Question
The rate constants $${k_1}$$ and $${k_2}$$ for two different reactions are $${10^{16}} \cdot {e^{\frac{{ - 2000}}{T}}}$$ and $${10^{15}} \cdot {e^{\frac{{ - 1000}}{T}}},$$ respectively. The temperature at which $${k_1} = {k_2}$$ is
A.
$$1000\,K$$
B.
$$\frac{{2000}}{{2.303}}\,K$$
C.
$$2000\,K$$
D.
$$\frac{{1000}}{{2.303}}\,K$$
Answer :
$$\frac{{1000}}{{2.303}}\,K$$
Solution :
$$\eqalign{
& {\text{Given,}}\,{k_1} = {10^{16}} \cdot {e^{ - \frac{{2000}}{T}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{k_2} = {10^{15}} \cdot {e^{ - \frac{{1000}}{T}}} \cr} $$
On taking $$\log $$ of both the equations we get
$$\eqalign{
& {\text{log}}\,{k_1} = 16 - \frac{{2000}}{{2.303T}} \cr
& {\text{log}}\,{k_2} = 15 - \frac{{1000}}{{2.303T}} \cr
& {\text{At}}\,\,{k_1} = {k_2} \cr
& 16 - \frac{{2000}}{{2.303T}} = 15 - \frac{{1000}}{{2.303T}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,T = \frac{{1000}}{{2.303}}K \cr} $$