Question

For a gaseous reaction, following data is given:
$A\to B, k_1= 10^{15}e^{-2000/T}$
$C \to D, k_2 = 10^{14}e^{-1000/T}$
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)

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