Calculate the pressure at which graphite & diamond will be at equilibrium at a temperature of 300 K. [Given : ΔfGºdiamond at 300 K = 3.00 kJ, dgraphite = 2.4 g/mL, ddiamond = 3.6 g/mL] Express your answer in multiples of 109 Pa.

Engineering

Chemistry

Gibbs Free Energy Calculations and Third Law

Question

Calculate the pressure at which graphite & diamond will be at equilibrium at a temperature of 300 K.
[Given : Δ_fGº_diamond at 300 K = 3.00 kJ, d_graphite = 2.4 g/mL, d_diamond = 3.6 g/mL]
Express your answer in multiples of 10⁹ Pa.

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C(s) → C(s)
graphit diamond
dU = VdP – SdT
$\int d (ΔG) = ΔV \int dP$
ΔG₂ – ΔG° = ΔV (P₂ – 10⁵)
3.02 × 10³ = $(\frac{12}{3 .5} - \frac{12}{2 .3})$ × 10^–6 (P₂ – 10^–5)
3.02 × 10⁹ = 12 $(\frac{2 .3 - 3 .5}{2 .3 \times 3 .5})$ (P₂ – 10^–5)
$\frac{2 .3 \times 3 .5 \times 3 .02 \times 10^{9}}{12 \times 1 .2} = (P_{2} - 10^{5})$

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