JEE (Advanced) 2017 Paper-2 Question 21

Question 21

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
$$\triangle_{f}G^{0}[C(graphite)]=0 kJ\ mol^{-1}$$
$$\triangle_{f}G^{0}[C(diamond)]=2.9 kJ\ mol^{-1}$$
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 $$2\times10^{-6}m^{3}mol^{-1}$$ If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with
[Useful information: 1 J = 1 kg $$m^{2}s^{-2}$$;1 Pa=1kg $$m^{-1}s^{-2}$$; $$1 bar = 10^{5} Pa$$] C(diamond), is

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