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← JEE Advanced 2017 · Paper 2
ChemistrySingle correct+3 / -1Question 21 of 54

JEE Advanced 2017 · Paper 2, Chemistry question 3

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

ΔfGo[C(graphite)]=0 kJ mol−1\Delta_f G^o[\mathrm{C(graphite)}] = 0\ \mathrm{kJ\,mol^{-1}}
ΔfGo[C(diamond)]=2.9 kJ mol−1.\Delta_f G^o[\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 \times 10^{-6}\ \mathrm{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 m2^2 s−2^{-2}; 1 Pa = 1 kg m−1^{-1} s−2^{-2}; 1 bar = 10510^5 Pa]

Choose one option
Answer and worked solution

Official answer: A

At constant TT, (∂ΔG∂P)T=ΔV\left(\dfrac{\partial\Delta G}{\partial P}\right)_T=\Delta V.

Equilibrium condition.
0=ΔG∘+ΔV(P−1 bar)  ⟹  2900=2×10−6(P−105) (SI units).0=\Delta G^\circ+\Delta V(P-1\text{ bar})\implies2900=2\times10^{-6}(P-10^5)\ \text{(SI units)}.

Solve.
P−105=1.45×109 Pa=14500 bar  ⟹  P=14501 bar.P-10^5=1.45\times10^9\text{ Pa}=14500\text{ bar}\implies P=14501\text{ bar}.

(A)

Topics:Gibbs energy and spontaneityEquilibrium and Gibbs energy
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