Question

The rate constant of a particular first order reaction is 5.45 x 10^-2 sec^-1 at 40.0 oC. What is the rate constant of this reaction at 65.0 oC if the energy of activation, Ea for this reaction is 65.5 kJ/mol

Answer 1

Answer: The rate constant for the reaction at 65°C is $0.350s^{-1}$

Explanation:

To calculate rate constant at 65°C of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{65^oC}}{K_{40^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{65^oC}$ = equilibrium constant at 65°C = ?

$K_{40^oC}$ = equilibrium constant at 40°C = $5.45\times 10^{-2}s^{-1}$

$E_a$ = Activation energy of the reaction = 65.5 kJ/mol = 65500 J/mol    (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $40^oC=[40+273]K=313K$

$T_2$ = final temperature = $65^oC=[65+273]K=338K$

Putting values in above equation, we get:

$\ln(\frac{K_{65^oC}}{5.45\times 10^{-2}})=\frac{65500J/mol}{8.314J/mol.K}[\frac{1}{313}-\frac{1}{338}]\\\\K_{65^oC}=0.350s^{-1}$

Hence, the rate constant for the reaction at 65°C is $0.350s^{-1}$