Looking for classes? Ksquare Career Institute, Bengaluru →
Rate constants of a reaction at 500 K and 700 K are 0.04 $s^{-1}$ and 0.14 $s^{-1}$, respectively; then, activation energy of the reaction is: (Given: $\log3.5=0.5441$, $R=8.31\ J\,K^{-1}mol^{-1}$)
A
182310 J
B
18500 J
C
18219 J
D
18030 J
Detailed Solution
$\ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\left[\dfrac{1}{T_1}-\dfrac{1}{T_2}\right]$. $\ln\dfrac{0.14}{0.04}=\log3.5\times2.303=0.5441\times2.303$. So $0.5441\times2.303=\dfrac{E_a}{8.31}\left[\dfrac{1}{500}-\dfrac{1}{700}\right]=\dfrac{E_a}{8.31}\times\dfrac{1}{1750}$. Solving, $E_a=0.5441\times2.303\times8.31\times1750\approx18219$ J.