Question

What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300K to 400K? (R = 8.314 J /mol K)
(A) 68.8
(B) 6.88
(C) 34.4
(D) 3.44

Answer 1

<>bHint: Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.

Complete step by step solution:
Given in the question:
Rate constant doubles for the reaction
This means that the rate constant for the first reaction will be equal to twice the rate constant for the second reaction.
Temperature given ${{T}_{1}}$$= 300 K$
${{T}_{2}}$= 400 K
The activation energy given formula:
\[\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}(\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}})\]
\[\log \dfrac{2}{1}=\dfrac{{{E}_{a}}}{(2.33)(8.31)}(\dfrac{1}{300}-\dfrac{1}{400})\]
\[{{E}_{a}}=(0.3)(2.3)(8.31)(3)(400)=6.88\]
The value of activation energy = 6.88KJ

Hence the correct answer is option (B).

Additional information:
Activation energy can be calculated using various methods. It can be calculated using the Arrhenius equation and also when then two temperatures and the rate constant at both temperatures are known. The temperature should be converted to kelvin while calculating activation energy using the Arrhenius equation.

Note: If the reaction is a third order reaction, the unit for third order reaction is ${{M}^{-2}}h{{r}^{-1}}or\text{ mo}{{\text{l}}^{-2}}{{L}^{2}}h{{r}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing