For a reaction, Δ ^ ^ = -5.7 kJ mol^ -1 at 298 K. Calculate K. ( R = 8.314 J K^ -1 mol^ -1 )

Correct answer: B. 10 The relationship between standard Gibbs free energy change and the equilibrium constant is given by: Δ G^ ^ = -2.303 RT K Given: Δ G^ ^ = -5.7 kJ mol^ -1 = -5700 J mol^ -1 R = 8.314 J K^ -1 mol^ -1 T = 298 K Substituting the values into the equation: -5700 = -2.303 × 8.314 × 298 × K -5700 = -5705.8 K K = (-5700)/(-5705.8) 1 K = 10^1 = 10 Answer: 10

MHT CET 202622 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 22 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

MHT CET 2026 · 22 April · Morning Shift · Chemistry · Thermodynamics (C)

For a reaction, Δ ^ ^ = -5.7 kJ mol^ -1 at 298 K. Calculate K. ( R...

For a reaction,

\Delta\textG^{}^{\circ} = -5.7

kJ mol^{-1}

at 298 K. Calculate K.<br>( R = 8.314 J K^{-1}

mol^{-1}

)

Options

A $1$
B $10$
C $10^2$
D $10^3$

Correct answer

B. $10$

Step-by-step solution

The relationship between standard Gibbs free energy change and the equilibrium constant is given by:

\Delta G^{}^{\circ} = -2.303 RT \log K

Given:

\Delta G^{}^{\circ} = -5.7 \text{ kJ mol}^{-1} = -5700 \text{ J mol}^{-1}

R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}

T = 298 \text{ K}

Substituting the values into the equation:

-5700 = -2.303 \times 8.314 \times 298 \times \log K

-5700 = -5705.8 \log K

\log K = \dfrac{-5700}{-5705.8} \approx 1

K = 10^1 = 10

Answer:

10

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