Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?

1 Answer
Jan 28, 2017

Answer:

The Arrhenius equation will tell us that the activation energy in this case must be 52.4 kJ/mol.

Explanation:

The Arrhenius equation says that the dependence of the rate constant #k# on temperature reads

#k=Ae^(-(E_a)/(RT))#

For the sake of this problem, let us say that #k_1# is the initial value of the constant at a temperature #T_1#, and #k_2# is the value at temperature #T_2#

To double the reaction rate, we must double the value of k, i.e. the ratio #k_2/k_1# must equal 2.

This will require the right hand side of the Arrhenius equation to change to

#Ae^(-(E_a)/(RT_2))# from #Ae^(-(E_a)/(RT_1))#

all of which can be written in ratio forms as

#k_2/k_1=e^((-E_a/R)(1/T_2-1/T_1))#

Taking the natural log of each side:

#ln2 = -(E_a/R)(1/T_2-1/T_1)#

or

#E_a= (ln2 (R))/(1/T_2-1/T_1)#

Using 308K for #T_2# and 298K for #T_1#

#E_a= (0.693 (8.314))/(1/308-1/298)#

#E_a=52.4 kJ/(mol)#