MediumChemistryHigh School (11–12)Chemical Thermodynamics › Entropy, Gibbs Energy and SpontaneityNumerical

For N₂(g) + 3H₂(g) → 2NH₃(g), Δ H° = -92.4 kJ/mol and Δ S° = -198.7 J/(K mol).

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For $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ , $\Delta H^\circ = -92.4\ \text{kJ/mol}$ and $\Delta S^\circ = -198.7\ \text{J/(K mol)}$ . Find $\Delta G^\circ$ at $298\ \text{K}$ . Above what temperature does the reaction stop being spontaneous?

Need a reminder? The key ideaShow

Gibbs energy: $G = H - TS$ , and at constant temperature $\Delta G = \Delta H - T\Delta S$ . $\Delta G < 0$ : spontaneous; $\Delta G = 0$ : equilibrium; $\Delta G > 0$ : not spontaneous. When $\Delta H$ and $\Delta S$ have the same sign, $\Delta G$ changes sign at $T = \Delta H/\Delta S$ . Put $\Delta S$ in kJ/(K mol), or $\Delta H$ in J/mol, before combining them.

The worked solution

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For N₂(g) + 3H₂(g) → 2NH₃(g), Δ H° = -92.4 kJ/mol and Δ S° = -198.7 J/(K mol).

The worked solution

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