A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k. what is the ratio of f at the higher temperature to f at the lower temperature?

Question

Chemistry

Stephen Goodwin

21 December, 07:46

A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k. what is the ratio of f at the higher temperature to f at the lower temperature?

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Deon Hester · Answer 1

Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency ...

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A * e ^ (-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ { (-Ea / R) * (1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k) * (1 / 505K - 1 / 485K) }

=> K1 / K2 = e ^ (2.0134494) ≈ 7.5

Answer: 7.5