Question

The activation energy of a chemical reaction can be determined by:
a.) Evaluating rate constants at two different temperatures
b.) Changing the concentration of reactants
c.) Evaluating the concentration of reactants at two different temperatures
d.) Evaluating rate constant at standard temperature

Answer 1

Hint: Activation energy is extra energy required by the reactants and by using the Arrhenius concept we can easily determine the activation energy of the chemical reaction.

Complete step by step answer:
First, of all, we should know what activation energy is. The excess energy which must be supplied to the reactants to bring their energy equal to the threshold energy and undergo chemical reactions is called activation energy because there are some reactions which require energy to proceed with the reaction like the burning of coal gas in the air. It is equal to the difference between the threshold energy needed for the reaction and the average kinetic energy of all the retracting molecules. That is,
Activation energy ${{E}_{a}}$ = threshold energy - average kinetic energy of the reacting molecules
${{E}_{a}}$ = E(Threshold)-E(reactants)
The activation energy of a chemical reaction can be calculated by the Arrhenius equation which gave the relation between the rate constant (constant of proportionality and depends on the order of reaction and is equal to the rate of reaction when the concentration of reactants is unity) and temperature which is as follows:
k=$A{{e}^{-{{E}_{{}^\circ }}/RT}}$ -----------(1)
where k is rate constant, A is known as Arrhenius factor, ${{E}_{a}}$ is the activation energy, ${{e}^{-{{E}_{{}^\circ }}/RT}}$corresponds to the fraction of molecules having the energy greater than the ${{E}_{a}}$and T is the absolute temperature and R is the gas constant.
taking log on both sides of equation (1), we get:
$ ln k = ln A - \dfrac{{{E}_{a}}}{R}×\dfrac{1}{T}$
Converting to the common logarithm, we get:
2.303 log k = 2.303 log A -$\dfrac{{{E}_{a}}}{RT}$
log k = 2.303 log A - $\dfrac{{{E}_{a}}}{2.303RT}$ --------(2)
now, the activation energy can be determined by measuring the values of rate constant at two different temperatures. Let ${{k}_{1}}$ and ${{k}_{2}}$ are the rate constants for the reaction at two different temperatures ${{T}_{1}}$ and ${{T}_{2}}$ then:
 log ${{k}_{1}}$ = 2.303 log A - $\dfrac{{{E}_{a}}}{2.303R{{T}_{1}}}$ ------------(3)
 log ${{k}_{2}}$ = 2.303 log A - $\dfrac{{{E}_{a}}}{2.303R{{T}_{2}}}$ ------------(4)
subtracting equation (3) from (4), we get
log${{k}_{2}}$ – log${{k}_{1}}$ =$\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right]$
log$\dfrac{{{k}_{2}}}{{{k}_{1}}}$= $\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right]$
log$\dfrac{{{k}_{2}}}{{{k}_{1}}}$= $\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2-}}{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$
thus, the activation energy can be calculated by evaluating the value of rate constants at different temperatures ${{T}_{1}}$ and ${{T}_{2}}$.
So, the correct answer is “Option D”.

Note: Don’t mix the terms activation energy and threshold energy. Activation energy is the excess energy possess by the reactants to undergo the chemical whereas threshold energy is the minimum energy required by the reactants to undergo a chemical reaction.