Chemical Equilibrium

2NO₂ (g) $\buildrel \, \over \longrightarrow $ N₂O₄ (g); $\Delta $H = $-$ ve

At equilibrium, N₂O₄ $\rightleftharpoons$ 2NO₂ ; $\Delta $H = + ve

On adding the inert gas at constant pressure, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Therefore, in above case, reaction will move towards forward direction which will lead to the decomposition of dinitrogen tetraoxide (N₂O₄).

	CO	+	Cl2	$\rightleftharpoons$	COCl2
Initially number of moles	2		3		0
At equilibrium number of moles	1		2		1

The equilibrium constant,

K_c = ${{\left[ {COC{l_2}} \right]} \over {\left[ {CO} \right]\left[ {C{l_2}} \right]}}$

=  ${{\left( {{1 \over 5}} \right)} \over {\left( {{1 \over 5}} \right) \times \left( {{2 \over 5}} \right)}}$

=   ${5 \over 2}$

=  2.5

From Boyle's law, we know when volume increase then pressure decreases, and to keep the pressure on original value the reaction should proceeds in the direction where the number of moles of gases increases.

In this reaction, only satisfy this.

2 NO₂(g) $\rightleftharpoons$ 2 NO(g) + O₂ (g)

$\therefore\,\,\,\,$ $\Delta $n_g = (2 + 1) $-$ 2 = 1

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

$a=0.81$

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

On decreasing total pressure, the reaction will move in the forward direction where number of gaseous molecules is less.

Hence, option (A) is correct.

At the start of reaction Q = 0. From $\Delta$G = $\Delta$G^o + RT lnQ, we have that for Q = 0 at the start of reaction $\Delta$_rG is negative, this causes dissociation of X₂ to take place spontaneously.

Hence, option (B) is correct.

If b_equilibrium = 0.7, then value of equilibrium constant is

${K_p} = {{8\beta _{eq}^2} \over {4 - \beta _{eq}^2}} = {{8{{(0.7)}^2}} \over {7 - {{(0.7)}^2}}}$

The value of K_p is greater than 1, which is not possible as given that $\Delta$G^o > 0 for the reaction.

Hence, option (C) is incorrect.

As $\Delta$G^o > 0 and $\Delta$G^o = $-$RT ln(K_p)

$\Delta$G^o > 1, so K_p should be less than 1.

We know that

${K_p} = {K_c}{(RT)^{\Delta {n_g}}}$

${K_c} = {{{K_p}} \over {{{(RT)}^{\Delta {n_g}}}}}$

Therefore, K_c < K_p . As K_p is less than 1, so K_c is also less than 1.

Hence, option (D) is correct.

Total number of moles at equilibrium = (1 + $\alpha$) = 1 + (b_eq / 2)

Therefore,

${p_x} = \left( {{{{\beta _{eq}}} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)$

${p_{{x_2}}} = \left( {{{1 - ({\beta _{eq}}/2)} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)$

Therefore, the equilibrium constant K_p is given by

${K_p} = {{p_x^2} \over {{p_{{x_2}}}}} = {{\left( {{{{\beta _{eq}}} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)} \over {\left( {{{1 - ({\beta _{eq}}/2)} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)}}$

$ = {{\beta _{eq}^2} \over {1 - (\beta _{eq}^2/4)}}{p_T} = {{\beta _{eq}^2} \over {1 - (\beta _{eq}^2/4)}} \times 2 = {{8\beta _{eq}^2} \over {4 - \beta _{eq}^2}}$

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

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

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

$e = 2.718$

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

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

$ = {{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}$

The thermal decomposition of limestone (CaCO₃) is represented as:

$ \mathrm{CaCO}_3(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}), \Delta_r \mathrm{H} = \text {positive} $

The reaction is endothermic, meaning it requires heat to proceed.

(i) The forward reaction is endothermic (ΔH > 0), indicating heat absorption, while the reverse reaction is exothermic (ΔH < 0), indicating heat release.

Therefore, increasing the temperature will favor the forward reaction, leading to more CaCO₃ dissociation. Conversely, decreasing the temperature will favor the reverse reaction. Option (A) is correct.

The equilibrium constant (K) varies with temperature but does not depend on the concentration of any species in an aqueous solution or the pressure of gases such as CO₂. Option (B) is correct.

(ii) The equilibrium constant (K) for the reaction is expressed as:

$ \mathrm{K}_{\mathrm{C}} = [\mathrm{CO}_2] $

Because CaCO₃ and CaO are solids, their concentrations do not affect the equilibrium constant. Option (B) is correct.

(iii) At a given temperature, the equilibrium constant in terms of pressure (K_P) depends on the vapour pressure of CO₂.

$ \mathrm{K}_{\mathrm{P}} = \mathrm{P}_{\mathrm{CO}_2} $

The equilibrium constant (K_C) also depends on the pressure of CO₂:

$ \mathrm{K}_{\mathrm{C}} = \mathrm{P}_{\mathrm{CO}_2} (\mathrm{RT})^{-\Delta n} $

Option (C) is correct.

The temperature dependence of the Arrhenius constant is given by:

$ \ln \frac{\mathrm{K}_2}{\mathrm{K}_1} = \frac{\Delta_r \mathrm{H}}{\mathrm{R}} \left[ \frac{1}{\mathrm{T}_2} - \frac{1}{\mathrm{T}_1} \right] $

(iv) A catalyst alters the activation energy of the reaction, thus affecting its speed. However, it does not change the enthalpy or heat of the reactants and products. As a result, the enthalpy of the reaction ($\Delta_r H$) remains unchanged. Option (D) is correct.

Δ_rH is the enthalpy of the reaction.

