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
\(\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\) [C(graphite)] = 0 kJ mol–1
\(\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\)[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–1 s–2; 1 bar = 105 Pa] (2017)
\(\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\) [C(graphite)] = 0 kJ mol–1
\(\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\)[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–1 s–2; 1 bar = 105 Pa] (2017)
- 58001 bar
- 1450 bar
- 14501 bar
- 29001 bar
Answer
(C) 14501 bar
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