Question

For the reaction  
\(\text{A}_2(\text{g}) \rightleftharpoons 2\text{A}(\text{g})\)
\(\Delta G^o(\text{A}_2) = -100 \, \text{kJ/mol}\)
\(\Delta G^o(\text{A}) = -50.8625 \, \text{kJ/mol}\)
At 300 K and 1 atm pressure, the degree of dissociation of A$_2$ gas at equilibrium is \(x \times 10^{-2}\). Find \(x\).
\(\text{[R = 8.3 J/molK]}\)
 

Hint:

The equilibrium constant expression for dissociation reactions can be used to determine the degree of dissociation from Gibbs free energy values.

Answer 1

Correct Answer: 58

Step 1: Using the formula for Gibbs Free Energy.
For the reaction: \[ \Delta G = \Delta G^o + RT \ln K \] Where \(K\) is the equilibrium constant. Using the given data: \[ \Delta G^o = -100 \, \text{kJ/mol} = -100,000 \, \text{J/mol} \] \[ \Delta G = -50.8625 \, \text{kJ/mol} = -50,862.5 \, \text{J/mol} \] \[ R = 8.3 \, \text{J/mol K}, \, T = 300 \, \text{K} \]
Step 2: Calculating K.
Substituting into the Gibbs free energy equation: \[ -50,862.5 = -100,000 + (8.3)(300) \ln K \] \[ \ln K = \frac{-50,862.5 + 100,000}{8.3 \times 300} = 0.693 \] Thus, \[ K = e^{0.693} = 2 \]
Step 3: Determining the degree of dissociation.
For the dissociation reaction: \[ A_2(g) \rightleftharpoons 2A(g) \] The equilibrium constant \( K \) can also be expressed as: \[ K = \frac{[A]^2}{[A_2]} \] Let the initial concentration of A$_2$ be 1 mole (for simplicity), and the concentration of A at equilibrium be \(2x\). The degree of dissociation is given by: \[ K = \frac{(2x)^2}{1 - x} \] Substituting the value of \(K\): \[ 2 = \frac{4x^2}{1 - x} \] Solving for \(x\), we get: \[ x = \frac{1}{\sqrt{3}} \quad \text{and} \quad x \approx 0.577 \]
Step 4: Conclusion.
Thus, the degree of dissociation is \( x \times 10^{-2} = 57.736 \times 10^{-2} \), or \( x = 5.7736 \times 10^{-2} \). The correct value of \(x\) is approximately \( 5.77 \times 10^{-2} \), which corresponds to the correct answer (58).