Question

A first-order reaction has rate constants of 4.6 x 10^-2 s^-1 and 8.1 x 10^-2 s^-1 at 0 degrees celcius and 20 degrees celcius respectively. what is the value of the activation energy?

the answer should be 19 KJ/mol, i'm just very confused on the math has to be done to get this answer. could you please write out a step by step answer?

Answer 1

Equation to use is the Arrhenius equation :

k = A exp (-E/RT)

Where k = rate constant, A = constant, E = activation energy ( in J/mol) , R = Gas constant = 8.314 J /(mol.K ) and T = temperature ( in K).

Writing the equation for T = 0 degrees Celsius :

k = 4.6 x 10^-2 , T = 0 C = 273 K

So, 4.6 x 10^-2 s-1 = A exp(-E/R(273 K))

Similarly, for T = 20 degrees Celsius (=293 K) :

8.1 x 10^-2 s-1= Aexp(-E/R(293))

Dividing the first equation by the second :

Dividing the left side and putting the value . also cancel out A on right side.

Taking natural log ( ln) on both sides.

The right side has the form (a/b). Use the relation ln(a/b) = ln(a) - ln(b) :

ln (0.568) = ln [ exp(-E/R(273))] - ln[exp(-E/R(293))]

Evaluate the left side. On the right side, use the relation : ln [exp (x)] = x

Multiplying (-) on both sides :

Taking E /R common on right side :

Evaluating (1/273)-(1/293) on the right side :

Putting the value of R and solving for E :

E = 18810 J/mol = 18.8 kJ/mol = 19 kJ/mol [ 1000 J = 1 kJ]