Question

Calculate $\Delta_r G$ for the reaction at $27^{\circ} C$
$H _2( g )+2 Ag ^{+}( aq ) \rightleftharpoons 2 A ( s )+2 H ^{+} \text {(aq) }$
Given : $P_{ H _2}=0.5$ bar; $\left[ Ag ^{+}\right]=10^{-5} M$;
$\left[ H ^{+}\right]=10^{-3} M ; \Delta G^{\circ}\left[ Ag ^{+}( aq )\right]=77.1 \,kJ / mol$

• A. $-154.2 \,kJ / mol$
• B. $-178.9 \,kJ / mol$
• C. $-129.5 \,kJ / mol$
• D. None of these

Answer 1

Answer: (C)

$\Delta_r G^{\circ}=\Delta G_f$ (Products) $-\Delta G_f$ (Reactants)
$=0-77.1 \times 2=-154.2 \,kJ / mol$
$Q=\frac{\left[ H ^{+}\right]^2}{R_{ H _2} \cdot\left[ Ag ^{+}\right]^2}=\frac{10^{-6}}{0.5 \times\left(10^{-10}\right)}=2 \times 10^4$
$\Delta_n G=\Delta_r G^{\circ}+R T \ln Q$
$\Delta_r G=-154.2+\frac{8.314 \times 300 \ln \left(2 \times 10^4\right)}{1000}=-129.5\, kJ / mol$