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)

$\frac{1000}{2.303} K$

Answer 2

Given, $\, \, \, \, \, k_1=10^{16}.e^{-2000/T}$
\hspace15mm k_2=10^{15}.e^{-1000/T}
On taking log of both the equations we get
\hspace15mm log \, k_1=16-\frac{2000}{2.303 T}
\hspace15mm log \, k_2=15-\frac{1000}{2.303 T}
At $k_1=k_2$
\hspace10mm 16-\frac{2000}{2.303 T}=15-\frac{1000}{2.303 T}
\hspace30mm T=\frac{1000}{2.303} K