JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Chemical Thermodynamics

The standard state Gibbs free energies of formation of C (graphite) and C (diamond) at

T = 298 K

are

Δ_{f} G^{o} [C (graphite)] = 0 kJ mol^{- 1} Δ_{f} G^{o} [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 \times 1 0^{- 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^{2} s^{- 2}; 1 Pa = 1 kg m^{- 1} s^{- 2}; 1

bar

= 1 0^{5} Pa

]

(A)14501 bar✓ Correct
(B)58001 bar
(C)1450 bar
(D)29001 bar

Answer: (A)

Solution

At equilibrium,

Δ G = 0

. For the conversion at constant temperature:

Δ G = Δ G^{o} + P Δ V = 0

Δ G^{o} = - P Δ V

where

Δ G^{o} = Δ_{f} G^{o} (diamond) - Δ_{f} G^{o} (graphite) = 2.9 kJ/mol = 2900 J/mol

and

Δ V = - 2 \times 1 0^{- 6} m^{3} /mol

(volume decreases)

2900 = P \times 2 \times 1 0^{- 6}

P = \frac{2900}{2 \times 1 0 ^{- 6}} = 1.45 \times 1 0^{9} Pa = 1.45 \times 1 0^{4} bar = 14500 bar

The closest answer is 14501 bar.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Chemical Thermodynamics

Solution

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