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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Chemical Thermodynamics

For a certain reaction at

300 K, K = 10

, then

Δ G^{\circ}

for the same reaction is - ____

\times 1 0^{- 1} kJ mol^{- 1}

.
(Given

R = 8.314 JK^{- 1} mol^{- 1}

)

Answer: 57

Solution

Δ G^{\circ} = - R T ln K

. With

R = 8.314 J K^{- 1} mo l^{- 1}

,

T = 300 K

, and

K = 10

:

Δ G^{\circ} = - 8.314 \times 300 \times ln 10 = - 8.314 \times 300 \times 2.303 = - 5744 J/mol = - 57.44 \times 1 0^{- 1} kJ/mol

Rounded to the nearest integer, the magnitude is 57.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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