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The effect of temperature on the equilibrium constant is expressed as, T 2 > T 1   logK 2 / logK 1 = - ΔH 2 . 303 1   T 2 - 1   T 1 . For endothermic reactions, false statement is:

Options

  1. A1 T 2 - 1 T 1 = positive
  2. BΔH = positive
  3. Clog   K 2 > log   K 1
  4. DK 2 > K 1

Correct answer

A. 1 T 2 - 1 T 1 = positive

Step-by-step solution

Given that: log K 2 K 1 = - ΔH 2 . 303 1   T 2 - 1   T 1 . For endothermic reactions, ∆ H > 0 . Now, for T 2 > T 1 : ⇒ log K 2 K 1 = - ΔH 2 . 303 T 2 - T 1 T 1 T 2 ⇒ log K 2 K 1 > 0 ⇒ log   K 2 - log   K 1 > 0 ⇒ log   K 2 > log   K 1 ∴ K 2 > K 1 . So, 1 T 2 - 1 T 1 < 0 . ( ∴ T 2 > T 1 )

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