Q.Calculate $\Delta G$ for the reaction at 25 $^0$C $\mathrm{CO(g) + 2\,H_2(g) \rightarrow CH_3OH(g)}$, $\Delta G^0$ = -24.8 kJ mol$^{-1}$ The partial pressures of gases are $P_{CO}$ = 4 bar, $P_{H_2}$ = 2 bar and $P_{CH_3OH}$ = 2 bar

Under nonstandard pressures, $\Delta G = \Delta G^0 + 2.303\,RT \log_{10} Q_p$ with $Q_p = P_{CH_3OH}/(P_{CO} \times P_{H_2}^2) = 2/16 = 0.125$. …

Step 1. $\Delta G = \Delta G^0 + RT \ln Q_p = \Delta G^0 + 2.303\,RT \log_{10} Q_p$.

Step 2. $Q_p = \dfrac{P_{CH_3OH}}{P_{CO} \times P_{H_2}^2} = \dfrac{2}{4 \times 2^2} = \dfrac{2}{4 \times 4} = \dfrac{1}{8} = 0.125$; $\log_{10} 0.125 = -0.903$.

Step 3. $2.303\,RT = 2.303 \times 8.314 \times 10^{-3}\ \mathrm{kJ\ K^{-1}\ mol^{-1}} \times 298\ \mathrm{K} = 5.706$ kJ mol$^{-1}$. …

Form the reaction quotient Qp from the partial pressures (squaring H2's for its coefficient of 2), then evaluate Delta …

• Forgetting to SQUARE the H2 partial pressure for its coefficient of 2.
• Inverting Qp (it is products over reactants). …