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The Gibbs free energy of formation at 298 K is ΔG°f = 0 kJ mol⁻¹ for graphite and ΔG°f = 2.9 kJ mol⁻¹ for diamond. Under standard conditions (pure substances at 1 bar), the transformation of graphite to diamond reduces the molar volume by 2 × 10⁻⁶ m³ mol⁻¹. If this conversion takes place isothermally at 298 K, what pressure is required for graphite and diamond to be in equilibrium?

1. 29001 bar
2. 1450 bar
3. 14501 bar
4. 58001 bar

Correct answer: 14501 bar

Solution

The correct pressure is obtained by using the equation ΔG° = ΔP(ΔV), where ΔG° is the Gibbs free energy of formation, ΔP is the change in pressure, and ΔV is the change in volume. Given that ΔG°f for diamond is 2.9 kJ mol⁻¹ and the transformation reduces the molar volume by 2 × 10⁻⁶ m³ mol⁻¹, we can calculate the required pressure.

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