Question

Calculate the activation energy for a reaction of which rate constants becomes 4 times when temperature changes from ${30}^{0}C$ to ${50}^{0}C$ [R= 8.314 $Jmo{l}^{-1}{K}^{-1}$]

Answer 1

$ln\frac{k^{\prime }}{k}=\frac{E_a}{R}\times \left[\frac{T^{\prime }-T}{T{T}^{\prime }}\right]$

$T=30{}^{o}C=30+273K=303K$

$T^{\prime }=50{}^{o}C=50+273K=323K$

$ln\frac{k^{\prime }}{k}=\frac{E_a}{R}\times \left[\frac{T^{\prime }-T}{T{T}^{\prime }}\right]$

$ln\frac{4k}{k}=\frac{E_a}{8.314J/mol/K}\times \left[\frac{323-303}{303\times 323}\right]$

$1.386=2.46\times {10}^{-5}\times E_a$
$E_a=56400J/mol$
$E_a=56.4kJ/mol$
because $1kJ=1000J$
Note: Since rate constant becomes 4 times, $k^{\prime }=4k$