Question

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

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

Answer 1

Answer: (B)

Given $k_1={10}^{10}{e}^{-20,000/T}$
$k_2={10}^{12}{e}^{-24,606/T}$
$k_1=k_2$
${10}^{10}{e}^{-20,000/T}={10}^{12}{e}^{-24,606/T}$
$e^{\frac{-20,000}{T}+\frac{24,606}{T}}={10}^2$
$e^{\frac{4,606}{T}}={10}^2$
On taking log both sides
$\frac{4606}{2.303T}=\log {10}^2$
$2\log 10\times T=\frac{4606}{2.303}$
$T=\frac{4606}{2.303\times 2}$
$=\frac{4606}{4.606}=1000K$