Thermodynamics and Thermochemistry 1 Question 19

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

ΔfG[C( graphite )]=0 kJ mol1

ΔfG[C( diamond )]=2.9 kJ 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 2×106 m3 mol1. 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

(2017 Adv.)

Thermodynamics and Thermochemistry 105

[Useful information : 1 J=1 kg m2 s2,

1 Pa=1 kg m1 s2;1bar=105 Pa]

(a) 58001 bar

(b) 1450 bar

(c) 14501 bar

(d) 29001 bar

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Solution:

  1. G=HTS=U+pVTS

dG=dU+pdV+VdpTdSSdT=VdpSdT

dG=Vdp if isothermal process (dT=0)

[dU+pdV=dq=TdS]

ΔG=VΔp

Now taking initial state as standard state

Now (ii)-(i) gives,

GgrGgr=VgrΔp GdGd=VdΔp

(VdVgr)Δp=GdGgr+(GgrGd

At equilibrium, Gd=Ggr

(VgrVd)Δp=GdGgr=2.9×103 J Δp=2.9×1032×106 Pa=292×108 Pa=290002bar p=p0+290002=1+290002=14501bar