JEE Advanced 2017 Chemistry question
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
Δ_C(graphite)]=0\ kJ mol^{-1}
Δ_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 m mol. 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 s; 1 Pa = 1 kg m s; 1 bar = Pa]
Answer: (A)
Solution
At constant temperature . For graphite → diamond, J mol at 1 bar and $Δ
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