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

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Q 2958312204.     The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are

`Delta_fG° ` [C(graphite)] = 0 kJ `mol^-1`

`Delta_f G°` [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 xx 10^-6 m^3` `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 `m^2s^-2`; 1 Pa = 1 kg `m^-1 s^-2`; 1 bar = `10^5` Pa]

JEE 2017 Advanced Paper 2
A

14501 bar

B

29001 bar

C

1450 bar

D

58001 bar

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Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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