Question

Consider the following reaction approaching equilibrium at \(27^\circ \text{C}\) and 1 atm pressure:
\[\text{A + B} \rightleftharpoons \text{C + D}\]
\[K_f = 10^3, \quad K_r = 10^2\]
The standard Gibbs energy change (\(\Delta_r G^\circ\)) at \(27^\circ \text{C}\) is — kJ mol\(^{-1}\) (Nearest integer).

Hint:

For Gibbs free energy calculations:
       Use the relationship \(\Delta_r G^\circ = -RT \ln K\).
       Ensure the equilibrium constant is correctly derived from forward and reverse reaction constants.

Answer 1

Correct Answer: 6

1. Relationship Between \(\Delta_r G^\circ\) and Equilibrium Constant:
The standard Gibbs free energy change is related to the equilibrium constant (\(K\)) as:
\[\Delta_r G^\circ = -RT \ln K.\]
2. Calculate the Overall Equilibrium Constant (\(K\)):
The equilibrium constant for the reaction is:
\[K = \frac{K_f}{K_r} = \frac{10^3}{10^2} = 10.\]
3.Substitute Values:
Since \(K = 10\), \(\ln K = \ln 10\). Therefore:
\[\Delta_r G^\circ = -RT \ln K = -(8.3 \times 10^{-3}~\text{kJ mol}^{-1}~\text{K}^{-1}) \times 300~\text{K} \times 2.3.\]
\[\Delta_r G^\circ = -6~\text{kJ mol}^{-1}.\]
4. Result:
The standard Gibbs energy change is \(6~\text{kJ mol}^{-1}\).