Questions: Question 1 2 pts Calculate ΔG° for the following reaction of ammonia with fluorine. 2 NH3(g) + 5 F2(g) → N2 F4(g) + 6 HF(g) Substance: NH3(g) F2(g) N2 F4(g) HF(g) ΔG° f(kJ / mol): -16.4 0 79.9 -275.4 179.1 kJ -179.1 kJ 1539.7 kJ -1539.7 kJ None of these choices are correct.

Solution Steps

The balanced chemical equation for the reaction is: \[ 2 \mathrm{NH}_{3}(\mathrm{~g}) + 5 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{~F}_{4}(\mathrm{~g}) + 6 \mathrm{HF}(\mathrm{g}) \]

From the given data: \[ \Delta \mathrm{G}^{\circ} \mathrm{f}(\mathrm{kJ} / \mathrm{mol}): \begin{array}{lllll} \mathrm{NH}_{3}(\mathrm{~g}) & \mathrm{F}_{2}(\mathrm{~g}) & \mathrm{N}_{2} \mathrm{~F}_{4}(\mathrm{~g}) & \mathrm{HF}(\mathrm{g}) \\ -16.4 & 0 & 79.9 & -275.4 \end{array} \]

\[ \Delta \mathrm{G}^{\circ}_{\text{reactants}} = 2 \times \Delta \mathrm{G}^{\circ}_{\mathrm{f}, \mathrm{NH}_{3}} + 5 \times \Delta \mathrm{G}^{\circ}_{\mathrm{f}, \mathrm{F}_{2}} \] \[ \Delta \mathrm{G}^{\circ}_{\text{reactants}} = 2 \times (-16.4) + 5 \times 0 = -32.8 \, \mathrm{kJ} \]

\[ \Delta \mathrm{G}^{\circ}_{\text{products}} = \Delta \mathrm{G}^{\circ}_{\mathrm{f}, \mathrm{N}_{2} \mathrm{~F}_{4}} + 6 \times \Delta \mathrm{G}^{\circ}_{\mathrm{f}, \mathrm{HF}} \] \[ \Delta \mathrm{G}^{\circ}_{\text{products}} = 79.9 + 6 \times (-275.4) = 79.9 - 1652.4 = -1572.5 \, \mathrm{kJ} \]

\[ \Delta \mathrm{G}^{\circ}_{\text{reaction}} = \Delta \mathrm{G}^{\circ}_{\text{products}} - \Delta \mathrm{G}^{\circ}_{\text{reactants}} \] \[ \Delta \mathrm{G}^{\circ}_{\text{reaction}} = -1572.5 - (-32.8) = -1572.5 + 32.8 = -1539.7 \, \mathrm{kJ} \]

Final Answer