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Arrhenius Equation

The specific ...

Question

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

55.33 k J m o l e^{- 1}

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64.67 k J m o l e^{- 1}

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68.34 k J m o l e^{- 1}

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70.03 k J m o l e^{- 1}

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Solution

The correct option is A $55.33 k J m o l e^{- 1}$
As we know,
$k = A e^{- E_{a} / R T}$
$k_{1} = A e^{- E_{a} / R * 300}$
$k_{2} = A e^{- E_{a} / R * 320} = 4 k_{1}$
by solving
$E_{a} = 55.33 k J /$ mole

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