Question

How do you solve for K using the Arrhenius equation? A first order reaction has an activation energy of ${E_a} = 65.7KJ/mol$ and a frequency factor (pre-exponential factor, A) of $1.31 \times {10^{12}}{s^{ - 1}}$. Calculate the rate constant at $19^\circ C$.

Answer 1

Hint: In this question the value of activation energy and the value of pre-exponential factor is given. The rate constant is calculated by the formula of Arrhenius equation. The expression of Arrhenius equation is given by $k = A{e^{ - Ea/RT}}$ where K is the rate constant.

Complete answer:
It is given that for the first order reaction the activation energy is 65.7KJ/mol and the frequency factor is $1.31 \times {10^{12}}{s^{ - 1}}$.
The Arrhenius equation is the expression used to represent the relation between rate constant, absolute temperature, and A factor.
The expression for Arrhenius equation is shown below.
$k = A{e^{ - Ea/RT}}$
Where,
k is the rate constant of the chemical reaction.
A is the pre-exponential factor.
e is the base of natural logarithm
${E_a}$ is the activation energy
R is the universal gas constant.
T denotes the absolute temperature.
The value of universal gas constant R is $8.314J/(mol.k)$
Convert the temperature given in degree Celsius to kelvin.
$ \Rightarrow T = 19 + 273$
$ \Rightarrow T = 292$
Convert the value of activation energy given in KJ/mol into J.
$ \Rightarrow {E_a} = 65.7KJ = 65.7 \times {10^3}J$
First calculate the exponent part of the equation, so that it will be easy to calculate the rate constant value.
$ \Rightarrow - \dfrac{{{E_a}}}{{RT}} = \dfrac{{65.7 \times {{10}^3}J.mo{l^{ - 1}}}}{{8.314J.mo{l^{ - 1}}.{K^{ - 1}}.292K}}$
$ \Rightarrow - 27.0$
To calculate the value of the rate constant, substitute the values in the equation.
$ \Rightarrow K = 1.31 \times {10^{12}}{s^{ - 1}} \times {e^{ - 27.0}}$
$ \Rightarrow K = 2.34{s^{ - 1}}$
Therefore, the rate constant is $2.34{s^{ - 1}}$.

Note:
Make sure to convert the value of temperature given in Celsius scale to kelvin scale as the Arrhenius equation requires the temperature value in kelvin scale not in Celsius scale. To convert degree Celsius to kelvin add the value 273 to the Celsius value. $T(K) = ^\circ C + 273$.