Question

For the two gaseous reactions, following data.
$A\to B;k_1={10}^{10}{e}^{-20,000/T}$
$C\to D;k_2={10}^{12}{e}^{-24,606/T}$
the temperature at which $k_1$ becomes equal to $k_2$ is:

• A. $400$K
• B. $1000$K
• C. $800$K
• D. $1500$K
• E. $500$K

Answer 1

The correct option is B $1000$K

Given, $k_1={10}^{10}{e}^{-20,000/T}$.

$k_2={10}^{12}{e}^{-24,606/T}$

$k_1=k_2\Rightarrow {10}^{10}{e}^{-20,000/T}={10}^{12}{e}^{-24,606/T}$

$e^{-\frac{20,000}{T}+\frac{24606}{T}}={10}^2$

$e^{\frac{4606}{T}}={10}^2$

On taking log both sides,

$\frac{4606}{2.303T}=log{10}^2$

$2log10\times T=\frac{4606}{2.303}$

$T=\frac{4606}{2.303\times 2}=\frac{4606}{4.606}=1000$K.

Option B is correct.