Why is the negative sign in Δ G° = -2.303 RT log K retained?

Because standard free energy and equilibrium constant are fundamentally related by Δ G° = -RT ln K. A spontaneous standard state reaction (; Delta G° 1, so log K must be positive.

JEE MainChemistry

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Calculating Equilibrium Constant from Standard Enthalpy and Entropy Changes

The value of log K for the reaction A leftrightharpoons B at 298K is....... (Nearest integer) Given: ΔH° = -54.07kJ mol^(-1) ΔS° = 10JK^(-1)mol^(-1) (Take 2.303 × 8.314 × 298 = 5705)

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Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why is the negative sign in

Δ G^{\circ} = - 2.303 R T \log K

retained?

Because standard free energy and equilibrium constant are fundamentally related by $Δ G^{\circ} = - R T \ln K$ . A spontaneous standard state reaction ( $D e l t a G^{\circ} < 0$ ) yields $K > 1$ , so $\log K$ must be positive.

Answer

Accepted answer: 10

The value of $\log K$ for the reaction at $298 K$ is $10$ .

Why each option works or fails

Step-by-step solution

Given: $T = 298 K$ , $Δ H^{\circ} = - 54.07 {kJ mol}^{- 1} = - 54070 {J mol}^{- 1}$ , $Δ S^{\circ} = 10 {J K}^{- 1} {mol}^{- 1}$ , and $2.303 \times R \times T = 2.303 \times 8.314 \times 298 = 5705 {J mol}^{- 1}$ .
Find: Value of $\log K$ for the equilibrium reaction $A ⇌ B$ rounded to the nearest integer.
Strategise: Use Gibbs-Helmholtz relation $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ , and relate standard free energy to the equilibrium constant via $Δ G^{\circ} = - 2.303 R T \log K$ . Equating gives $\log K = - \frac{Δ G^{\circ}}{2.303 R T}$ .
Execute: Calculate $Δ G^{\circ}$ : $Δ G^{\circ} = - 54070 - (298 \times 10) = - 54070 - 2980 = - 57050 {J mol}^{- 1}$ Then calculate $\log K$ : $\log K = \frac{- Δ G^{\circ}}{2.303 R T} = \frac{- (- 57050)}{5705} = 10$
Verify: Since $Δ H^{\circ} < 0$ and $Δ S^{\circ} > 0$ , the reaction is strongly exergonic (spontaneous), which means $Δ G^{\circ} < 0$ and therefore $Δ G^{\circ} = - 2.303 R T \log K$ , making $\log K > 0$ . The integer $10$ matches the pre-factored arithmetic cleanly.

Calculating Equilibrium Constant from Standard Enthalpy and Entropy Changes

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The value of $\log K$ for the reaction at $298 K$ is $10$ .

Why each option works or fails

Solution

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