The rate constant for a reaction at 320 K is found to be fou

The rate constant for a reaction at 320 K is found to be four times the value at 300 K. Calculate the activation energy.

Solution

We know that for a chemical reaction, we have the Arrhenius equation given as:

K = Ae^-E/RT where k is the rate constant, E is the activation energy, R is gas constant, T is the temperature.

Now, for two different temperatures, we have:

Ka = Ae^-E/RTa

and Kb = Ae^-E/RTb

Taking logarithms for the expression above we can write:

lnKa = lnA - E/RT_a

lnKb = lnA - E/RT_b

Now, subtracting the two equations ,we get:

ln(Ka/Kb) = (E/R)[1/T_b - 1/T_a]

For the given problem, we know that the rate constant becomes four times for change in temperature from 300 to 320, hence we get:

ln4 = (E/R)[1/300 - 1/320] [Using R = 8.314]

or, E = 55323.12633 J/mol = 55.3231 KJ/mol