Question

If the activation energy for a given compound is found to be 42.0 kJ/mol, with a frequency factor of 8.0 × 1010 s-1, what is the rate constant for this reaction at 298 K?

Answer 1

Answer:3480s⁻¹

Step-by-step explanation:We can solve the following problem using the Arrhenius equation.

Arrhenius equation is given by:

`K=Aexp[-Ea/RT]`

A=Pre-exponential factor or frequency factor

Ea=Activation energy

R=Ideal gas constant

T=Temperature

K=Rate constant

From the Arrhenius equation we can see that the rate constant K is related with the activation energy and frequency factor.

In the question we are given with the following data:

Ea=42KJ/mol=42x 1000 J/mol

A=8.0×10¹ per second

T=298K

R=8.314J/K mol

when we substitute these given values in Arrhenius equation

`K=A{exp[-Ea / RT]}\\K=8\ *10^(10) s^(-1){exp[-42000/ 8.314*298]}\\K=8\ *10^(10) s^(-1){exp[-16.95]}\\K=4.35*10^{^(-8)}*8.0*10^{^(10)}s^(-1)\\K=34.8*10^(2)s^(-1)\\`

K=3480s⁻¹

The value of rate constant obtained is 3480s⁻¹.