Chemical Equilibrium

Given,

$ \begin{gathered} 2 \mathrm{AO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{AO}_3(g) ; \\ K_p=4 \times 10^{10} \end{gathered} $

Divide above reaction by 2

$ \begin{aligned} A \mathrm{O}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons A \mathrm{O}_3(g) & \\ K_p^{\prime}=\sqrt{K_p}=2 \times 1.0^5 & \end{aligned} $

$ \begin{aligned} &\text { Multiply above by } 3 \text {, }\\ &\begin{aligned} & 3 \mathrm{AO}_2+\frac{3}{2} \mathrm{O}_2 \rightleftharpoons 3 \mathrm{AO}_3 ; \\ & K_p^{\prime}=\left(2 \times 10^5\right)^3=8 \times 10^{15} \end{aligned} \end{aligned} $

Given reaction,

$ \mathrm{CO}_2(g)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) ; K=0.53 $

At equilibrium, $[\mathrm{CO}]=0.25 \mathrm{~mol} / \mathrm{L}$,

$ \begin{aligned} {\left[\mathrm{CO}_2\right] } & =0.5 \mathrm{~mol} / \mathrm{L} \\ {\left[\mathrm{H}_2\right] } & =0.6 \mathrm{~mol} / \mathrm{L},\left[\mathrm{H}_2 \mathrm{O}\right]=x \end{aligned} $

Rate reaction; $K=\frac{[\mathrm{CO}]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]}=\frac{0.25 \times x}{0.5 \times 0.6}$

$ x=0.636 $

$ \text { Given, reaction is } $

Partial pressure is given by

$ \begin{aligned} p_{\mathrm{N}_2 \mathrm{O}_4} & =\left(\frac{1-\alpha}{1+\alpha}\right) p ; \quad p_{\mathrm{NO}_2}=\frac{2 \alpha}{1+\alpha} \cdot p \\ K_p & =\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}} \end{aligned} $

Substituting the values,

$ \alpha=\left(\frac{\frac{K_p}{p}}{4+\frac{K_p}{p}}\right)^{1 / 2} . $

Given the equilibrium constants for the reactions:

$2A(g) \rightleftharpoons B(g) + C(g)$ with $K_1 = 16$

$2B(g) + C(g) \rightleftharpoons 2D(g)$ with $K_2 = 25$

We need to find the equilibrium constant $K$ for the reaction:

$ A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g) $

To determine this, we first manipulate the given reactions as follows:

Divide both sides of the first reaction by 2:

$ A(g) \rightleftharpoons \frac{1}{2}B(g) + \frac{1}{2}C(g) $

When you perform this division, the equilibrium constant becomes the square root of the original. Therefore, the equilibrium constant for this modified reaction is $\sqrt{K_1} = \sqrt{16} = 4$.

Next, divide both sides of the second reaction by 2:

$ B(g) + \frac{1}{2}C(g) \rightleftharpoons D(g) $

Similarly, the equilibrium constant for this modified reaction is $\sqrt{K_2} = \sqrt{25} = 5$.

By adding these two modified reactions, we obtain the target reaction:

$ A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g) $

The equilibrium constant for the resulting reaction is the product of the equilibrium constants of the individual modified reactions:

$ K = \sqrt{K_1} \cdot \sqrt{K_2} = 4 \cdot 5 = 20 $

Thus, the equilibrium constant $K$ for the reaction $A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g)$ at temperature $T(K)$ is 20.

