Question

how many times does a reaction speed up if the T rises from 10*C to 40* C?

Answer 1

Arrhenius explained the variation of rate constants with temperature for several elementary reactions using the relationship

k = A exp(-E_a/RT)

where the rate constant k is the total frequency of collisions between reaction molecules A times the fraction of collisions exp(-E_a/RT) that have an energy that exceeds a threshold "activation energy" E_a at a temperature of T (in kelvins). R is the universal gas constant.

To see what temperature rise is required to change the rate constant from k₁ (at T₁) to k₂ (at T₂), take the ratio of the Arrhenius equations for each of the two temperatures:

Let's assume an activation energy of 50 kJ mol^-1. In the equation, we have to write that as 50000 J mol^-1. The value of the gas constant, R, is 8.31 J K^-1 mol^-1.

At 40°C (313 K) the value of the fraction is:

exp(-E_a/RT) = exp(-50000/8.314*313) = 4.5 X 10-9

At 10°C (283 K) the value of the fraction is:

exp(-E_a/RT) = exp(-50000/8.314*283) = 0.59 X 10-9

hence ratio is 4.5/0.59 = 7.63

7.63 times reaction speed up if the T rises from 10*C to 40* C