Question

At $25{{\text{ }}^{\circ }}C$ , the values of rate constant, activation energy and Arrhenius constant of a reaction are $3\times {{10}^{-4}}{{\sec }^{-1}}$, $129KJ\text{ }mol{{e}^{-1}}$ and $2\times {{10}^{15}}{{\sec }^{-1}}$ respectively.
The value of rate constant ( in ${{\sec }^{-1}}$) as $T\to \infty $is:
(a) 0
(b) $2\times {{10}^{15}}$
(c) $3\times {{10}^{-4}}$
(d) $6\times {{10}^{11}}$

Answer 1

Hint: Activation energy is extra energy required by the reactants and by using the concept of Arrhenius concept we can easily, determine the rate constant of the chemical reaction by using the formula as: $k=A{{e}^{\frac{^{-{{E}_{{}^\circ }}}}{^{RT}}}}$

Complete answer:
First of all we should know what is activation energy. 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 the reaction like burning of coal gas in 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)
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 concentration of reactants is unity) and temperature which is as follows:
$k=A{{e}^{\frac{^{-{{E}_{{}^\circ }}}}{^{RT}}}}$ (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.
Now considering the statement;
As we know that;
$T\to \infty $(given)
Then ; ${{e}^{\frac{^{-{{E}_{{}^\circ }}}}{^{RT}}}}=1$
So, the equation(1) will become as;
K=A (2)
And we know that
$A=2\times {{10}^{15}}{{\sec }^{-1}}$
Then, put this value in equation (2), we get;
$k=2\times {{10}^{15}}{{\sec }^{-1}}$
So, thus at $25{{\text{ }}^{\circ }}C$ if the values of rate constant, activation energy and Arrhenius constant of a reaction are $3\times {{10}^{-4}}{{\sec }^{-1}}$, $129KJ\text{ }mol{{e}^{-1}}$ and $2\times {{10}^{15}}{{\sec }^{-1}}$ respectively, then The value of rate constant ( in ${{\sec }^{-1}}$) as $T\to \infty $ is $k=2\times {{10}^{15}}{{\sec }^{-1}}$.

Hence, option (b) is correct.

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