The rate constant of a reaction is 6 × 10^-3 s^-1at 50° and 9 × 10^-3s^-1at 100° C. Calculate the energy of activation of the reaction.(a) 6.123 kJ mol^-1(b) 8.124 kJ mol^-1(c) 12.357 kJ mol^-1(d) 18.256 kJ mol^-1This question was posed to me during an internship interview.I'd like to ask this question from Chemical Kinetics topic in division Chemical Kinetics of Chemistry – Class 12

The rate constant of a reaction is 6 × 10^-3 s^-1at 50° and 9 × 10^

1.	The rate constant of a reaction is 6 × 10^-3 s^-1at 50° and 9 × 10^-3s^-1at 100° C. Calculate the energy of activation of the reaction.(a) 6.123 kJ mol^-1(b) 8.124 kJ mol^-1(c) 12.357 kJ mol^-1(d) 18.256 kJ mol^-1This question was posed to me during an internship interview.I'd like to ask this question from Chemical Kinetics topic in division Chemical Kinetics of Chemistry – Class 12
Answer» The correct choice is (B) 8.124 kJ mol^-1 Easiest EXPLANATION: Given, K1=6 × 10^-3s^-1T1=50 + 273=323 K k2=9 × 10^-3s^-1T2=100 + 273=373 K Substituting these values in the equation: log (k2 / k1 ) = (EA / 2.303 R) × ((T2 – T1) / T1 T2), we get log (9 × 10^-3s^-1 / 6 × 10^-3 s^-1) = ((Ea / (2.303 × 8.314)) × ((373 – 323) / (373 × 323)) log 9 / 6 = ((Ea / (2.303 × 8.314)) × (50 / (373 × 323)) Ea = 8.124 kJ mol^-1.

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