Question

The frequency factor and activation energy for a chemical reaction are A = 4.23 x 10–12 cm3/(molecule·s) and Ea = 12.9 kJ/mol at 384.7 K, respectively. Determine the rate constant for this reaction at 384.7 K.

Answer 1

Answer : The rate constant for this reaction at 384.7 K is, $7.493\times 10^{-14}cm^3mole^{-1}$

Explanation :

The relation between frequency factor, rate constant and activation energy for a chemical reaction is,

$k=A\times e^{[\frac{-Ea}{RT}]}$

where,

k = rate constant = ?

A = frequency factor = $4.23\times 10^{-12}cm^3\text{ molecule}^{-1}s^{-1}$

Ea = activation energy = 12.9 kJ/mol

T = temperature = 384.7 K

Now put all the given values in this formula, we get:

$k=4.23\times 10^{-12}cm^3\text{ molecule}^{-1}s^{-1}\times e^{[\frac{-12.9kJ/mol}{(8.314J/mole.K)\times (384.7K)}]}$

$k=7.493\times 10^{-14}cm^3mole^{-1}$

Therefore, the rate constant for this reaction at 384.7 K is, $7.493\times 10^{-14}cm^3mole^{-1}$