What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{\circ} C$ to $35^{\circ} C$ ? ( $R = 8.314 J m o l^{- 1} K^{- 1}$ )

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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15.1 k J m o l^{- 1}

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Solution

The correct option is C $34.7 k J m o l^{- 1}$
$log \frac{K_{2}}{K_{1}} = - \frac{E_{a}}{2.303 R} [\frac{T_{1} - T_{2}}{T_{1} . T_{2}}]$

$log 2 = - \frac{E_{a}}{2.303 \times 8.314} [\frac{293 - 308}{293 \times 308}]$

$E_{a} = \frac{0.301 \times 2.303 \times 8.314 \times 293 \times 308}{15}$

$= 34.67 k J m o l^{- 1} \approx 34.7 k J m o l^{- 1}$

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Similar questions

20 ° C to 35 ° C

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

(Given: l o g 2 = 0.3010)

20^{o}

35^{o}

= 8.314

K^{- 1}

m o l^{- 1}

20^{0} C

35^{0} C

m o l^{- 1} K^{- 1}

20^{\circ} C

35^{\circ} C

20^{o} C

35^{o} C

(R = 8.134 J m o l^{- 1} K^{- 1})

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