Question

For a reaction \[A \to B\] , Arrhenius equation is given as ${\log _e}k = 4 - \dfrac{{1000}}{T}$ ​; what will be the activation energy in \[J/mol\] for above reaction?
(A) \[8314\]
(B) \[2000\]
(C) \[2814\]
(D) \[3412\]

Answer 1

Hint: We need to know the Arrhenius equation and its relation to activation energy. It is obvious that chemical reactions speed up when the temperature is increased. In other words, if the temperature changes, the rate constant also changes and lower the activation energy, faster the rate of the reaction. The same theory applies when the temperature is decreased ( \[{E_a}\] increases). This can be mathematically shown with the help of Arrhenius Equation as:
$k = A{e^{\dfrac{{ - Ea}}{{RT}}}}$
$k$ is the rate constant
$T$ is the temperature in Kelvin
$R$ is the Gas Constant $ = 8314J{K^{ - 1}}mo{l^{ - 1}}$
\[{E_a}\] is the activation energy expressed in \[J/mol\]
$e$ is a mathematical quantity
$A$ is called the pre-exponential factor. $A$ in the Arrhenius equation, has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation.

Complete step by step answer:
Given in the question that ${\log _e}k = 4 - \dfrac{{1000}}{T}$ . We are to find the activation energy in \[J/mol\].
As we can see that the value of ${\log _e}k$ is given, we take log on both sides of the Arrhenius equation.
Therefore,
\[\log k = \log \left( {A{e^{\dfrac{{ - Ea}}{{RT}}}}} \right)\]
$\log k = \log A + \dfrac{{ - {E_a}}}{{RT}}$
On substituting the given value we get,
$ \Rightarrow 4 - \dfrac{{1000}}{T} = \log A + \dfrac{{ - {E_a}}}{{RT}}$ (log A corresponds to 4 which can be ignored)
Hence, $\dfrac{{ - {E_a}}}{{RT}} = \dfrac{{ - 1000}}{T}$
${E_a} = 8.314J{K^{ - 1}}mo{l^{ - 1}} \times 1000$
On multiplying we get,
\[ \Rightarrow {E_a} = 8314J/mol\]
Hence the correct option is option (A).

Note:
We must be noted that the Arrhenius equation can also be written as $\dfrac{{ - {E_a}}}{R}\dfrac{1}{T} + \log A$ which is the equation of a straight line (y=mx+c) whose slope is $\dfrac{{ - {E_a}}}{R}$ . This is another simple way of determining the activation energy from values of k observed at different temperatures, by plotting log k as a function of $\dfrac{1}{T}$.