Exercises — Coupling reactions — driving unfavorable reactions

WHAT we do: just add the two arrows on the energy number line. $\Delta G_{\text{total}}=\Delta G_1+\Delta G_2=(+25)+(-40)=-15\text{ kJ}.$ WHY: $\Delta G$ is a state function, so when the intermediate cancels, only the endpoints survive and the changes add. Since $-15<0$, the coupled process is spontaneous. ✓

They multiply: $K_{\text{total}}=K_1\times K_2$. WHY: $\Delta G^{\circ }=-RT\ln K$, so an additive $\Delta G^{\circ }$ becomes a $\ln (K_1{K}_2)=\ln K_1+\ln K_2$ — sum of logs equals log of a product.

Only (a). The intermediate $B$ made in the first is consumed by the second — a real molecular link. In (b) nothing is shared, so there is no pathway connecting them; adding their $\Delta G$ would be meaningless arithmetic.

$\Delta G_{\text{net}}^{\circ }=(-43.0)+(+30.5)=-12.5\text{ kJ/mol}.$ The favorable magnitude $43.0>30.5$, so the surplus drives ATP synthesis. Spontaneous. ✓ (This is exactly how muscle rapidly regenerates ATP during a sprint.)

$K_{\text{total}}=10^{-5}\times 10^9=10^4.$ Since $10^4\gg 1$, the net reaction strongly favors products. A tiny $K_1$ was rescued by a huge $K_2$.

WHY this tool: we use $\Delta G^{\circ }=-RT\ln K$ because it is the bridge between the energy number and the equilibrium constant. $\ln K=-\frac{\Delta G^{\circ }}{RT}=-\frac{-22800}{8.314\times 298}=\frac{22800}{2477.6}=\mathrm{9.203.}$ $K_{\text{total}}=e^{9.203}\approx 9.94\times 10^3\approx 1.0\times 10^4.$

We need $\Delta G_{\text{total}}<0$: $318+\Delta G_2<0⟹\Delta G_2<-318\text{ kJ}⟹|\Delta G_2|>318\text{ kJ}.$ So carbon's oxidation must release more than 318 kJ. On an Ellingham diagram this is the temperature where the carbon line drops below the ZnO line — the crossover point. The shared intermediate is $O_2$: ZnO releases it, carbon grabs it.

No. Coupling requires a genuine shared molecule so the two become one pathway. With no link, reaction 1 still has its own $\Delta G_1=+20>0$ and does not run. The "sum" $+5$ describes no real process. Coupling is chemistry, not arithmetic.

WHY this tool: $\Delta G=\Delta H-T\Delta S$ shows when an unfavorable reaction flips sign, telling us the temperature range where coupling is actually required. Set $\Delta G=0$ to find the switch-over: $T=\frac{\Delta H}{\Delta S}=\frac{30000\text{ J}}{150{\text{ J K}}^{-1}}=200\text{ K}.$

• Below $200\text{ K}$: $\Delta G>0$ → unfavorable → needs coupling.
• Above $200\text{ K}$: $\Delta G<0$ → favorable on its own.

(a) $\Delta G_{\text{net}}^{\circ }=+13.8+(-30.5)=-16.7\text{ kJ/mol}.$ (b) Convert to joules: $-16700\text{ J/mol}$. $\ln K=-\frac{-16700}{8.314\times 310}=\frac{16700}{2577.3}=6.479,$ $K_{\text{net}}=e^{6.479}\approx 6.5\times 10^2\approx 650.$ So coupling turns a barely-happening step into one with $K\approx 650$ — glucose is trapped inside the cell as Glucose-6-P.

Need $55+n(-30.5)<0⟹n>\frac{55}{30.5}=1.80$. So $n=2$ ATP. $\Delta G_{\text{net}}^{\circ }=55+2(-30.5)=55-61=-6\text{ kJ/mol}<0.✓$ One ATP ($55-30.5=+24.5$) is not enough; two are required.

(a) $K_{\text{total}}=(2\times 10^{-6})(5\times 10^4)=10\times 10^{-2}=\mathrm{0.10.}$ Note $K_{\text{total}}=0.10<1$: still net unfavorable! The huge $K_2$ was not quite enough to overcome the tiny $K_1$. (b) Hess's Law: $\Delta G$ of a combined path is the sum of step $\Delta G$'s (state function). Since each step's $\Delta G^{\circ }=-RT\ln K$, adding them gives $-RT\ln K_{\text{tot}}=-RT(\ln K_1+\ln K_2)=-RT\ln (K_1{K}_2)$, so the $K$'s multiply.

First combine the net reaction $\text{MO}+C\to \text{M}+CO$: $\Delta H_{\text{net}}=290+(-110)=+180\text{ kJ}=180000\text{ J},$ $\Delta S_{\text{net}}=80+90=+170{\text{ J K}}^{-1}.$ Spontaneous when $\Delta G_{\text{net}}=\Delta H_{\text{net}}-T\Delta S_{\text{net}}<0$: $T>\frac{\Delta H_{\text{net}}}{\Delta S_{\text{net}}}=\frac{180000}{170}=1058.8\text{ K}.$ So above $\approx 1059\text{ K}$ the reduction runs — the carbon line has crossed below the MO line on the Ellingham Diagram.

(a) $\Delta G_{\text{net}}^{\circ }=18+12+(-45)=-15\text{ kJ}<0.$ Overall spontaneous. ✓ (b) Step 3 ($-45\text{ kJ}$) is the favorable "falling weight" that drags the two uphill steps (total $+30\text{ kJ}$) through. Surplus $=45-30=15\text{ kJ}$, matching the net.

(a) $\Delta G_{\text{net}}^{\circ }=-30.5+25.0=-5.5\text{ kJ/mol}=-5500\text{ J/mol}.$ $\ln K=\frac{5500}{8.314\times 310}=\frac{5500}{2577.3}=2.134,K_{\text{net}}=e^{2.134}\approx \mathrm{8.45.}$ (b) With the real (non-standard) hydrolysis value: $\Delta G_{\text{net}}=-50+25.0=-25.0\text{ kJ/mol}.$ Cells maintain a large ATP:ADP ratio far from equilibrium, so the actual $\Delta G$ of hydrolysis is much more negative than the standard $\Delta G^{\circ }$. This extra push makes coupled syntheses run strongly forward — the cell deliberately stays far from equilibrium.