Question

For a given exothermic reaction, Kₚ and K'ₚ are the equilibrium
constants at temperatures T₁ and T₂, respectively. Assuming
that heat of reaction is constant in temperature range between
T₁ and T₂, it is readily observed that: [2014]
(a) Kₚ > K'ₚ (b) Kₚ < K'ₚ
(c) Kₚ = K'ₚ (d) Kₚ = 1/K'ₚ

Answer 1

answer : option (a) Kₚ > K'ₚ

we know, relation between temperature and rate constant for a thermodynamic process is given by, log(k₂/k₁) = ∆H°/2.303R [1/T₁ - 1/T₂]

at equilibrium,

rate constant = equilibrium constant

so, K₂ = K'ₚ and K₁ = Kₚ

now log( K'ₚ/Kₚ ) = ∆H°/2.303R [1/T₁ - 1/T₂]

for exothermic reaction, T₁ < T₂ and ∆H° < 0

so, [1/T₁ - 1/T₂ ] >0

and ∆H°/2.303R [1/T₁ - 1/T₂] < 0

⇒log(K'ₚ/Kₚ) < 0

⇒K'ₚ/Kₚ < 1

⇒Kₚ > K'ₚ

hence option (a) is correct choice.

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