The rate of the reaction gets doubled when the temperature changes from 7°C to 17°C. By what factor will it change for the temperature change from 17°C to 27°C? [Antilog(0.30) = 2.0, Antilog(0.280) =

Question

The rate of the reaction gets doubled when the temperature changes from 7°C to 17°C. By what factor will it change for the temperature change from 17°C to 27°C? [Antilog(0.30) = 2.0, Antilog(0.280) = 1.91, Antilog(0.322) = 2.1]

Prashant Kotian · Accepted Answer

Correct answer: C. ln(K2/K1) = (Ea/R) × (10 / (280 × 290)) = ln 2. For 17 to 27°C, ln(K3/K2) = (Ea/R) × (10 / (290 × 300)). Thus ln(K3/K2) = ln 2 × (280 / 300) = 0.301 × (28/30) ≈ 0.281. Factor = 10^0.281 = Antilog(0.281) ≈ 1.91. Therefore, correct answer is C.

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