Rate constants of a reaction at 500 K and 700 mid K are 0.04 ~s^-1 and 0.14 ~s^-1, respectively; then, activation energy of the reaction is : (Given:…

$\mathrm{K}=A e^{-E_a / R T}$K=Ae−Ea​/RT After taking In both side $\ln K=\ln A-\frac{E_a}{R T}$lnK=lnA−RTEa​​ $\operatorname{In} \mathrm{K}_1=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_1}$InK1​=lnA−RT1​Ea​​ at temp. $\mathrm{T}_1$T1​...(i) $\operatorname{In} \mathrm{K}_2=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}$InK2​=lnA−RT2​Ea​​ at temp. $\mathrm{T}_2$T2​,,(ii) (ii) - (i) $\operatorname{In} K_2-\operatorname{InK} K_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$InK2​−InKK1​=REa​​[T1​1​−T2​1​] $\ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{500}-\frac{1}{700}\right]$ln K1​K2​​=REa​​[5001​−7001​] $\ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500 \times 700}\right]$ln0.040.14​=REa​​[500×700700−500​] $\ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500 \times 700}\right]$ln414​=REa​​[500×700200​] $\log 3.5=\frac{E_a}{2.303 \times R}\left[\frac{1}{250 \times 7}\right]$log3.5=2.303×REa​​[250×71​] $0.5441=\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{250 \times 7}\right]$0.5441=2.303×8.31Ea​​[250×71​] $E_a=0.5441 \times 8.31 \times 250 \times 7 \times 2.303$Ea​=0.5441×8.31×250×7×2.303 $=0.5441 \times 83.1 \times 25 \times 7 \times 2.303$=0.5441×83.1×25×7×2.303 $=18222.65$=18222.65 $\approx 18219 \mathrm{~J}$≈18219 J