Question

A reaction has a rate constant of 2.08 × 10−4 s−1 at 26 oC and 0.394 s−1 at 79 oC . Determine the activation barrier for the reaction. Enter your answer numerically and in terms of kJ/mol

Answer 1

Answer: The activation energy of the reaction is 124.6 kJ/mol

Explanation:

To calculate activation energy of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{79^oC}$ = equilibrium constant at 79°C = $0.394s^{-1}$

$K_{26^oC}$ = equilibrium constant at 26°C = $2.08\times 10^{-4}s^{-1}$

$E_a$ = Activation energy of the reaction = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $26^oC=[26+273]K=299K$

$T_2$ = final temperature = $79^oC=[79+273]K=352K$

Putting values in above equation, we get:

$\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol$

Hence, the activation energy of the reaction is 124.6 kJ/mol