Question
Calculate the standard free energy change for the ionization HF(aq) → H+ (aq) + F– (aq) from the following data.
HF(aq) → HF(g): ΔG° = 23.9 kJ
HF(g) → H(g) + F(g): ΔG° = 555.1 kJ
H(g) → H+ (g) + e: ΔG° = 1320.2 kJ
F(g) + e → F– (g): ΔG° = –347.5 kJ
H+ (g) + F– (g) $\overset{\quad aq.\quad}{\rightarrow}$H+ (aq) + F– (aq): ΔG° = –1513.6 kJ
Options
Solution
$\Delta_{r}H^{o} = \left\lbrack \Delta_{f}H_{CO_{2}(g)} + 2 \times \Delta_{f}H_{H_{2}O(l)} \right\rbrack - \left\lbrack \Delta_{f}H_{CH_{4}(g)} + 2 \times \Delta_{f}H_{O_{2}(g)} \right\rbrack$
$= \left\lbrack ( - 393.5) + 2( - 286) \right\rbrack - \left\lbrack ( - 74.5) + 2x0 \right\rbrack = - 891\text{ kJ}$
$\Delta_{r}S^{o} = \left\lbrack S_{CO_{2}(g)} + 2 \times S_{H_{2}O(l)} \right\rbrack - \left\lbrack \Delta_{f}S_{CH_{4}(g)} + 2 \times S_{O_{2}(l)} \right\rbrack $$${= \lbrack 216 + 2 \times 70\rbrack - \lbrack 186 + 2 \times 205\rbrack = - 240\text{ J/K} }{\text{Now, }\Delta G^{o} = \Delta H^{o} - T.\Delta S^{o} = ( - 891) - 300 \times \left( \frac{- 240}{1000} \right) = - 819\text{ kJ}}$$
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