Consider the reaction A $$\rightleftharpoons$$ B at 1000 K. At time t’, the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at 1 bar. Given below is the plot of the partial pressure of B with time.
What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?

At every moment the partial pressure of A is kept constant at $$P_A = 1\;\text{bar}$$. For the reaction $$A \rightleftharpoons B$$ the equilibrium constant (in terms of pressure) is therefore

$$K_p = \frac{P_B}{P_A} = P_B$$ because $$P_A = 1\;\text{bar}$$.

From the graph (region before time $$t'$$, i.e. at 1000 K) the equilibrium value of the partial pressure of B is read as

$$P_{B,\,1000\ \text{K}} = 2\;\text{bar}$$.

Hence, at 1000 K,

$$K_{p,\,1000\ \text{K}} = 2 \qquad -(1)$$

Immediately after the temperature is raised to 2000 K and the system is allowed to re-equilibrate, the graph shows the new steady value

$$P_{B,\,2000\ \text{K}} = 4\;\text{bar}$$.

Therefore, at 2000 K,

$$K_{p,\,2000\ \text{K}} = 4 \qquad -(2)$$

The standard Gibbs free energy change is related to the equilibrium constant by

$$\Delta G^\circ = -RT \ln K_p$$.

Taking the ratio of the standard Gibbs energies at the two temperatures:

$$\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = \frac{-R(1000)\ln K_{p,\,1000}}{-R(2000)\ln K_{p,\,2000}} = \frac{1000\,\ln 2}{2000\,\ln 4}$$

$$= \frac{1}{2}\cdot\frac{\ln 2}{2\ln 2} = \frac{1}{2}\cdot\frac{1}{2} = 0.25$$

Thus, the required ratio of the standard Gibbs energies is 0.25.