For the reaction $A \rightleftharpoons B$ with standard-state convention in pressure units, $K_p = \frac{p_B}{p_A}$ because stoichiometric coefficients are 1:1.

Given: throughout the experiment, $p_A = 1\ \text{bar}$ Hence at equilibrium, $K_p = p_B$ numerically (when pressure is read in bar relative to standard state 1 bar).

From the graph of partial pressure of $B$ vs time:

• before heating, at $1000\,\text{K}$, the equilibrium partial pressure of $B$ is 2 bar,
• after heating to $2000\,\text{K}$ and re-establishing equilibrium, the equilibrium partial pressure of $B$ is 4 bar.

Therefore, $K_{p,1000} = 2, \qquad K_{p,2000} = 4$

Use the relation between standard Gibbs energy and equilibrium constant $\Delta G^\circ = -RT\ln K$

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

Now compute the ratio: $\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = \frac{-1000R\ln 2}{-2000R\ln 4}$

Since $\ln 4 = 2\ln 2$ we get $\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = \frac{1000\ln 2}{2000\cdot 2\ln 2} = \frac{1}{4}$

Hence, $\boxed{\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = 0.25}$