7. In two trials with identical starting concentrations of reactants, the rate increases by a factor...
7. In two trials with identical starting concentrations of reactants, the rate increases by a factor of 3.27 when the temperature is increased from 400 K to 670 K. What is the activation energy for this reaction? Express your answer in kJ / mol.
8. The activation energy for a particular reaction is 4.8 kJ mol-1. The temperature in one trial reaction is 350 K. To what temperature must the reaction be heated to increase the rate constant by a factor of 2.5?
Homework Answers
7)
ln (k2 / k1) = Ea / R [1 / T1 - 1/ T2]
ln (3.27) = Ea / 8.314 x 10^-3 [1/ 400 - 1/ 670]
Ea = 9.77
activation energy Ea = 9.77 kJ/mol
8)
ln (k2 / k1) = Ea / R [1 / T1 - 1/ T2]
ln (2.5) = 4.8 / 8.314 x 10^-3 [1/ 350 - 1/ T2]
T2 = 787 K
temperature = 787 K
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