The given reaction is,

$ \begin{array}{ccc} & \mathrm{PCl}_5 & \rightleftharpoons \mathrm{PCl}_3(g) & +\mathrm{Cl}_2(g) \\\\ \text { Moles at initial } & x & 0 & 0 \\\\ \text { Moles at equilibrium } & x-0.2 & 0.2 & 0.2 \end{array} $

Equilibrium constant of above reaction is,

$ \begin{aligned} K_C&=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} \\\\ 2 \times 10^{-2}&=\frac{[0.2][0.2]}{[x-0.2]} \\\\ \Rightarrow \quad x&=2.2 \mathrm{~mol} \end{aligned} $

To determine the equilibrium concentration of $ BO(g) $, we need to consider the reaction and the given equilibrium constant:

$ AO_2(g) + BO_2(g) \rightleftharpoons AO_3(g) + BO(g) $

The reaction quotient, $ Q_c $, is given by:

$ Q_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} $

Given $ K_c = 16 $ at equilibrium, the system satisfies:

$ K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = 16 $

Let $ x $ be the change in concentration of products and reactants at equilibrium. The initial concentration for all species is $ 1 \, \text{mol/L} $, thus:

Initial concentrations: $[AO_2] = [BO_2] = [AO_3] = [BO] = 1 \, \text{mol/L}$

Change in concentrations:

$[AO_2] = [BO_2] = 1 - x $

$[AO_3] = [BO] = 1 + x $

At equilibrium:

$ K_c = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = 16 $

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

Taking the square root of both sides:

$ \frac{1 + x}{1 - x} = 4 $

Solving for $ x $:

$ 1 + x = 4 - 4x $

$ 1 + 5x = 4 $

$ 5x = 3 $

$ x = 0.6 $

Thus, the equilibrium concentration of $ BO $ is:

$ [BO] = 1 + x = 1 + 0.6 = 1.6 \, \text{mol/L} $

Therefore, the correct option is:

Option A: 1.6

The $K_C$ for the reaction, $\frac{1}{3} \mathrm{~N}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \frac{2}{3} \mathrm{NH}_3(\mathrm{~g})$ is 50 (at $T(\mathrm{~K}))$.

$ \therefore \quad K_C=\frac{\left[\mathrm{NH}_3\right]^{2 / 3}}{\left[\mathrm{~N}_2\right]^{1 / 3}\left[\mathrm{H}_2\right]}=50 $

Thus, for reaction,

$ \begin{aligned} 2 \mathrm{NH}_3(g) & \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \\ K_{C_1} & =\frac{\left[\mathrm{H}_2\right]^3\left[\mathrm{~N}_2\right]}{\left[\mathrm{NH}_3\right]^2} \\ K_{C_1} & =\left[\frac{1}{K_C}\right]^3=\left(\frac{1}{50}\right)^3 \\ & =8 \times 10^{-6} \end{aligned} $

Note ' $A$ ' will be formed in same amount as ' $Z$ '.

$ \begin{aligned} & \therefore \text { Equilibrium constant, } K_C=\frac{[Z][A]}{[X][Y]} \\ & \quad=\frac{0.5 \times 0.5}{(1-0.5)(1-0.5)}=1 \end{aligned} $

At 750 K and 10 atm , the equilibrium constant of reaction $2 A(g) \rightleftharpoons B(g)+C(g)$ is 3.52 . It will be same at 750 K and 7.04 atm pressure as equilibrium constant doesn't depend on pressure, catalyst and concentration. It just depends on absolute temperature.

$ \begin{aligned} & \text { } 4 \mathrm{NH}_3+5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g) \begin{aligned} & +6 \mathrm{H}_2 \mathrm{O}(g) \end{aligned} \\ & \begin{aligned} K_p & =\frac{\left(p_{\mathrm{NO}}\right)^4\left(p_{\mathrm{H}_2 \mathrm{O}}\right)^6}{\left(p_{\mathrm{O}_2}\right)^5\left(p_{\mathrm{NH}_3}\right)^4} \Rightarrow \frac{(0.5)^4(0.5)^6}{(0.5)^5(0.5)^4} \\ & =0.5 \mathrm{~atm} \end{aligned} \end{aligned} $

