Chemical Equilibrium

Total number of moles $=0.5+1=1.5 \mathrm{~mol}$

$\begin{aligned} & p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{0.5}{1.5} \times 1 \mathrm{~atm}=\frac{1}{3} \mathrm{~atm} \\ & p_{\mathrm{NO}_2}=\frac{1}{1.5} \times 1 \mathrm{~atm}=\frac{2}{3} \mathrm{~atm} \end{aligned}$

According to law of chemical equilibrium

$K_p=\frac{p^2 \mathrm{NO}_2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{\left(\frac{2}{3}\right)^2}{\left(\frac{1}{3}\right)}=1.33 \mathrm{~atm}$

We know,

$\Delta G \Upsilon=-2.303 R T \log K_p$ ...... (i)

Substituting value in Eq. (i), we get

$\begin{aligned} \Delta G \Upsilon & =-2.303 \times 0.082 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 333 \mathrm{~K} \\ & =-7.79 \mathrm{~atm} \text { litre } \quad \end{aligned}$

$\begin{aligned} & 1 \mathrm{~atm} \text { litre }=101.32500 \text { joules } \\ & \begin{aligned} -7.79 \mathrm{~atm} \text { litre } & =-7.79 \times 101.32 \text { joules } \\ & =-789.2 \mathrm{~J} \mathrm{~mol}^{-1} \approx-790 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} \end{aligned}$.

Le-Chatelier principle is not applicable to pure solids and liquids because of the reason that they experience negligible change in concentration during chemical equilibrium which means that adding or removing a solid form system at equilibrium has no effect on equilibrium.

$\mathrm{Fe}(s)+\mathrm{S}(s) \rightleftharpoons \mathrm{FeS}(s)$ is in solid equilibrium phase, hence Le-Chatelier principle is not applicable.

$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$

Given, $\Delta_f H^{\circ}[\mathrm{NO}(g)]=90.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\begin{aligned} & \Delta_f H^{\circ}\left[\mathrm{NO}_2(g)\right]=32.48 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta H^{\circ}=\Sigma \Delta H^{\circ} \text { product }-\Sigma \Delta_f H^{\circ}(\mathrm{NO}) \\ & =3248-90.4 \\ & =-57.92 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \end{aligned}$

We know that,

Standard Gibb's free energy, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$\begin{gathered} =-57.92-298 \times(-70.8) \\ =-57.92+21.098=-36.82 \mathrm{~kJ} / \mathrm{mol} \\ =-38.82 \times 10^3 \mathrm{~J} / \mathrm{mol} \end{gathered}$

We also know, $\Delta G^{\circ}=-2.303 R T \log K_{\text {eq }}$

Here, $K_{\mathrm{eq}}=$ Equilibrium constant

$\begin{aligned} -36.82 \times 10^3 & =-2303 \times 8.314 \times 298 \times \log K_{\mathrm{eq}} \\ \log K_{\mathrm{eq}} & =\frac{36.82}{5.70584} \\ \log K_{\mathrm{eq}} & =0.501 \\ K_{\mathrm{eq}} & =\text { antilog } 0.501 \\ & =3.162 \times 10^6 \end{aligned}$

$\text { } \begin{aligned} & \frac{1}{2} X_2+\frac{3}{2} Y_2 \rightleftharpoons X Y_3 ; \Delta H=-30 \mathrm{~kJ} \\ & \Delta S_{\text {reaction }}=\sum_\limits{i=1} \Delta S_{\text {product }}-\sum \Delta S_{\text {reactant }} \\ & X_2+3 Y_2 \rightleftharpoons 2 X Y_3 \\ & \Delta S_{\text {reaction }}=2 \times 50-3 \times 40-1 \times 60 \\ &=-80 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}$

Using Gibb's free energy, $\Delta G=\Delta H-T \Delta S$

$\begin{aligned} \Delta G & =0 \\ 0 & =\Delta H-T \Delta S \Rightarrow \Delta H=T \Delta S \\ 1000 \times(-60) & =T \times(-80) \Rightarrow T=750 \mathrm{~K} .\end{aligned}$

$\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)$

The above equilibrium reaction is exothermic reaction. When pressure is lowered, the forward reaction is favoured.

On going from $p_1$ to $p_3$ yield of $\mathrm{SO}_3$ is increasing because lower pressure favours forward reaction.

The relationship between K$_p$ and K$_c$ is

${K_p} = {K_c}{(RT)^{\Delta n}}$ ..... (i)

Here, $\Delta n$ = (number of moles of product of gas) $-$ (number of moles of reactant of gas)

K$_p$ = equilibrium constant at constant pressure

K$_c$ = equilibrium constant at constant concentration

R = rate constant

$2 A(g) \rightleftharpoons B(g)+C(g)$

The value of $\Delta n$ is $=2-2=0$

Put value of $\Delta n$ in Eq. (i) we get,

$K_p=K_C(R T)^0 \Rightarrow\left[K_p=K_C\right] $

To determine the correct relationship between the equilibrium constants $K_C$ and $K_p$ for the reaction:

$\text{Fe}_2\text{N(s)} + \frac{3}{2}\text{H}_2\text{(g)} \rightleftharpoons 2\text{Fe(s)} + \text{NH}_3\text{(g)}$

First, consider the general relationship between $K_C$ and $K_p$:

$ K_p = K_C(RT)^{\Delta n} $

where $\Delta n$ is the change in the number of moles of gas between the products and the reactants, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

For the given reaction, calculate $\Delta n$:

