The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at T = 298K are Δ _ f G ^ o [C ( graphite] = 0kJ mol ^ - 1 Δ _ f G ^ o [C(diamond)] = 2 .9 kJ mol ^ - 1 The standard state means that

Correct answer: A. 14501 bar C(graphite) → C(diamond); Δ G ^ o = Δ _ f G _ diamond ^ o - Δ _ f G _ graphite ^ o = 2 .9 kJ / mol at 1 bar As dG _ T = V . dP ∫ _ Δ G _ 1 ^ Δ G _ 2 d( Δ G _ T ) = ∫ _ P _ 1 ^ P _ 2 Δ V . dP Δ G _ 2 - Δ G _ 1 = Δ V . ( P _ 2 - P _ 1 ) (2 .9 × 10 ^ 3 - 0) = (- 2 × 10 ^ - 6 )(1 - P _ 2 ) P _ 2 - 1 = 2 .9 × 10 ^ 3 2 × 10 ^ - 6 Pa = 1 .45 × 10 ^ 4 bar P _ 2 = 14501bar .

JEE Advanced 2017Paper 2

Exam: JEE Advanced
Year: 2017
Paper: Paper 2
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Advanced 2017 · Paper 2 · Chemistry · Thermodynamics (C)

The standard state Gibb's free energies of formation of C...

The standard state Gibb's free energies of formation of

C

(graphite)

and C

(diamond) at

T = 298K

are

{\Delta}_{f}{G}^{o}\left[C ( graphite\right] = 0kJ{mol}^{- 1}

{\Delta}_{f}{G}^{o}\left[C\left(diamond\right)\right] = 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 {10}^{- 6}{m}^{3}{mol}^{- 1}

. If

C

(graphite) is converted to

C

(diamond) isothermally at

T = 298K

, the pressure at which

C

(graphite) is in equilibrium with

C

(diamond), is
[Useful information:

1J = 1kg{m}^{2}{s}^{- 2}, 1Pa = 1kg{m}^{- 1}{s}^{- 2}; 1bar = {10}^{5}Pa

]

Options

A $14501$ bar
B $29001$ bar
C $58001bar$
D $1450bar$

Correct answer

A. $14501$ bar

Step-by-step solution

C\left(graphite\right) \to C\left(diamond\right); {\Delta G}^{o} = {\Delta}_{f}{G}_{diamond}^{o} - {\Delta}_{f}{G}_{graphite}^{o} = 2 .9 kJ / mol $at$ 1 $barAs$ {dG}_{T} = V . dP $$ \mathop{∫}_{{\Delta G}_{1}}^{{\Delta G}_{2}}d\left({\Delta G}_{T}\right) = \mathop{∫}_{{P}_{1}}^{{P}_{2}}\Delta V . dP $$ {\Delta G}_{2} - {\Delta G}_{1} = \Delta V . \left({P}_{2} - {P}_{1}\right) $\left(2 .9 \times {10}^{3} - 0\right) = \left(- 2 \times {10}^{- 6}\right)\left(1 - {P}_{2}\right)$

${P}_{2} - 1 = \dfrac{2 .9 \times {10}^{3}}{2 \times {10}^{- 6}}Pa = 1 .45 \times {10}^{4}bar$

${P}_{2} = 14501bar .$

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