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Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are

ΔfG[C(graphite)]=0 kJmol1
ΔfG[C(diamond)]=2.9 kJmol1

The 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 2×106m3mol1. 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 kg m2s2; 1 Pa=1 kg m1s2; 1 bar=105 Pa]

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

The correct option is A 14501 bar
dG=VdPSdT
at constant temperature (298K),
SdT=0
dG=VdP
P1 dG= P1 VdP
So,
GG0= V (P-1) [Since solids are involved therefore V is almost constant]
So, we have
ΔrG=[G0diamond+Vd(P1)][G0graphite+Vg(P1)]
0=2.9×103+(P1)105(2×)106
Pi = 1 bar
P=14501 bar

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