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JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Chemical Thermodynamics

For a certain thermochemical reaction

M \to N

at

T = 400 K, Δ H^{⊖} = 77.2 kJ mol^{- 1}, Δ S = 122 JK^{- 1}

,

lo g

equilibrium constant

(lo g K)

is -

____

\times 1 0^{- 1}

.

Answer: 37

Solution

Δ G^{⊖} = Δ H^{⊖} - T Δ S^{⊖} = 77.2 \times 1 0^{3} - 400 \times 122 = 77200 - 48800 = 28400

J.

Using

Δ G^{⊖} = - 2.303 R T lo g K

:

28400 = - 2.303 \times 8.314 \times 400 \times lo g K

, so

lo g K = \frac{- 28400}{7663.7} = - 3.708 = - 37.08 \times 1 0^{- 1}

.

The magnitude required is 37.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer

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