Question

what is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K?(R=8.314)​

Answer 1

Answer:

I think it is 320 Kelvin

Answer 2

The activation energy for the reaction is 2.29766 kJ

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $200K$ = k

$K_2$ = rate constant at $400K$ = 2k

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 200 K

$T_2$ = final temperature = 400 K

Now put all the given values in this formula, we get :

$\log (\frac{2k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{200K}-\frac{1}{400K}]$

$Ea=J/mole=2297.66J/mol=2.29766kJ$

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