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Question

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

ΔfGC(graphite)=0kJ mol1
ΔfGC(diamond)=2.9kJ mol1
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 kgm2s2;1 Pa=1 kg m1s2;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 bar
CgraphiteCdiamand

ΔGor=ΔGodiamondΔGographite

2.9KJmol10KJmol1=2.9KJmol1

ΔGor=ΔHorTΔSor

ΔSor0

TΔSor0

ΔHor=ΔEor+Δ(PV)

Δor=0 because of isothermal process ΔT=0

ΔHor=2.9×103=Δ(PV)

At equilibrium pressure is constant

2.9×103Jmol1=P(V2V1)

P=2.9×103Jmol12×106m3mol1

=1450×106 pascal

=14500bar

Total pressure = equilibrium pressure + initial pressure
=14501bar

Hence, the correct option is A.

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