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(C) At equilibrium,the change in Gibbs free energy is given by $\Delta G = \Delta G^0 + P \Delta V = 0$.Therefore,$P \Delta V = -\Delta G^0$.Here,$\De

Vedclass · Accepted Answer

$14500$

Vedclass · Answer

(C) At equilibrium,the change in Gibbs free energy is given by $\Delta G = \Delta G^0 + P \Delta V = 0$.Therefore,$P \Delta V = -\Delta G^0$.Here,$\Delta G^0 = \Delta_f G^0[C(	ext{diamond})] - \Delta_f G^0[C(	ext{graphite})] = 2.9 \ kJ \ mol^{-1} - 0 = 2.9 	imes 10^3 \ J \ mol^{-1}$.The change in volume is $\Delta V = V_{	ext{diamond}} - V_{	ext{graphite}} = -2 	imes 10^{-6} \ m^3 \ mol^{-1}$.Substituting these values into $P \Delta V = -\Delta G^0$:$P 	imes (-2 	imes 10^{-6} \ m^3 \ mol^{-1}) = -2.9 	imes 10^3 \ J \ mol^{-1}$.$P = \frac{2.9 	imes 10^3}{2 	imes 10^{-6}} \ Pa = 1.45 	imes 10^9 \ Pa$.Since $1 \ bar = 10^5 \ Pa$,$P = \frac{1.45 	imes 10^9}{10^5} \ bar = 1.45 	imes 10^4 \ bar = 14500 \ bar$.

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