GATE 2025 BT Q.65: solution and answer

Official answer: -32.00 to -30.00Source: Official answer key, GATE 2025 BT (IIT Roorkee)

1. Add the two reactions. Glucose 6-phosphate and glucose appear on both sides and cancel, leaving $\text{ATP}+{\text{H}}_2\text{O}\to \text{ADP}+{\text{P}}_i$, which is ATP hydrolysis.

2. Standard free energy changes of reactions that add together are added: $\Delta G_{\text{hyd}}^{\circ }=\Delta G_1^{\circ }+\Delta G_2^{\circ }$. Equivalently, the equilibrium constants multiply: $K_{\text{hyd}}=K_1{K}_2$.

3. Use $\Delta G^{\circ }=-RT\ln K$ with $T=25+273.15=298.15\text{K}$: $RT=8.314\times 298.15\approx 2478.8{\text{J mol}}^{-1}$.

4. $K_{\text{hyd}}=360\times 800=2.88\times 10^5$, so $\ln K_{\text{hyd}}=\ln 2.88+5\ln 10\approx 1.0578+11.5129=12.5707$.

5. $\Delta G_{\text{hyd}}^{\circ }=-2478.8\times 12.5707\approx -31160{\text{J mol}}^{-1}\approx -31.16{\text{kJ mol}}^{-1}$.

6. Check by parts: $\Delta G_1^{\circ }=-2478.8\ln 360=-2478.8\times 5.8861\approx -14.59{\text{kJ mol}}^{-1}$ and $\Delta G_2^{\circ }=-2478.8\ln 800=-2478.8\times 6.6846\approx -16.57{\text{kJ mol}}^{-1}$. Their sum is $-31.16{\text{kJ mol}}^{-1}$, in agreement.