Question

The rate constants $k_1$ and $k_2$ for two different reactions are ${10}^{15}{e}^{-200/T}$ and ${10}^{16}{e}^{-1000/T}$ respectively.
The temperature at which $k_1=k_2$ is:

• A. $\frac{2000}{2.303}K$
• B. $\frac{800}{2.303}K$
  
• C. $\frac{1000}{2.303}K$
• D. $1000K$

Answer 1

Answer: (B)

$\frac{800}{2.303}K$

Given, $k_1={10}^{15}{e}^{-200T};$ $k_2={10}^{16}{e}^{-1000T}$

According to Arrhenius equation;

$k=A{e}^{-\frac{-E_a}{RT}}$

$\ln k=\ln A-\frac{E_a}{RT}$

For $k_1=k_2$

${10}^{15}{e}^{\frac{-200}{T}}={10}^{16}{e}^{\frac{-1000}{T}}$

$e^{800T}=10$

$800T=\ln 10$

$\Rightarrow 800T=2.303$

$\Rightarrow T=\frac{800}{2.303}K$