For a substitution reaction, rate constant increases by factor $10.6$ when temperature changes from $25^{o} C$ to $35^{o} C$ . Calculate energy of activation of the reaction.

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Solution

$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Rate increases by a factor 10.6. Substituting the values,

$l o g 10.6$ = $\frac{E_{a}}{2.303 \times 8.314} (\frac{1}{298} - \frac{1}{308})$

$E_{a} = 180.18 k J {m o l}^{- 1}$

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For a substitution reaction, rate constant increases by factor 10.6 when temperature changes from 25oC to 35oC. Calculate energy of activation of the reaction.

log k2k1 = Ea2.303×R(1T1−1T2)Rate increases by a factor 10.6. Substituting the values,log 10.6 = Ea2.303×8.314(1298−1308)Ea=180.18 kJ mol−1

For a substitution reaction, rate constant increases by factor $10.6$ when temperature changes from $25^{o} C$ to $35^{o} C$ . Calculate energy of activation of the reaction.

$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Rate increases by a factor 10.6. Substituting the values,

$l o g 10.6$ = $\frac{E_{a}}{2.303 \times 8.314} (\frac{1}{298} - \frac{1}{308})$

$E_{a} = 180.18 k J {m o l}^{- 1}$