Question

Dissociation of a gas \( A_2 \) takes place according to the following chemical reaction. At equilibrium, the total pressure is \( 1 \, \text{bar} \) at \( 300 \, \text{K} \).

\[ A_2(g) \rightleftharpoons 2A(g) \]
The standard Gibbs energy of formation of the involved substances is given below:

Substance	\( \Delta G_f^\circ \) (kJ mol\(^{-1}\))
\( A_2 \)	\(-100.00\)
\( A \)	\(-50.832\)

 

The degree of dissociation of \( A_2(g) \) is given by
\[ (x \times 10^{-2})^{1/2} \]
where \( x = \) ________ (Nearest integer).

[Given: \( R = 8 \, \text{J mol}^{-1}\text{K}^{-1} \), \( \log 2 = 0.3010 \), \( \log 3 = 0.48 \). Assume degree of dissociation is not negligible.]
 

Hint:

Always calculate \( \Delta G^\circ \) first to find \(K_p\), then relate \(K_p\) with degree of dissociation using partial pressures when dissociation is not negligible.

Answer 1

Correct Answer: 67

The dissociation of \( A_2(g) \) into \( 2A(g) \) can be analyzed to find the degree of dissociation. At equilibrium, let's denote the initial moles of \( A_2 \) as 1. If the degree of dissociation is \( \alpha \), we have:

• Initial moles of \( A_2 = 1 \)
• Moles of \( A_2 \) at equilibrium = \( 1-\alpha \)
• Moles of \( A \) at equilibrium = \( 2\alpha \)

Total moles at equilibrium = \( 1+\alpha \). The equilibrium constant \( K_p \) is given by:

\( K_p = \frac{(P_A)^2}{P_{A_2}} \) where \( P_i \) are partial pressures.

Using \( P_T = 1 \, \text{bar} \), \( P_{A_2} = (1-\alpha) \frac{P_T}{1+\alpha} \) and \( P_A = \frac{2\alpha P_T}{1+\alpha} \).

\( K_p = \frac{\left(\frac{2\alpha}{1+\alpha}\right)^2 \times P_T^2}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P_T} = \frac{4\alpha^2 P_T}{(1-\alpha)(1+\alpha)} \)

Now, calculate \( \Delta G_\text{rxn}^\circ \) using Gibbs free energy:

\( \Delta G_\text{rxn}^\circ = 2(-50.832) - (-100.00) = -1.664 \, \text{kJ/mol} = -1664 \, \text{J/mol} \).

\( \Delta G = \Delta G_\text{rxn}^\circ + RT \ln K_p \). At equilibrium, \(\Delta G = 0\):

\( 0 = -1664 + 8 \times 300 \ln K_p \)

\( \ln K_p = \frac{1664}{2400} \approx 0.6933 \) \( K_p = e^{0.6933} = 2 \)

Equate to the \( K_p \) expression:

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

\( 8\alpha^2 = 1-\alpha^2 \)

\( 9\alpha^2 = 1 \) thus \(\alpha = \frac{1}{3} \approx 0.33\)

The degree of dissociation \( \alpha = \sqrt{\frac{x \times 10^{-2}}{2}} \).

\( \alpha^2 = \frac{x \times 10^{-2}}{2} = (0.33)^2 = 0.1089 \)

\( x \approx 67 \). Verify: \( 67 \in [67,67] \). The range is satisfied.