Thermodynamics JEE 2023: Phase Transition Enthalpy at 300 K

What does the JEE Advanced 2023 phase-transition question ask?

The numerical answer to this JEE Advanced 2023 thermodynamics question is 300, in joules per mole. Calculate the enthalpy difference at 600 K first, then correct it to 300 K. The question is from JEE Advanced 2023, Paper 2, Chemistry, Thermodynamics, classified as medium difficulty by the question bank.

Phases alpha and beta have zero Gibbs free-energy difference at 600 K. The supplied transition entropy at this temperature and the heat-capacity difference are:

Δ S = 1 J K^{- 1} m o l^{- 1}, Δ C_{p} = 1 J K^{- 1} m o l^{- 1} .

Find the enthalpy of beta minus that of alpha at 300 K. This is a numerical-answer question; the supplied record contains no answer options.

Keep the phase order fixed throughout:

Δ H = H_{β} - H_{α}, Δ S = S_{β} - S_{α}, Δ C_{p} = C_{p, β} - C_{p, α} .

How do you find the enthalpy difference at 600 K?

At 600 K, the enthalpy difference is 600 joules per mole. The phases are in equilibrium at this transition temperature, so their Gibbs free-energy difference is zero. This lets us equate the enthalpy change to temperature times the supplied transition entropy at the stated equilibrium temperature.

Start with the Gibbs relation: $Δ G = Δ H - T Δ S .$

At the transition temperature: $Δ G_{600} = 0 .$

Substitute the given temperature and entropy, using the stated SI units: $0 = Δ H_{600} - 600 \times 1 .$

Rearrange:

Δ H_{600} = 600 \times 1 = 600 J m o l^{- 1} .

The units cancel:

K \times J K^{- 1} m o l^{- 1} = J m o l^{- 1} .

Zero Gibbs free-energy change does not mean zero enthalpy change. Here, the enthalpy and temperature-times-entropy terms cancel. The value at 600 K is an intermediate result, not the requested answer at 300 K.

How does Kirchhoff’s law give the answer at 300 K?

Kirchhoff’s law reduces the enthalpy difference from 600 to 300 joules per mole when the temperature falls from 600 K to 300 K. Following the official solution, treat the supplied heat-capacity difference as constant across this interval. Keep the temperatures in the same order in the numerator and denominator.

The official finite-difference equation is:

Δ C_{p} = \frac{Δ H_{600} - Δ H_{300}}{600 - 300} .

Substitute the known values, using the stated SI units:

1 = \frac{600 - Δ H_{300}}{300} .

Multiply both sides by the denominator: $300 = 600 - Δ H_{300} .$

Rearrange:

Δ H_{300} = 600 - 300 = 300 J m o l^{- 1} .

Thus:

{(H_{β} - H_{α})}_{300 K} = 300 J m o l^{- 1}

The numerical entry is 300. Kirchhoff’s law corrects the enthalpy difference between the phases, not the enthalpy of beta alone: the supplied heat-capacity difference is itself beta minus alpha.

How can you check the sign and units?

The enthalpy difference must decrease on cooling because the heat-capacity difference is positive. It increases as temperature increases, so lowering the temperature must reduce it.

The correction magnitude is:

(1 J K^{- 1} m o l^{- 1}) (300 K) = 300 J m o l^{- 1} .

The result passes the direction check:

Δ H_{300} = 300 J m o l^{- 1} < Δ H_{600} = 600 J m o l^{- 1} .

The question asks for beta minus alpha, so do not reverse the sign in the final line. Use this sequence: equilibrium first, temperature correction second, sign check last.

Why is 900 the wrong numerical result?

900 joules per mole is an illustrative wrong numerical result, not an official wrong option. It comes from adding a positive correction while cooling. The error is treating the positive magnitude of the temperature interval as the signed change from 600 K to 300 K.

The incorrect calculation is:

Δ H_{300} = 600 + 1 \times (600 - 300) = 900 J m o l^{- 1} (incorrect) .

The actual temperature change has the opposite sign:

T_{f i n a l} - T_{i n i t i a l} = 300 - 600 = - 300 K .

A positive heat-capacity difference therefore gives a negative enthalpy correction:

(1 J K^{- 1} m o l^{- 1}) (- 300 K) = - 300 J m o l^{- 1} .

This agrees with the official equation:

Δ H_{600} - Δ H_{300} = 300 J m o l^{- 1} .

The higher-temperature enthalpy difference must exceed the lower-temperature value by 300 joules per mole. A lower-temperature answer of 900 violates that equation.

How can you test your method on related thermodynamics questions?

Check it by reversing the phase direction, setting the heat-capacity difference to zero, and testing where equilibrium is established. These are original practice variations, not additional verified JEE PYQs.

What is the enthalpy change for beta to alpha at 300 K?

The reverse transition has an enthalpy change of negative 300 joules per mole. Reversing the phase order reverses the sign:

{(H_{α} - H_{β})}_{300 K} = - {(H_{β} - H_{α})}_{300 K} = - 300 J m o l^{- 1} .

What would the answer be if the heat-capacity difference were zero?

It would be 600 joules per mole, keeping the equilibrium and entropy data at 600 K unchanged. The equilibrium calculation stays the same, but the temperature correction vanishes:

Δ H_{600} = 600 J m o l^{- 1} .

Δ C_{p} = 0 \Rightarrow Δ H_{300} - Δ H_{600} = 0 .

Δ H_{300} = 600 J m o l^{- 1} .

Can you set the Gibbs free-energy change to zero at 300 K?

No. Equilibrium is specified at 600 K, not at 300 K. This relation cannot be extended to another temperature without establishing zero Gibbs free-energy change there: $Δ H = T Δ S .$

Substituting the lower temperature and the supplied entropy happens to reproduce the original answer: $300 \times 1 = 300 .$

But this is not a valid substitute for Kirchhoff’s law. The given entropy must not be assumed unchanged when the heat-capacity difference is nonzero.

Before using the equilibrium relation, write the temperature at which equilibrium is given. Then write the signed temperature interval before applying the correction.

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Frequently asked questions

What is the answer to the JEE Advanced 2023 phase-transition thermodynamics question?

The enthalpy difference, H_beta minus H_alpha, at 300 K is 300 J/mol, so the numerical entry is 300. First calculate the enthalpy difference at the equilibrium temperature of 600 K, then apply Kirchhoff's law to correct it to 300 K.

How do you calculate the enthalpy difference at 600 K?

At 600 K, the phases are in equilibrium, so ΔG = ΔH − TΔS = 0. With ΔS = 1 J/(K mol), the enthalpy difference is ΔH = 600 × 1 = 600 J/mol. Zero Gibbs free-energy difference does not mean zero enthalpy difference.

Why is 900 wrong in the JEE 2023 thermodynamics question?

The value 900 J/mol comes from adding a positive correction while cooling. The signed temperature change is 300 − 600 = −300 K, so a constant heat-capacity difference of 1 J/(K mol) gives a correction of −300 J/mol. The correct enthalpy difference is therefore 600 − 300 = 300 J/mol.

Can I use ΔH = TΔS directly at 300 K?

No, because equilibrium is specified at 600 K, not at 300 K. The supplied entropy difference also cannot be assumed unchanged when the heat-capacity difference is nonzero. Multiplying 300 by the supplied entropy happens to give the correct number here, but it is not a valid method.