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

Step 1: Calculate standard Gibbs energy change of reaction. 
\[ \Delta G^\circ = 2\Delta G_f^\circ(A) - \Delta G_f^\circ(A_2) \] \[ = 2(-50.832) - (-100.00) \] \[ = -101.664 + 100 = -1.664 \, \text{kJ mol}^{-1} \] Step 2: Calculate equilibrium constant \(K_p\). 
\[ \Delta G^\circ = -RT \ln K_p \] \[ -1664 = - (8)(300)\ln K_p \] \[ \ln K_p = \frac{1664}{2400} = 0.693 \] \[ K_p = e^{0.693} = 2 \] Step 3: Write expression for \(K_p\) in terms of degree of dissociation. 
Let degree of dissociation \(= \alpha\). 
Initial moles: \[ A_2 = 1, \quad A = 0 \] Equilibrium moles: \[ A_2 = 1 - \alpha, \quad A = 2\alpha \] Total moles: \[ = 1 + \alpha \] Partial pressures: \[ P_{A_2} = \frac{1 - \alpha}{1 + \alpha}, \quad P_A = \frac{2\alpha}{1 + \alpha} \] \[ K_p = \frac{P_A^2}{P_{A_2}} = \frac{(2\alpha)^2}{(1-\alpha)(1+\alpha)} \] \[ K_p = \frac{4\alpha^2}{1 - \alpha^2} \] Step 4: Substitute \(K_p = 2\). 
\[ 2 = \frac{4\alpha^2}{1 - \alpha^2} \] \[ 2 - 2\alpha^2 = 4\alpha^2 \] \[ 6\alpha^2 = 2 \] \[ \alpha^2 = \frac{1}{3} \] \[ \alpha = \sqrt{\frac{1}{3}} \approx 0.577 \] Step 5: Express in the required form. 
\[ \alpha = (x \times 10^{-2})^{1/2} \Rightarrow \alpha^2 = x \times 10^{-2} \] \[ x \times 10^{-2} = 0.333 \Rightarrow x = 33.3 \]  Final Answer: \[ \boxed{33} \]