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Question

# The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 K areΔfG∘C(graphite)=0kJ mol−1ΔfG∘C(diamond)=2.9kJ mol−1The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite C(graphite) to diamond C(diamond) reduces its volume by 2×10−6m3mol−1. If C(graphite) is converted to C(diamond) isothermally at T=298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is?[Useful information: 1 J=1 kgm2s−2;1 Pa=1 kg m−1s−2;1 bar=105Pa]

A
14501 bar
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B
29001 bar
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C
1450 bar
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D
58001 bar
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Solution

## The correct option is B 14501 barCgraphite→CdiamandΔGor=ΔGodiamond−ΔGographite2.9KJmol−1−0KJmol−1=2.9KJmol−1ΔGor=ΔHor−TΔSorΔSor≈0TΔSor≈0ΔHor=ΔEor+Δ(PV)Δor=0 because of isothermal process ΔT=0ΔHor=2.9×103=Δ(PV)At equilibrium pressure is constant2.9×103Jmol−1=P(V2−V1)P=2.9×103Jmol−12×10−6m3mol−1 =1450×106 pascal =14500barTotal pressure = equilibrium pressure + initial pressure =14501barHence, the correct option is A.

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