Given the same reactant concentrations, the reaction CO(g) + Cl2(g) →COCl2(g) at 243°C is 97.5 times...
Given the same reactant concentrations, the reaction
CO(g) + Cl2(g)
→COCl2(g)
at 243°C is 97.5 times as fast as the same reaction at 153°C. Calculate the activation energy for this reaction in kj/mol. Assume that the frequency factor is constant.
Homework Answers
CO(g) + Cl2(g) →COCl2(g)
T1 = 153 + 273 = 426K
T2 = 243+273 = 516K
lnK2/K1 = Ea/R[ 1/T1- 1/T2]
ln(97.5/1) = Ea/8.314[1/426-1/516]
4.5798 = Ea(0.00234-0.00194)/8.314
Ea = 4.5798*8.314/0.0004 = 95191J = 95.191KJ/mole
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