Moles of gas in products: NH_3(g) = 1 mole

Moles of gas in reactants: ${3 \over 2}$ H_2(g) = 1.5 moles

Therefore:

$ \Delta n = 1 - 1.5 = -0.5 $

Substitute this value into the equation relating $K_C$ and $K_p$:

$ K_p = K_C(RT)^{-0.5} $

Rearrange to express $K_C$:

$ K_C = K_p(RT)^{0.5} $

Thus, the correct option is:

Option A: $K_C = K_p(RT)^{1/2}$

We have two reactions:

$ A \;\rightleftharpoons\; B + C \quad\text{with } K_{eq}^{(1)}$

$ B + C \;\rightleftharpoons\; P \quad\text{with } K_{eq}^{(2)}$

We want the equilibrium constant for the overall reaction

$ A \;\rightleftharpoons\; P. $

How to combine the equilibrium constants

When two reactions are added to yield a net (overall) reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions. Formally:

$ \text{(Reaction 1)} + \text{(Reaction 2)} \;\Rightarrow\; \text{(Net Reaction)}, $

and

$ K_{\text{net}} \;=\; K_{\text{(Reaction 1)}} \times K_{\text{(Reaction 2)}}. $

Applying it here

Reaction 1: $A \to B + C$ has $K_{eq}^{(1)}.$

Reaction 2: $B + C \to P$ has $K_{eq}^{(2)}.$

When we add them,

$ \underbrace{A}_{\text{Reactant 1}} \;\longrightarrow\; \underbrace{B + C}_{\text{Products of Reaction 1}=\text{Reactants of Reaction 2}} \;\longrightarrow\; \underbrace{P}_{\text{Product of Reaction 2}}. $

So the net reaction is $A \to P$, and its equilibrium constant is

$ K_{eq}^{\text{(net)}} \;=\; K_{eq}^{(1)} \times K_{eq}^{(2)}. $

Thus, the correct answer is:

$ \boxed{K_{eq}^{(1)} \, K_{eq}^{(2)}}. $

Hence, the answer (Option D) is correct.

∵ When ferric nitrate is mixed with aqueous solution of potassium thiocyanate, i.e. (KSCN).

The intensity of red colour becomes constant on attaining equilibrium.

When, (I) oxalic acid, i.e. $\left(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4\right)$ or (II) mercuric chloride, i.e. $\left(\mathrm{HgCl}_2\right)$ is added at equilibrium.

Both (I) and (II) will decrease the intensity of red colour. Due to formation of complex.

Hence, option (a) is the correct answer.

$2 A \rightleftharpoons B+C$

Reaction quotient, $Q=\frac{[B][C]}{[A]^2}=\frac{\left(6 \times 10^{-5}\right) \times\left(6 \times 10^{-5}\right)}{\left(6 \times 10^{-5}\right)^2}$

$ \Rightarrow \quad Q=1>K_{\mathrm{eq}}=2 \times 10^{-3} $

So, the reaction will proceed in backward direction.

Given, at an instant,

$ \begin{aligned} {[A] } & =[B]=[C] \\ & =6 \times 10^{-5} \mathrm{M} \end{aligned} $

and equilibrium constant,

$ K_{\mathrm{eq}}=2 \times 10^{-3} $

$ \begin{aligned} & \text { For the forward reaction of }\\ &\begin{aligned} & A \rightleftharpoons B, \\ & \ln K=-\frac{\Delta H}{R T} \end{aligned} \end{aligned} $

$ \begin{aligned} & K=e^{[\text {van't Hoff isotherm equation }]} \\ &=e^{\frac{-(-41.5)}{8.314 \times 10^{-3} \times 500}}=e^{9.983} \simeq e^{10} \\ & {[\because \Delta H=\text { Enthalpy change of the }} \text { forward reaction }] \\ &=-41.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $

$ \begin{aligned} R & =8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \\ T & =500 \mathrm{~K}, \\ K & =\text { Equilibrium constant] } \end{aligned} $

$ \begin{aligned} &\text { Given, vapour density }(d)=40\\ &\begin{aligned} & \qquad \mathrm{N}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{NO}_2 \\ & \text { Molar mass } 92 \\ & \quad D=\frac{M\left(\mathrm{~N}_2 \mathrm{O}_4\right)}{2}=\frac{92}{2}=46 \\ & \therefore \alpha \text { (Degree of dissociation) }=\frac{D-d}{(n-1) d} \\ & \alpha=\frac{46-40}{(2-1) 40} \\ & \quad\left(n=\text { number of } \mathrm{NO}_2 \text { at equilibrium }\right) \\ & \alpha=\frac{6}{40}=0.15 \end{aligned} \end{aligned} $

$\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \quad K_C=$ ?

Given, at equilibrium concentration

$ \begin{aligned} \mathrm{N}_2 & =4 \times 10^{-3} \mathrm{M} \\ \mathrm{O}_2 & =3 \times 10^{-3} \mathrm{M} \\ \mathrm{NO} & =3 \times 10^{-3} \mathrm{M} \end{aligned} $

$K_C$ for the above reaction can be given as,

$ K_C=\frac{\left[\mathrm{NO}^2\right.}{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Put the concentration in the Eq. (i)

$ \begin{aligned} K_C & =\frac{\left[3 \times 10^{-3}\right]^2}{\left[4 \times 10^{-3}\right]\left[3 \times 10^{-3}\right]} \\ \Rightarrow \quad K_C & =\frac{3 \times 10^{-3}}{4 \times 10^{-3}}=0.75 \end{aligned} $

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