At room temperature $(20^{\circ} C)$ orange juice gets spoilt in about 64 hours. In a refrigerator at $3^{\circ} C$ juice can be stored three times as long before it gets spoilt. The activation energy of the reaction that causes the spoiling of juice is :

43.46 k J m o l^{- 1}

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46.76 k J m o l^{- 1}

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49.54 k J m o l^{- 1}

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50.23 k J m o l^{- 1}

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Solution

The correct option is A $43.46 k J m o l^{- 1}$
The Arrhenius expression is $k = A e^{- E a / R T}$
At two different temperatures, this expression takes the form $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
Substitute values in the above expression.
$log 3 = \frac{E_{a}}{2.303 \times 8.314} [\frac{1}{276} - \frac{1}{293}]$
Thus, the activation energy is $E_{a} = 43.46 k J m o l^{- 1}$ .

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At room temperature (20∘C) orange juice gets spoilt in about 64 hours. In a refrigerator at 3∘C juice can be stored three times as long before it gets spoilt. The activation energy of the reaction that causes the spoiling of juice is :

The correct option is A 43.46 kJmol−1The Arrhenius expression is k=Ae−Ea/RTAt two different temperatures, this expression takes the form logk2k1=Ea2.303R[1T1−1T2]Substitute values in the above expression.log3=Ea2.303×8.314[1276−1293]Thus, the activation energy is Ea=43.46 kJmol−1.

At room temperature $(20^{\circ} C)$ orange juice gets spoilt in about 64 hours. In a refrigerator at $3^{\circ} C$ juice can be stored three times as long before it gets spoilt. The activation energy of the reaction that causes the spoiling of juice is :