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} $

For the equilibrium reaction in aqueous medium :

$ 2 \mathrm{Cu}^{+} \rightleftharpoons \mathrm{Cu}^0+\mathrm{Cu}^{2+} $

(i) If concentration of copper (I) is reduced, then according to Le-chatelier's principle the equilibrium reaction will shift backward.

(ii) The copper (II) ion reacts with chloride $\left(\mathrm{Cl}^{-}\right)$ion forming copper (III) chloride, which reacts with copper to produce the precipitate of CuCl .

$ \begin{aligned} & \mathrm{Cu}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CuCl}_2 \\\\ & \mathrm{CuCl}_2+\mathrm{Cu} \rightarrow 2 \mathrm{CuCl} \end{aligned} $

Since, the end product involves consumption of copper (I) ion to form CuCl , hence, reactive shifts backward.

(iii) The copper (II) ion also reacts with cyanide $\left(\mathrm{CN}^{-}\right)$ion forming copper (II) cyanide which on further reactions form $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$ with copper in (I) oxidation state shifting reaction backward.

$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{CN}^{-} & \rightarrow \mathrm{Cu}(\mathrm{CO})_2 \\\\ 2 \mathrm{Cu}(\mathrm{CN})_2 & \rightarrow 2 \mathrm{CuCN}+(\mathrm{CN})_2 \\\\ \mathrm{CuCN}+3 \mathrm{CN}^{-} & \rightarrow\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-} \end{aligned} $

(iv) Reaction of copper (II) ion with thiocyanide $\left(\mathrm{SCN}^{-}\right)$gives a complex $\left(\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-}$ with copper in +1 oxidation state. This shifts reaction backward.

$ \mathrm{Cu}^{2+}+4 \mathrm{SCN}^{-} \rightarrow\left[\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-} $

Hence, cyanide, thiocyanide and chloride forms complexes with copper in I oxidation state. Hence, reaction shifts backward to generate more copper (I) ions.

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

The Gibbs free energy varies from standard value as a function of temperature and equilibrium constant.

$\Delta$G = $\Delta$G$^\circ$ + RT ln K$_{eq}$

The standard Gibbs energy for a reaction is given by $\Delta$G$^\circ$.

At equilibrium, $\Delta$G = 0whereas $\Delta$G$^\circ$ for a reaction may or may not be zero.

$\Delta$G$^\circ$ = $-$RT ln K$_{eq}$

For a spontaneous process, Gibbs energy for a reaction is always negative, $\Delta$G < 0.

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

${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\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$

In complex $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$, the central metal ion is Ag and $\mathrm{NH}_3$ groups are ligands. The formation constant is denoted by $\mathrm{K}_f$.

$ \mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} $

The chemical reactions involved in the formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$complex are given below:

Step 1: $\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)^{+}\right] ; \mathrm{K}_1=3.5 \times 10^{-3}$

Step 2: $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$;$ \mathrm{K}_2=1.7 \times 10^{-3} $

In the formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$, two steps. The value of $\mathrm{K}_f$ can be obtained by following Formula:

$ \mathrm{K}_f=k_1 \times k_2 $

Substitute the values in the above formula,

$ \begin{array}{ll} & \mathrm{K}_f=3.5 \times 10^{-3} \times 1.7 \times 10^{-3} \\ \therefore & \mathrm{~K}_f=5.95 \times 10^{-6} \end{array} $

The $\mathrm{K}_f$ value does not match with values in options A, B and C.

The chemical reaction for formation of $\mathrm{NH}_3$ is reversible in nature.

Forward reaction: $\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$

$ \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons \mathrm{~K}_{\mathrm{c}} 2 \mathrm{NH}_3 $

Backward reaction: $2 \mathrm{NH}_3 \rightarrow \mathrm{~N}_2+3 \mathrm{H}_2$

The equilibrium constant $K_c$ in terms of concentration of reactants and products for the reaction is given below:

$ \mathrm{K}_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} $

The equilibrium constant $\mathrm{K}_p$ in terms of partial pressure of reactants and products for the reaction is given below :

$ \mathrm{K}_p=\frac{\left(\mathrm{p}_{\mathrm{NH}_3}\right)^2}{\mathrm{p}_{\mathrm{N}_2}\left(\mathrm{p}_{\mathrm{H}_2}\right)^3} $

At constant temperature, the values for $\mathrm{K}_c$ and $\mathrm{K}_p$ remain constant.

When $\mathrm{N}_2$ is added at equilibrium condition during this reaction, lowering of activation energy for both forward and backward reaction occurs in the presence of catalyst. This leads to increases in the rate of both forward and backward to the same extent. Thus, addition of catalyst helps in establishing equilibrium in less time. This means that there is increase in the concentration of both reactants and products in same order by catalyst. Thus, the value of $\mathrm{K}_c$ and $\mathrm{K}_p$ will remain unaffected, keeping the equilibrium state of the reaction unaffected by catalyst.

As there is no change in equilibrium condition, the Gibbs free energy will be zero.

Hence, $\Delta \mathrm{G}=0$

Thus, $\Delta \mathrm{G}$ for complete reaction is,

$ \begin{aligned} \Delta \mathrm{G} & =\mathrm{G}_{\text {products }}-\mathrm{G}_{\text {reactants }} \\ \Delta \mathrm{G} & =\left(2 \mathrm{G}_{\mathrm{NH}_3}\right)-\left(3 \mathrm{G}_{\mathrm{H}_2}+\mathrm{G}_{\mathrm{N}_2}\right) \\ 2 \mathrm{G}_{\mathrm{NH}_3} & =\mathrm{G}_{\mathrm{N}_2}+3 \mathrm{G}_{\mathrm{H}_2} \end{aligned} $

Here, G is Gibbs free energy per mole of the gaseous species measured at that partial pressure.

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}.$

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