How much faster would a reaction proceed at $25^{0}$ C than at $0^{0}$ C if the activation energy is $65$ kJ?

2

times

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16

times

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11

times

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6

times

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Solution

The correct option is C $11$ times
$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g \frac{k_{2}}{k_{1}} = \frac{65000}{2.303 \times 8.314} (\frac{1}{273} - \frac{1}{298})$

$\frac{k_{2}}{k_{1}}$ = $11$

So, It is 11 times faster.

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Similar questions

How much faster would a reaction proceed at $25^{\circ} C$ than at $0^{\circ} C$ if the activation energy is 65 kJ?

25^{o}

0^{o}

65

25^{\circ} C

0^{\circ} C

65 kJ

R = 8.3 {J K}^{- 1} {mol}^{- 1}, 10^{1.043} = 11

65 k J m o l^{- 1}

60^{o}

30^{o}

(log 5 = 0.69990)

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