Question

The rate constants $k_1$ and $k_2$ for two different reactions are ${10}^{16}{e}^{-2000/T}$ and ${10}^{15}{e}^{-1000/T}$ respectively. The temperature at which $k_1=k_2$ is:

• A. $2000$ $K$
• B. $\frac{1000}{2.303}$ $K$
• C. $1000$ $K$
• D. $\frac{2000}{2.303}$ $K$

Answer 1

The correct option is B $\frac{1000}{2.303}$ $K$

${10}^{15}{e}^{-1000/T}={10}^{16}{e}^{-2000/T}$

$\frac{e^{-2000/T}}{e^{-1000/T}}=\frac{{10}^{15}}{{10}^{16}}$

$e^{-1000/T}={10}^{-1}$

${\log }_e{e}^{-1000/T}={\log }_e{10}^{-1}$

$\frac{-1000}{T}=2.303{\log }_{10}{10}^{-1}=-2.303$

$T=\frac{1000}{2.303}K$