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

Answer 1

Answer: (D)

$ k 1=10^{16} e ^{-2000 T } $
$k 2=10^{15} e ^{-2000 T } $
By taking log on both of them
$ \Longrightarrow \log k 1=\log 10^{16}-\frac{2000}{2.303} t $
$\Longrightarrow \log k 2=\log 10^{15}-\frac{1000}{2.303} t $
Now, $k 1= k 2$, Hence from equations $T =\frac{1000}{2.303} k$