What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{\circ} C$ to $35^{\circ} C$ ?

521 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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342 k J m o l^{- 1}

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Solution

The correct option is A $34.7 k J m o l^{- 1}$
$K_{2} = 2 K_{1}, T_{2} = 35 + 273 = 308$ $K, T_{1} = 20 + 273 = 293$ $K$

$ln \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$∴ ln 2 = \frac{E_{a}}{8.314} [\frac{1}{293} - \frac{1}{308}]$

$∴ \frac{8.314 \times ln 2}{(3.41 - 3.24) \times 10^{- 3}} = E_{a}$

$∴ E_{a} \approx 34.7$ $k J / m o l$

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Similar questions

20^{\circ} C

35^{\circ} C

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

20^{o} C

35^{o} C

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

20^{0} C

35^{0} C

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

20^{o}

35^{o}

= 8.314

K^{- 1}

m o l^{- 1}

20 ° C to 35 ° C

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

(Given: l o g 2 = 0.3010)

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