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Activation Energy

How much fast...

Question

How much faster would a reaction proceed at 298 K than at 273 K, if the activation energy is $65 k J m o l^{- 1}$ ?

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Solution

The correct option is C 11 times
$∵ k = A e^{- E a / R T}$
$⟹ \frac{k_{25}}{k_{0}} = \frac{A_{25} e^{65000 / 8.314 \times 298}}{A_{0} e^{65000 / 8.314 \times 275}}$
$⟹ \frac{e^{- 26.24}}{e^{- 28.64}}$
$⟹ e^{2.40} = 11$
Thus, the reaction will proceed 11 times faster.

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Similar questions

25^{0}

0^{0}

65

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

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

293

K

103

k J

{m o l}^{- 1}

273

K

7.87 \times 10^{- 7}

s^{- 1}

(R = 8.314 \times 10^{- 3} k J {m o l}^{- 1} K^{- 1})

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