Thermodynamics JEE 2017: Graphite

What is the graphite–diamond question in Thermodynamics JEE 2017?

The equilibrium pressure in this Thermodynamics JEE 2017 question is 14501 bar, not 14500 bar: the calculation first gives the increase above standard pressure. It is a single-correct MCQ from JEE Advanced 2017, Paper 2, Chemistry. This question bank classifies it as medium difficulty, with an expected solving time of 120 seconds, not an official exam allowance.

Find the pressure at which pure graphite and pure diamond coexist in equilibrium at 298 K. Their standard molar Gibbs energies of formation, at a standard pressure of 1 bar, are:

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}

Converting one mole of graphite to diamond decreases the molar volume by:

2 \times 10^{- 6} m^{3} m o l^{- 1}

The answer options are:

Use the supplied unit relations:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}

1 b a r = 10^{5} P a

How do you choose the reaction direction and signs?

Use diamond minus graphite for both the Gibbs energy change and the volume change. The question specifies conversion from graphite to diamond, so keep that direction fixed. At 298 K and 1 bar, graphite has the lower molar Gibbs energy and is thermodynamically more stable. $C (g r a p h i t e) ⟶ C (d i a m o n d)$

The standard Gibbs energy change is:

Δ G^{\circ} & = Δ_{f} G^{\circ} [C (d i a m o n d)] - Δ_{f} G^{\circ} [C (g r a p h i t e)] & = (2.9 - 0) k J m o l^{- 1} & = 2.9 k J m o l^{- 1} & = 2.9 \times 10^{3} J m o l^{- 1}

The stated volume reduction gives:

Δ V = V_{d i a m o n d} - V_{g r a p h i t e} = - 2 \times 10^{- 6} m^{3} m o l^{- 1}

“Decreases” supplies the negative sign. The given magnitude alone is not the reaction volume change. Retain that negative sign when substituting: it is what allows increasing pressure to favour diamond.

How does integration give the pressure-dependent Gibbs energy?

Integrating the volume difference over the pressure interval gives the pressure correction to the conversion Gibbs energy. At constant temperature, each allotrope’s molar Gibbs energy changes with pressure according to its molar volume. Both Gibbs energy and volume are used on a molar basis throughout this calculation.

For each allotrope: $d G = V d P$

d G_{d i a m o n d} = V_{d i a m o n d} d P, d G_{g r a p h i t e} = V_{g r a p h i t e} d P

Subtract graphite from diamond: $d (Δ G) = Δ V d P$

The official solution assumes that the supplied molar volume change remains constant over the pressure interval. Integrate from the standard pressure to the required pressure:

Δ G (P) - Δ G^{\circ} = \int_{P^{\circ}}^{P} Δ V d P = Δ V (P - P^{\circ})

Therefore:

Δ G (P) = Δ G^{\circ} + Δ V (P - P^{\circ})

$P^{\circ} = 1 b a r$ Because the volume change is negative, raising pressure lowers the Gibbs energy change for graphite becoming diamond. The difference between the solids decreases, not each solid’s own Gibbs energy.

At coexistence, the two molar Gibbs energies are equal. Set the conversion Gibbs energy change to zero, not either solid’s individual Gibbs energy:

G_{d i a m o n d} (P) = G_{g r a p h i t e} (P) \Rightarrow Δ G (P) = 0

Why is the equilibrium pressure 14501 bar rather than 14500 bar?

The calculation gives an increase of 14500 bar above 1 bar, so adding the reference pressure gives 14501 bar. Keep pressure in pascals while using joules and cubic metres. Substituting pressure in bar directly would make the energy and pressure-volume terms inconsistent.

At equilibrium, substitute the signed values:

0 = 2.9 \times 10^{3} + (- 2 \times 10^{- 6}) (P - P^{\circ})

Rearrange without dropping the reference pressure:

2 \times 10^{- 6} (P - P^{\circ}) = 2.9 \times 10^{3}

P - P^{\circ} = \frac{2.9 \times 10^{3}}{2 \times 10^{- 6}} = 1.45 \times 10^{9} P a

Check the dimensions before converting units. Moles cancel, and energy per volume is pressure:

\frac{J m o l^{- 1}}{m^{3} m o l^{- 1}} = J m^{- 3} = k g m^{- 1} s^{- 2} = P a

Convert the pressure difference to bar:

P - P^{\circ} = \frac{1.45 \times 10^{9}}{10^{5}} b a r = 1.45 \times 10^{4} b a r = 14500 b a r

Recover the pressure asked for:

P = (14500 + 1) b a r = 14501 b a r

Correct option: A. The value 14500 bar is the increase above standard pressure; 14501 bar is the equilibrium pressure. Retain the final 1 bar rather than rounding it away.

