Question

For a gaseous reaction, following data is given:
$A \, \rightarrow \, B, \, k_{1}=10^{15} \, e^{- 2000 / T}$
$C \, \rightarrow \, D, \, k_{2}=10^{14} \, e^{- 1000 / T}$
The temperature at which $k_{1}=k_{2}$ is :

• A. $1000 \, K$
• B. $2000 \, K$
• C. $868.82 \, K$
• D. $434.2 \, K$

Answer 1

Answer: (D)

When $k_{1}=k_{2}$ ,
$10^{15} \, e^{- 2000 / T}=10^{14} \, e^{- 1000 / T}$
$10=e^{1000 / T}$
$2.303 \, log \, 10=\frac{1000}{T}$
$T=434.2 \, K$