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Arrhenius Equation

The specific ...

Question

The specific rate constant for a reaction increases by a factor of 4, if the temperature is change from $27^{\circ} C$ to $47^{\circ} C$ . Find the activation energy for the reaction?

43.46 k J m o l^{- 1}

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5.533 k J m o l^{- 1}

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27.9 k J m o l^{- 1}

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None of these

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Solution

The correct option is A $43.46 k J m o l^{- 1}$
The activation energy
$E_{a} = \frac{2.303 R T T^{'}}{T^{'} - T} l o g \frac{k^{'}}{k}$
$E_{a} = \frac{2.303 \times 8.314 \times 300 \times 320}{320 - 300} l o g \frac{4}{1}$
$E_{a} = 43457 J / m o l = 43.46 k J / m o l$

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Similar questions

27^{\circ}

47^{\circ}

27^{\circ} C

47^{\circ} C

55.776 k J m o l^{- 1}

l n 2 = 0.7, R (g a s c o n s t) = 8.3 J / (m o l . K)^{- 1})

20^{\circ} C

55.776 k J m o l^{- 1}

l n 2 = 0.7

R = 8.3 J m o l^{- 1} K^{- 1})

27^{\circ} C

47^{\circ} C

55.776 k J m o l^{- 1}

l n 2 = 0.7, R (g a s c o n s t) = 8.3 J / (m o l . K)^{- 1})

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52^{\circ} C

k J m o l^{- 1}

l n 2 = 0.7

\frac{25}{3} J / m o l K

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