4.8     The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K Calculate E_{a}.

Answers (1)
M manish

Given data 

T_{1}(initial temperature) = 298K and T_{2} (final temperature)= 308K

And we know that rate of reaction is nearly doubled when temperature rise 10-degree

So, \frac{k_{2}}{k_{1}}=2 and R = 8.314 J/mol/K

now, \log\frac{k_{2}}{k_{1}}=\frac{ E_{a}}{2.303}[\frac{T_{2}-T_{1}}{T_{1}T_{2}}]

On putting the value of given data we get,

  \log2=\frac{ E_{a}}{2.303}[\frac{10}{298\times 308}]

Activation energy (E_{a}) = \frac{2.303\times 8.314\times 298\times 308\times \log2}{10}Jmol^{-1}

                                      =52.9 KJ/mol(approx)

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