Question

(i) \(X ( g ) \rightleftharpoons Y ( g )+ Z\) (g) \(K _{ p 1}=3\)
(ii)\(A ( g ) \rightleftharpoons 2 B ( g )\,\, K _{ p 2}=1\)
If the degree of dissociation and initial concentration of both the reactants \(X ( g )\) and \(A ( g )\) are equal, then the ratio of the total pressure at equilibrium \(\left(\frac{p_1}{p_2}\right)\) is equal to \(x : 1\). The value of \(x\) is _________ (Nearest integer)

Hint:

The degree of dissociation (α) can be calculated by relating the equilibrium constant to the initial and equilibrium concentrations.

Answer 1

Correct Answer: 12

\( \begin{array}{ccc} & \text{x(g)} \rightleftharpoons \text{y(g)} + \text{z(g)} & k_{p_1} = 3 \\ \text{Initial moles} & n & - & - \\ \text{at equilibrium} & n-\alpha n & \alpha n & \alpha n \\ \end{array} \)

\(k_{p_1} = \frac{\left(\frac{\alpha}{1+\alpha} \times p_1\right)^2}{\frac{1-\alpha}{1+\alpha}p_1} \)

\(3 = \frac{\alpha^2 \times p_1}{1-\alpha^2}\)

\( \begin{array}{ccc} & \text{A(g)} \rightleftharpoons 2\text{B(g)} & k_{p_2} = 1 \\ \text{Initial mole} & n & - \\ \text{at equilibrium} & x-\alpha n & 2 \alpha n & p_{\text{total}} = p_2\\ \end{array} \)

\(k_{p_2} = \frac{\left(\frac{2\alpha}{1+\alpha} \times p_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times p_2} \)

\(1 = \frac{4\alpha^2 \times p_2}{1-\alpha^2} \)

\( \frac{k_{p_1}}{k_{p_2}} = \frac{p_1}{4p_2} \)

\( \frac{3}{1} = \frac{p_1}{4p_2} \)

\( \therefore p_1 : p_2 = 12 : 1 \)

\( x = 12 \)