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 \(x\). \(\text{[R = 8.3 J/molK]}\)
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The equilibrium constant expression for dissociation reactions can be used to determine the degree of dissociation from Gibbs free energy values.
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).
