Question

Calculate value of activation energy if rate of reaction doubles when rise in temp from 295k to 305k takes place.

Answer 1

$\large\underline{\bigstar \: \:{\sf Given-}}$

• Rate of activation Energy double
• Temperature rise from ${\sf (T_1)=295K\: to \: (T_2)\: = 305K}$

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Activation Energy ${\sf (E_a)}$

$\large\underline{\bigstar \: \:{\sf Formula \: Used -}}$

$\color{violet}\bullet\underline{\boxed{\sf log\dfrac{K_2}{K_1}=\dfrac{E_a}{2.303R}\left[\dfrac{T_2-T_1}{T_1T_2}\right]}}$

$\large\underline{\bigstar \: \: {\sf Solution -}}$

$\implies{\sf log2=\dfrac{E_a}{2.303R}\left[\dfrac{305-285}{295×305}\right] }$

R = Gas Constant = 8.314${\sf Jmol^{-1}K^{-1}}$

$\implies{\sf 0.30=\dfrac{E_a}{2.303×8.314}\left[\dfrac{20}{89975}\right]}$

$\implies{\sf E_a=\dfrac{0.30×2.303×8.314×89975}{20}}$

$\implies{\sf E_a=\dfrac{516829.2}{20} }$

$\implies{\sf E_a=25841.4J }$

$\color{red}\implies{\sf E_a=25.84kJ }$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Activation Energy is $\color{red}{\sf 25.84kJ}$.