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).
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.
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 \(6~\text{kJ mol}^{-1}\).
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
Match List-I with List-II: List-I