a certain reaction proceeds three times faster at 585K than it did at 535K. Calculate the activation energy, in KJ/mol, for this reaction.

Answer 1

Use the Arrhenius equation: ln k2/k1 = -Ea/R (1/T2 - 1/T1)

We can set T1 = 535K and T2 = 585K

Since the rate of reaction is 3x faster at T2 than at T1, we can then set k2 = 3k1

We can now solve for Ea

ln 3k1/k1 = -Ea/8.314 J/mol-K (1/585K - 1/535K)

1.099 = -Ea/8.314 (0.00171 - 0.00187)

1.099 = -Ea/8.314 (-0.00016)

Ea = 57,107 J/mol = 57.1 kJ/mol