Rate constants of a reaction at 500 K and $700 \mid \mathrm{K}$ are $0.04 \mathrm{~s}^{-1}$ and $0.14…
Rate constants of a reaction at 500 K and $700 \mid \mathrm{K}$ are $0.04 \mathrm{~s}^{-1}$ and $0.14 \mathrm{~s}^{-1}$, respectively; then, activation energy of the reaction is :
(Given: $\log 3.5=0.5441, \mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
182310 J
18500 J
18219 J
18030 J
Solution
$\mathrm{K}=A e^{-E_a / R T}$
After taking In both side
$\ln K=\ln A-\frac{E_a}{R T}$
$\operatorname{In} \mathrm{K}_1=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_1}$ at temp. $\mathrm{T}_1$...(i)
$\operatorname{In} \mathrm{K}_2=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}$ at temp. $\mathrm{T}_2$,,(ii)
(ii) - (i)
$\operatorname{In} K_2-\operatorname{InK} K_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{500}-\frac{1}{700}\right]$
$\ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500 \times 700}\right]$
$\ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500 \times 700}\right]$
$\log 3.5=\frac{E_a}{2.303 \times R}\left[\frac{1}{250 \times 7}\right]$
$0.5441=\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{250 \times 7}\right]$
$E_a=0.5441 \times 8.31 \times 250 \times 7 \times 2.303$
$=0.5441 \times 83.1 \times 25 \times 7 \times 2.303$
$=18222.65$
$\approx 18219 \mathrm{~J}$