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MCQEasyJEE 2024Gibbs Free Energy & Equilibrium Constant

JEE Chemistry 2024 Question with Solution

For a reaction, if the equilibrium constant at 500 K500 \, \text{K} is 44, then the value of the standard Gibbs free energy ΔG∘\Delta G^\circ at this temperature is:

  • A

    −1155 J-1155 \, \text{J}

  • B

    1386 J1386 \, \text{J}

  • C

    −1386 J-1386 \, \text{J}

  • D

    1155 J1155 \, \text{J}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The equilibrium constant is K=4K = 4 at T=500 KT = 500 \, \text{K}.

Find: The standard Gibbs free energy ΔG∘\Delta G^\circ.

Use the relation

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K

Substituting the values:

ΔG∘=−(8.314)(500)ln⁡(4)\Delta G^\circ = -(8.314)(500)\ln(4)

Using ln⁡(4)≈1.386\ln(4) \approx 1.386,

ΔG∘≈−(8.314)(500)(1.386)\Delta G^\circ \approx -(8.314)(500)(1.386)

This gives the final value reported in the solution as −1155 J-1155 \, \text{J}.

Therefore, the correct option is A.

Common mistakes

  • Using ΔG∘=RTln⁡K\Delta G^\circ = RT \ln K without the negative sign gives the wrong sign of Gibbs free energy. Always use ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K.

  • Using logarithm base 1010 instead of natural logarithm is incorrect here because the formula requires ln⁡K\ln K. Use natural log unless the formula is explicitly rewritten for log⁡\log.

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