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

Question

vladimir2022 [97]

3 years ago

15

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?

Chemistry

1 answer:

Sveta_85 [38] · Answer 1 · 2022-04-04T10:45:45+03:00

Answer:3480s⁻¹

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

$K=A{exp[-Ea \div RT]}\\K=8\ \times10^{10} s^{-1}{exp[-42000\div 8.314\times298]}\\K=8\ \times10^{10} s^{-1}{exp[-16.95]}\\K=4.35\times10^{^{-8}}\times8.0\times10^{^{10}}s^{-1}\\K=34.8\times10^{2}s^{-1}\\$

K=3480s⁻¹

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