $(x)$

and from second step $\left( y \right) = \left( {x + y} \right)$

[As $\,\,\,x > > y\,\,$ so $\,\,\,x + y \simeq x\,\,\,$ and $x - y \simeq x$]

So, $\,\,\,{K_2} \simeq {{y \times x} \over x} = y$

$ \Rightarrow {K_2} = 4.8 \times {10^{ - 11}}$

$ = y = \left[ {CO_3^{{2^ - }}} \right]$

So the concentration of $\,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = \,\,\,$ concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations.

Thus the final concentrations are

$\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}$

$\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}$

Reaction $(c)$ can be obtained by adding reactions $(a)$ and $(b)$ therefore ${K_3} = {K_1}.{K_2}$

Hence $(c)$ is the correct answer.

Let the initial moles of $X$ be $'a'$

and that of $Z$ be $'b'$ the for the given reactions,

we have $X\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y$

$\eqalign{ & Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \cr & At\,\,equi.\,\,\,\,\,\,\,a\left( {1 - \alpha } \right)\,\,\,\,\,\,\,2a\alpha \cr & (moles) \cr} $

Total no. of moles $ = a\left( {1 - \alpha } \right) + 2a\alpha $

$ = a - a\alpha + 2a\alpha $ $ = a\left( {1 + \alpha } \right)$

Now, $\,\,\,{K_{{p_1}}} = {{{{\left( {{n_y}} \right)}^2}} \over {{n_x}}} \times {\left( {{{{P_{{T_1}}}} \over {\sum n }}} \right)^{\Delta n}}$

or, $\,\,\,{K_{{p_1}}} = {{{{\left( {2a\alpha } \right)}^2}.{P_{{T_1}}}} \over {\left[ {a\left( {1 - \alpha } \right)} \right]\left[ {a\left( {1 + \alpha } \right)} \right]}}$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $Z\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,P+Q$

$\eqalign{ & Initial\,\,\,\,\,\,\,\,\,\,\,b\,\,moles\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \cr & Ar\,\,equi.\,\,\,\,\,\,b\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,b\alpha \,\,\,\,\,\,\,b\alpha \cr & (moles) \cr} $

Total no. of moles

$ = b\left( {1 - \alpha } \right) + b\alpha + b\alpha $

$ = b - b\alpha + b\alpha + b\alpha $

$ = b\left( {1 + \alpha } \right)$

Now $\,\,\,{K_{{P_2}}} = {{{n_Q} \times {n_P}} \over {{n_z}}} \times {\left[ {{{{P_{{T_2}}}} \over {\sum\nolimits_n \, }}} \right]^{\Delta n}}$

or$\,\,\,{K_{{P_2}}} = {{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}} \over {\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}}$

or$\,\,\,$ $\,\,\,{{{K_{{P_1}}}} \over {{K_{P2}}}} = {{4{\alpha ^2}.{P_{{T_1}}}} \over {\left( {1 - {\alpha ^2}} \right)}} \times {{{{\left( {1 - \alpha } \right)}^2}} \over {{P_{{T_2}}}.{\alpha ^2}}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {{4{P_{{T_1}}}} \over {{P_{{T_2}}}}}$
or$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over 9}\,\,\,\,\,$ [ as $\,\,\,{{{K_{{P_1}}}} \over {{K_{{P_1}}}}} = {1 \over 9}\,\,$ given ]

or$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over {36}}$

or$\,\,\,1:36$

$S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)$

${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$

$ = 4.9 \times {10^{ - 2}};$

On taking the square of the above reaction

${{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}$

$ = 24.01 \times {10^{ - 4}}$

now $K{'_C}$

for $\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}$

$ = {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$

$ = {1 \over {24.01 \times {{10}^{ - 4}}}}$

$ = 416$

$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x $

Total moles after dissociation

$1 - x + x + x = 1 + x$

${P_{PC{l_3}}} = $ mole fraction of $PCl{}_3 \times $ Total pressure

$ = \left( {{x \over {1 + x}}} \right)P$

For the reaction : -

$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$

Given ${K_c} = 1.8 \times {10^{ - 6}}\,\,$ at $\,\,{184^ \circ }C$

$R=0.0831$ $\,\,kJ/mol.k$

${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$

$ = 6.836 \times {10^{ - 6}}$

$\left[ {} \right.$ as $\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]$

Hence it is clear that ${K_p} > {K_c}$

The reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of ${F_2}$ and or $C{l_2}$ favour the for ward direction and hence the production of $Cl{F_3}.$

$\mathop {N{H_4}HS\left( s \right)\,\rightleftharpoons\,}\limits_{\matrix{ {start} \cr {At\,\,equib.} \cr } } \,\,\mathop {N{H_3}\left( g \right)}\limits_{\matrix{ {0.5\,\,atm} \cr {0.5 + x\,\,atm} \cr } } \,\, + \mathop {{H_2}S\left( g \right)}\limits_{\matrix{ {0\,\,atm} \cr {x\,\,atm} \cr } } $

Then $0.5 + x + x = 2x + 0.5 = 0.84\,\,$ (Given)

$ \Rightarrow x = 0.17\,\,atm.$

${P_{N{H_3}}} = 0.5 + 0.17 = 0.67\,atm;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ${P_H}_{_2S} = 0.17\,atm$

$K = {P_{N{H_3}}} \times {P_{{H_2}S}}$

$ = 0.67 \times 0.17\,at{m^2}$

$ = 0.1139 = 0.11$

For ${P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)$

${K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.$ The solids have concentration unity

${K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}$

$K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}$

$ = {1 \over {\sqrt {{K_c}} }}$

$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$

$ = 50$

${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$

Here $\Delta n = 1 - 2 = - 1$

$\therefore$ ${{{k_p}} \over {{K_c}}} = {1 \over {RT}}$

${K_c} = {{{{\left[ {N{O_2}} \right]}^2}} \over {\left[ {{N_2}{O_4}} \right]}} = {{\left[ {1.2 \times {{10}^{ - 2}}} \right]^2} \over {\left[ {4.8 \times {{10}^{ - 2}}} \right]}}$

$ = 3 \times {10^{ - 3}}\,mol/L$

Due to exothermicity of reaction low or optimum temperature will be required. Since $3$ moles are changing to $2$ moles.

$\therefore$ High pressure will be required.

In this reaction the ratio of number of moles of reactants to products is same $i.e.\,\,2:2,$ hence change in volume will not alter the number of moles.

${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$

$\Delta n = 1 - \left( {1 + {1 \over 2}} \right)$

$ = 1 - {3 \over 2} = - {1 \over 2}.$

$\therefore$ $\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}$