JEE Advanced 2017 Paper 2 Chemistry Q3

Chemistry · Free Energy & Chemical Equilibrium · Phase Equilibria & the Phase Rule

Problem

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 KT=298\,\mathrm{K} are ΔfG∘[C(graphite)]=0 kJ mol−1\Delta_fG^{\circ}[\mathrm{C(graphite)}]=0\,\mathrm{kJ\,mol^{-1}} and ΔfG∘[C(diamond)]=2.9 kJ mol−1\Delta_fG^{\circ}[\mathrm{C(diamond)}]=2.9\,\mathrm{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−12\times10^{-6}\,\mathrm{m^3\,mol^{-1}}. If C(graphite) is converted to C(diamond) isothermally at T=298 KT=298\,\mathrm{K}, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: 1 J=1 kg m2 s−21\,\mathrm{J}=1\,\mathrm{kg\,m^2\,s^{-2}}; 1 Pa=1 kg m−1 s−21\,\mathrm{Pa}=1\,\mathrm{kg\,m^{-1}\,s^{-2}}; 1 bar=105 Pa1\,\mathrm{bar}=10^5\,\mathrm{Pa}]

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