The standard state Gibbs free energies of formation of C (graphite) and C (diamond) at T…
Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
ΔfG°[C(graphite)] = 0 kJ mol–1;
fG°[C(diamond)] = 2.9 kJ mol–1
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 × 10–6 m3 mol–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 kg m2s–2; 1 Pa = 1 kg m–1s–2;
1 bar = 105 Pa]
ΔfG°[C(graphite)] = 0 kJ mol–1;
fG°[C(diamond)] = 2.9 kJ mol–1
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 × 10–6 m3 mol–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 kg m2s–2; 1 Pa = 1 kg m–1s–2;
1 bar = 105 Pa]
- 29001 bar
- 58001 bar
- 14501 bar
- 1450 bar
Answer
(B) 58001 bar
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