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Question

Chemistry Question on Chemical Kinetics

The rate constants for a reaction at 400K400\, K and 500K500\, K are 2.60×105s12.60 \times 10^{-5} s ^{-1} and 2.60×103s12.60 \times 10^{-3} s ^{-1} respectively. The activation energy of the reaction in kJmol1kJ\, mol ^{-1} is

A

38.3

B

57.4

C

114.9

D

76.6

Answer

76.6

Explanation

Solution

Given, k1=2.60×105s1k_{1}=2.60 \times 10^{-5} s ^{-1} k2=2.60×103s1k_{2}=2.60 \times 10^{-3} s ^{-1} T1=400KT_{1}=400 \,K T2=500KT_{2}=500 \,K According to Arrhenius equation, logk2k1=Ea2.303R[T2T1T1T2]\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right] log2.60×1032.60×105\Rightarrow \log \frac{2.60 \times 10^{-3}}{2.60 \times 10^{-5}} =Ea2.303×8.314[500400500×400]=\frac{E_{a}}{2.303 \times 8.314}\left[\frac{500-400}{500 \times 400}\right] Ea=2×2.303×8.314×2×105100E_{a} =\frac{2 \times 2.303 \times 8.314 \times 2 \times 10^{5}}{100} Ea=76.6×103J/molE_{a} =76.6 \times 10^{3} J / mol Ea=76.6kJ/mol \therefore E_{a} =76.6\, kJ / mol