Chemical Thermodynamics NEET 2003: Graphite to Diamond

What is the answer to the Chemical Thermodynamics NEET 2003 graphite–diamond question?

Option A is correct: high pressure favours diamond because it has the smaller molar volume. The Chemical Thermodynamics NEET 2003 question asks for the pressure at which graphite and diamond reach the transformation threshold at 298 K, given:

ρ_{graphite} = 2.25 g c m^{- 3}, ρ_{diamond} = 3.51 g c m^{- 3}

Δ G^{\circ} = + 1895 J m o l^{- 1} for graphite \to diamond

The answer choices are:

A $9.92 \times 10^{8} P a$
B $9.92 \times 10^{7} P a$
C $9.92 \times 10^{6} P a$
D $9.92 \times 10^{5} P a$

The supplied question bank rates this problem hard, with an expected solving time of 120 seconds. That is a practice benchmark, not an official exam time allowance.

Use three checks: pressure must favour the smaller-volume phase, the signed reaction volume must be negative, and the final units must be pascals.

Why do we set Gibbs free-energy change to zero, and which volume sign is correct?

At the transformation threshold, the two phases are in equilibrium, so the Gibbs free-energy change is zero. Diamond is denser than graphite, so the same amount of carbon occupies less volume as diamond. Increasing pressure therefore lowers the free-energy change for the forward transformation.

Write the process and its signed reaction-volume change: $graphite \to diamond$

Δ V_{r x n} = V_{diamond} - V_{graphite} < 0

The official method treats the solid molar volumes as constant:

Δ G = Δ G^{\circ} + Δ V_{r x n} (P - P^{\circ})

Here, standard pressure is the reference pressure. Since the standard free-energy change is positive, the pressure term must become negative enough to cancel it.

At equilibrium:

0 = Δ G^{\circ} + Δ V_{r x n} (P - P^{\circ})

Rearranging gives:

P - P^{\circ} = - \frac{Δ G^{\circ}}{Δ V_{r x n}} = \frac{Δ G^{\circ}}{V_{graphite} - V_{diamond}}

The two volume expressions have opposite signs. Final volume minus initial volume is the negative reaction-volume change. Initial volume minus final volume is the positive volume decrease used in the denominator.

Keep those definitions separate. Using a negative denominator in the final expression while dropping the minus sign would incorrectly predict a negative pressure increase.

How do the densities give the molar-volume decrease?

Divide carbon’s molar mass by each density, then subtract diamond’s molar volume from graphite’s. Do not subtract the densities themselves: the pressure correction needs a volume change per mole. Retain the official solution’s rounded intermediate values.

Using the carbon molar mass from the supplied solution:

M = 12 g m o l^{- 1}, V_{m} = \frac{M}{ρ}

For graphite:

V_{graphite} = \frac{12 g m o l^{- 1}}{2.25 g c m^{- 3}} = 5.33 c m^{3} m o l^{- 1}

For diamond:

V_{diamond} = \frac{12 g m o l^{- 1}}{3.51 g c m^{- 3}} = 3.42 c m^{3} m o l^{- 1}

The positive volume decrease is:

V_{graphite} - V_{diamond} = (5.33 - 3.42) c m^{3} m o l^{- 1} = 1.91 c m^{3} m o l^{- 1}

The grams cancel during the density division, leaving volume per mole. Convert that volume to SI before substituting into the pressure formula.

The length conversion must be cubed:

1 c m^{3} = (10^{- 2} m)^{3} = 10^{- 6} m^{3}

Therefore:

V_{graphite} - V_{diamond} = 1.91 \times 10^{- 6} m^{3} m o l^{- 1}

How does the calculation give option A, and what does that pressure mean?

Dividing the positive standard free-energy change by the positive molar-volume decrease gives option A, following the supplied official rounding. The official calculation neglects standard pressure because the required pressure is much larger, as the result below confirms. $P - P^{\circ} \approx P$

Substitute the values:

P \approx \frac{1895}{1.91 \times 10^{- 6}} = (\frac{1895}{1.91}) \times 10^{6} = 9.92 \times 10^{8} P a

Check the dimensions independently:

\frac{J m o l^{- 1}}{m^{3} m o l^{- 1}} = J m^{- 3} = P a

Thus the answer is:

Option A: 9.92 \times 10^{8} P a

The calculated pressure is of order: $10^{9} P a$

Standard pressure is only of order: $10^{5} P a$

That separation justifies neglecting standard pressure at the precision used here. The result identifies the equilibrium threshold, not a reaction rate.

