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At 298 K a reaction has Δ G° = -17.1 kJ/mol.

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At $298\ \text{K}$ a reaction has $\Delta G^\circ = -17.1\ \text{kJ/mol}$ . Find $\log K$ and $K$ . Are the products or the reactants favoured?

Need a reminder? The key ideaShow

At equilibrium $\Delta G = 0$ and $\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$ , with $R = 8.314\ \text{J/(K mol)}$ and $\Delta G^\circ$ in J/mol. A negative $\Delta G^\circ$ gives $K > 1$ , a positive one $K < 1$ . At another temperature, find $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ first.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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← Four reactions have these signs. (a) Δ H < 0, Δ S > 0.A reaction has K = 2.5 × 10⁻³ at 500 K. Find Δ G° at this temperature. →

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At 298 K a reaction has Δ G° = -17.1 kJ/mol.

The worked solution

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