Question

Given that k is the rate constant for some order of a reaction at temperature T, then the value of $\mathop {\lim }\limits_{T \to \infty } \log k$ (where A is the Arrhenius constant) is:
(A) A / 2.303
(B) A
(C) 2.303A
(D) logA

Answer 1

First take the Arrhenius equation and do logs on both sides. Then to the obtained equation put limits of T tends to ∞. This will lead you to your answer.

Complete step by step answer:
- First we need to see what the Arrhenius equation is.
It was experimentally proved that with rise in temperature by ${10^ \circ }$, the rate constantly gets doubled of its initial value. So, the Arrhenius equation was formulated to accurately explain the temperature dependence of the rate of a chemical reaction.
$k = A{e^{\frac{{ - {E_a}}}{{RT}}}}$
Or by taking natural logarithms on both sides: $\ln k = \ln A - \frac{{{E_a}}}{{RT}}$
Where A = Arrhenius factor or frequency factor or pre exponential factor;
R = gas constant;
${E_a}$ = activation energy
- Above equation can also be written as:
$\log k = \log A - \frac{{{E_a}}}{{2.303RT}}$
Now we will be taking limits on both sides:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \mathop {\lim }\limits_{T \to \infty } \left[ {\log A - \frac{{{E_a}}}{{RT}}} \right]$
Since there is no T variable in term logA so the limit will not apply to that. The equation now becomes:
$\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A - \mathop {\lim }\limits_{T \to \infty } \left[ {\frac{{{E_a}}}{{RT}}} \right]$
Now putting the values: we know that if T = ∞, $\frac{1}{T}$ will be = 0.
And so: $\frac{{{E_a}}}{{RT}}$ term will be = 0.
So, the equation becomes: $\mathop {\lim }\limits_{T \to \infty } (\log k) = \log A$

So, the correct option will be: (D) logA.

Note: According to the Arrhenius equation, the lower the activation energy faster will be the rate of the reaction. Even a small amount of catalyst can catalyse a large number of reactants and so to increase the rate of a reaction a catalyst can be added which provides an alternate pathway for the reaction by reducing the activation energy and thus reducing the potential energy barrier.