The rate of a reaction becomes four times when the temperature changes from $293 K$ to $313 K$ . Calculate the energy of activation $(E_{2})$ of the reaction assuming that it does not change with temperature.
$[R = 8.314 J / K m o l^{- 1}, log 4 = 0.6021]$

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Solution

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

$l o g 4 = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$E_{a} = 0.6021 \times 2.303 \times 8.314 \times 293 \times \frac{313}{20}$

$E_{a} = 52.863 k J$

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The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E2) of the reaction assuming that it does not change with temperature.[R=8.314 J/K mol−1,log4=0.6021]

logK2K1=Ea2.303R(1T1−1T2)log4=Ea2.303×8.314(1T1−1T2)Ea=0.6021×2.303×8.314×293×31320Ea=52.863kJ

The rate of a reaction becomes four times when the temperature changes from $293 K$ to $313 K$ . Calculate the energy of activation $(E_{2})$ of the reaction assuming that it does not change with temperature.
$[R = 8.314 J / K m o l^{- 1}, log 4 = 0.6021]$

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

$l o g 4 = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$E_{a} = 0.6021 \times 2.303 \times 8.314 \times 293 \times \frac{313}{20}$

$E_{a} = 52.863 k J$