What is the activation energy for a reaction if the rate doubles when the temperature is raised from $20^{o} C$ to $35^{o} C$ ? $(R = 8.134 J m o l^{- 1} K^{- 1})$

15.1 k J m o l^{- 1}

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342 k J m o l^{- 1}

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269 k J m o l^{- 1}

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34.7 k J m o l^{- 1}

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Solution

The correct option is B $34.7 k J m o l^{- 1}$
Given : $T_{1} = 20 + 273 = 293 K$

$T_{2} = 35 + 273 = 308 K$

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

Substituting the values, we get

$l o g 2 = \frac{E_{a}}{2.303 \times 8.314} \times (\frac{1}{293} - \frac{1}{308})$

$E_{a}$ = $34.7 k J {m o l}^{- 1}$

Therefore, the option is D.

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20^{\circ} C

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R = 8.314 J m o l^{- 1} K^{- 1}

20^{0} C

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(R = 8.314 J m o l^{- 1} K^{- 1})

(Given: l o g 2 = 0.3010)

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