Equilibrium does not guarantee rapid diamond formation. At the calculated threshold, the free-energy change is zero. Above it, this constant-volume model favours diamond thermodynamically.

Conversion at 298 K can still be extremely slow because the carbon structure must rearrange. Gibbs free energy determines the thermodynamic direction; kinetics determines how quickly the change occurs.

How can a cubic-unit conversion error produce option B?

One possible route to option B is making the converted volume denominator ten times too large. All the options have the same coefficient, so the exponent is the deciding check.

Suppose someone incorrectly writes:

1.91 c m^{3} m o l^{- 1} \overset{incorrect}{=} 1.91 \times 10^{- 5} m^{3} m o l^{- 1}

Their pressure calculation becomes:

P_{wrong} = \frac{1895}{1.91 \times 10^{- 5}} = 9.92 \times 10^{7} P a

That is option B. Pressure is inversely proportional to the volume denominator here, so a denominator ten times too large gives a pressure ten times too small.

Before substituting, write the conversion factor beside the volume:

(10^{- 2})^{3} = 10^{- 6}

Cube the conversion factor, not just the unit label. This catches the error before otherwise correct division leads to the wrong option.

Which three practice questions test the same reasoning?

These three practice questions are not additional verified PYQs. They test the standard-pressure sign, the signed reaction volume and the effect of raising pressure beyond equilibrium. Try each before reading its explanation; all use the supplied quantities and the same constant-volume model.

Is graphite to diamond spontaneous at 298 K and standard pressure?

No. At standard pressure, the pressure correction vanishes: $P - P^{\circ} = 0$

Δ G = Δ G^{\circ} = + 1895 J m o l^{- 1}

The positive free-energy change means the forward transformation is not spontaneous under those conditions. Diamond’s smaller volume does not, by itself, make the forward change favourable at standard pressure.

What is the signed molar-volume change for graphite to diamond?

It is negative, because reaction volume means final minus initial, not larger minus smaller. Use diamond’s molar volume first:

Δ V_{r x n} = (3.42 - 5.33) c m^{3} m o l^{- 1} = - 1.91 c m^{3} m o l^{- 1} = - 1.91 \times 10^{- 6} m^{3} m o l^{- 1}

What happens to Gibbs free-energy change above the equilibrium pressure?

It becomes negative within the constant-volume model. Relative to the equilibrium pressure:

Δ G (P) = Δ V_{r x n} (P - P_{e q})

Above equilibrium, the pressure difference is positive while the signed reaction volume is negative. Their product is negative, so diamond is thermodynamically favoured.

Before selecting an option, write final volume minus initial volume once. That line fixes the pressure sign before any arithmetic begins.

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For a worked example of the same idea, see Dual Nature NEET 2026: Electron-to-Photon Momentum Ratio.

Frequently asked questions

What is the answer to the NEET 2003 graphite–diamond question?

Option A is correct: the equilibrium pressure at 298 K is approximately 9.92 × 10⁸ Pa. Using the supplied rounded molar volumes, divide 1895 J/mol by the positive volume decrease of 1.91 × 10⁻⁶ m³/mol, neglecting standard pressure.

Why does high pressure favour diamond over graphite?

Diamond is denser than graphite, so its molar volume is smaller. For graphite converting to diamond, the signed reaction-volume change is negative, so increasing pressure lowers the Gibbs free-energy change and favours diamond.

Why is Gibbs free-energy change zero at the transformation pressure?

At the transformation threshold, graphite and diamond are in equilibrium, so the Gibbs free-energy change for conversion is zero. Above this pressure, the constant-volume model favours diamond thermodynamically. This does not guarantee rapid conversion at 298 K, because the carbon structure must rearrange.

How do I convert cm³/mol to m³/mol in this question?

Cube the length conversion: 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³. Therefore, the positive molar-volume decrease is 1.91 × 10⁻⁶ m³/mol. Using 10⁻⁵ instead makes the denominator ten times too large and gives option B, 9.92 × 10⁷ Pa.