Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from?

Answer 1

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C ? (R = 8.314 Jmol-1K-1) Option 1)

Answer 2

This is a incomplete question. The complete question is:

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^0$ to $35^0$ (R=8.314J/molK)

The activation energy for a reaction if its rate doubles when the temperature is raised from $20^0$ to $35^0$ is 79846.6 J/mole

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 $20^oC$ = k[/tex]

$K_2$ = rate constant at $35^oC$ = $2k$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/moleK

$T_1$ = initial temperature = $20^oC=273+20=293K$

$T_2$ = final temperature = $35^oC=273+35=308K$

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

$\log (\frac{2k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{308K}]$

$Ea=79846.6J/mole$

Learn More about Arrhenius equation

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