Question

For the reaction given below at 25°C: \[ \text{A}_2 \rightleftharpoons 2\text{A} \] Find \( K_p \). Given \( \Delta G^\circ_A = -50.384 \, \text{kJ/mol} \) and \( \Delta G^\circ_{A_2} = -100 \, \text{kJ/mol} \).

• A. 0.43
• B. 0.23
• C. 0.31
• D. 0.53

Hint:

The equilibrium constant can be calculated from the standard Gibbs free energy change using the relation \( \Delta G^\circ = -RT \ln K \).

Answer 1

The Correct Option is C

To find \( K_p \) for the reaction:

\[ \text{A}_2 \rightleftharpoons 2\text{A} \]

we use the relationship between the standard Gibbs free energy change (\( \Delta G^\circ \)) and the equilibrium constant (\( K \)), which is given by:

\(\Delta G^\circ = -RT \ln K_p\)

where:

• \( R \) is the universal gas constant, \( 8.314 \, \text{J/mol K} \).
• \( T \) is the temperature in Kelvin, which is 298.15 K at 25°C.

The reaction involves the dissociation of \(\text{A}_2\) into 2\(\text{A}\). Therefore, the change in standard Gibbs free energy for the reaction can be calculated as:

\(\Delta G^\circ = 2\Delta G^\circ_A - \Delta G^\circ_{A_2}\)

Substituting the given values:

\(\Delta G^\circ = 2(-50.384) - (-100)\)

\(\Delta G^\circ = -100.768 + 100\)

\(\Delta G^\circ = -0.768 \, \text{kJ/mol}\)

Convert \(\Delta G^\circ\) from kJ/mol to J/mol:

\(\Delta G^\circ = -0.768 \times 1000 = -768 \, \text{J/mol}\)

Now, use the formula to find \( K_p \):

\(-768 = -8.314 \times 298.15 \ln K_p\)

Simplifying this equation:

\(\ln K_p = \frac{-768}{-8.314 \times 298.15}\)

\(\ln K_p = \frac{-768}{-2477.48} \approx 0.3101\)

Solve for \( K_p \) by taking the exponential of both sides:

\(K_p = e^{0.3101} \approx 0.31\)

Thus, the correct answer is 0.31.