$\mathrm{K},\mathrm{A}$ and ${\mathrm{E}}_{\mathrm{a}}$ of a process at $25°\mathrm{c}$ respectively are $5\times {10}^{-4}{ \mathrm{s}}^{-1},6\times {10}^{14}{ \mathrm{s}}^{-1}\&108\mathrm{kj}/\mathrm{mol}$. Then the value of rate constant as time $\to \infty$ will be ( in $\left.{\mathrm{s}}^{-1}\right)$

$$\begin{gathered} \mathrm{Given} \\ \mathrm{k}=5\times {10}^{-4} \mathrm{s}-1,\mathrm{A}=6\times {10}^{14}{ \mathrm{s}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=108\mathrm{k} \\ \mathrm{By} \mathrm{as} \mathrm{henius} \mathrm{eqn} \\ \mathrm{K}={\mathrm{Ae}}^{-\mathrm{Ea}\mid \mathrm{RT}} \\ \mathrm{but} \mathrm{at} \mathrm{T}\to \infty \\ \begin{array}{r}\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}=0\\ \Rightarrow {\mathrm{e}}^0=0\\ \end{array} \\ \mathrm{so}, \mathrm{remains} \mathrm{same} \\ \mathrm{Rate} \mathrm{const}. \mathrm{at} \mathrm{T}\to \infty \\ =5\times {10}^{-4} \end{gathered}$$

Therefore, the correct option is (D).