Formation of ammonia is given by

$ \text {} \begin{aligned} Now, \,\, {\left[\mathrm{N}_2\right] } & =\frac{n_{\mathrm{N}_2}}{V}=\frac{(1-x)}{V} \\ {\left[\mathrm{H}_2\right] } & =\frac{n_{\mathrm{H}_2}}{V}=\frac{(3-3 x)}{V} \\ {\left[\mathrm{NH}_3\right] } & =\frac{n_{\mathrm{NH}_3}}{V}=\frac{2 x}{V} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} K_C & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{\left(\frac{2 x}{V}\right)^2}{\left(\frac{1-x}{V}\right)\left(\frac{3-3 x}{V}\right)^3} \\ & =\frac{4 x^2}{V^2} \times \frac{V^4}{(1-x)(3-3 x)^3}=\frac{4 x^2 V^2}{27(1-x)^4} . \end{aligned} \end{aligned} $

The given reaction is

$ 2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3 $

$k_p$ can be calculated as

$ k_p=\frac{\left[p_{\mathrm{SO}_3}\right]^2}{\left[p_{\mathrm{SO}_2}\right]^2\left[p_{\mathrm{O}_2}\right]} $

[ $p_{\mathrm{SO}_3}=$ partial pressure of $\mathrm{SO}_3$,

$p_{\mathrm{SO}_2}=$ partial pressure of $\mathrm{SO}_2$,

$p_{\mathrm{O}_2}=$ partial pressure of $\mathrm{O}_2$ ]

As, $\quad\left[p_{\mathrm{SO}_3}\right]=\left[p_{\mathrm{SO}_2}\right]$

$ \begin{aligned}So,\,\,\,\, 5 & =\frac{1}{\left[p_{\mathrm{O}_2}\right]} \\ {\left[p_{\mathrm{O}_2}\right] } & =\frac{1}{5}=0.2 \end{aligned} $

So, partial pressure of oxygen will be 0.2 atm .

On increasing pressure, equilibrium shifts towards lesser number of gaseous moles. It means gaseous moles on product are less than gaseous moles on reactant side, i.e. $\Delta n_g<0$.

(i) $X_2(g)+3 Y_2(g) \rightleftharpoons 2 X Y_3(g) \Delta n_g=2-4=-2$;

(ii) $X_2(g)+Y_2(g) \rightleftharpoons 2 X Y(g)\Delta n_g=2-2=0$;

(iii) $X_2(g)+Z_2(g) \rightleftharpoons 2 X Z(g)$; \Delta n_g=2-2=0

(iv) $X_2(g)+Y_4(g) \rightleftharpoons 2 X Y_2(g)\Delta n_g=2-2=0 $;

$\Delta n_g<0$ only for reaction (i) hence, position of equilibrium shifts towards the products if total pressure is increased.

$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$

$ K_p=K_C(R T)^{\Delta n_g} $

$\Delta n_g=$ moles of gaseous products moles of gaseous reactants.

$ \begin{aligned} \Delta n_g & =n_p(g)-n_r(g)=2-(1+3)=-2 \\ K_p & =K_C(R T)^{-2} \Rightarrow \frac{K_p}{K_C}=\frac{1}{(R T)^2} \end{aligned} $

$ \begin{aligned} &\text {For the reaction at equilibrium, }\\ &\begin{aligned} & A(s) \rightleftharpoons B(s)+C(g) \\ & K=\frac{[B][C]}{[A]} \end{aligned} \end{aligned} $

The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough amount is present to reach equilibrium. Therefore, the expression is written as, $K_C=[C] K_p=p C(g)$, where $p C(g)$ is the partial pressure of $C$.

$\Delta H$ is dependent on temperature, if the reaction is exothermic, and the temperature is increased, the reaction enthalpy becomes more negative (more heat energy is released). If the reaction is endothermic, and the temperature is increased, the reaction enthalpy becomes more positive (more heat energy is absorbed).

$\Delta H$ is independent of catalyst addition. The catalyst only decreases the activation energy of a reaction, thereby increases

the reaction rate but does not affect the enthalpy of reaction.

So, statements (A), (B) and (D) are only 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} $