This is the result within the solution’s constant-volume-change model. It is not a detailed high-pressure phase-diagram calculation.

How can wrongly averaging the volume change produce option D?

One possible incorrect calculation is to halve the supplied volume difference because two allotropes appear. This is an illustrative method error, not a documented common mistake or a claim about why the examiner chose option D.

The unjustified step would be:

Δ V_{w r o n g} = \frac{- 2 \times 10^{- 6}}{2} = - 1 \times 10^{- 6} m^{3} m o l^{- 1}

That gives:

P - P^{\circ} = \frac{2900}{1 \times 10^{- 6}} = 2.9 \times 10^{9} P a = 29000 b a r

P = (29000 + 1) b a r = 29001 b a r (option D)

A reaction volume change is a stoichiometric difference, not an average over the substances named. For one mole of graphite becoming one mole of diamond, the supplied negative volume change already compares the final and initial molar volumes. No division by two is warranted.

How do you apply the method to related thermodynamics questions?

Keep the same three checks: reaction direction, consistent units and reference pressure. Use the constant-volume-change model in each calculation below. These are original practice questions, not additional verified PYQs, with a worked answer immediately after each question.

Using the original data, find the Gibbs energy change for graphite becoming diamond at 10001 bar. Which allotrope is more stable?

Subtract the standard pressure before calculating the pressure correction:

P - P^{\circ} = (10001 - 1) b a r = 10000 b a r = 10^{9} P a

Δ G = 2900 + (- 2 \times 10^{- 6}) (10^{9}) = 2900 - 2000 = + 900 J m o l^{- 1} = + 0.9 k J m o l^{- 1}

The conversion still has a positive Gibbs energy change. Graphite remains more stable.

In a hypothetical variant, keep the original standard Gibbs energy change but use the following volume change. What is the equilibrium pressure?

Δ V = - 1 \times 10^{- 6} m^{3} m o l^{- 1}

P - P^{\circ} = \frac{2900}{1 \times 10^{- 6}} = 2.9 \times 10^{9} P a = 29000 b a r

$P = 29001 b a r$ The pressure increase doubles. Here the input is legitimately changed, unlike the unjustified halving in the wrong-option calculation.

If both the standard Gibbs energy change and constant volume change are positive, can raising pressure above standard pressure establish equilibrium?

No. Above the reference pressure, both terms are positive:

Δ G (P) = \underset{> 0}{\underset{⏟}{Δ G^{\circ}}} + \underset{> 0}{\underset{⏟}{Δ V (P - P^{\circ})}} > 0

Increasing pressure cannot make the conversion Gibbs energy zero within this model. Before substituting, write final minus initial, convert the units, and label the calculated pressure as either an increase or an absolute pressure.

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Frequently asked questions

Why is the equilibrium pressure 14501 bar and not 14500 bar?

The calculation gives P − P° = 14500 bar, which is the pressure increase above the standard pressure. Since P° = 1 bar, the equilibrium pressure is 14501 bar, corresponding to option A.

Why is the volume change negative for graphite to diamond?

The reaction volume change is the molar volume of diamond minus that of graphite. The question states that converting one mole of graphite to diamond decreases the volume by 2 × 10⁻⁶ m³, so ΔV = −2 × 10⁻⁶ m³ mol⁻¹. This negative sign makes increasing pressure favour diamond.

Which Gibbs energy formula is used in the JEE 2017 graphite–diamond question?

Use ΔG(P) = ΔG° + ΔV(P − P°), obtained by integrating d(ΔG) = ΔV dP at constant temperature. This assumes the supplied molar volume change remains constant over the pressure interval. At equilibrium, set the conversion Gibbs energy change to zero because graphite and diamond have equal molar Gibbs energies.

Why should we not divide the graphite–diamond volume change by two?

The supplied volume change already represents the final molar volume minus the initial molar volume for one mole of graphite becoming diamond. It is a reaction difference, not an average over two allotropes. Halving it incorrectly gives 29001 bar, which is option D.

Thermodynamics JEE 2017: Graphite–Diamond Equilibrium