What is the activation energy for a reaction whose rate constant doubles when temperature changes from $30^{o} C$ to $40^{o} C$ ?

72.56 K J m o l^{- 1}

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45.45 K J m o l^{- 1}

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54.66 k J m o l^{- 1}

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None of these

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Solution

The correct option is A $54.66 k J m o l^{- 1}$
Let $K_{1} = K$
then $K_{2} = 2 K$
$T_{1} = 30^{0} C + 273 = 303 K$ $T_{2} = 40^{0} + 273 = 313 K$
$log (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$o r, log (\frac{2 K}{K}) = \frac{E_{a}}{2.303 X 8.3} [\frac{313 - 303}{313 X 303}]$
$E_{a} = 54571 K J / m o l = 54.5 K J / m o l \approx 54.66 K J / m o l$

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m o l^{- 1}

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