The rate of a reaction doubles when its temperature changes from $300 K$ to $310 K$ . Activation energy of such a reaction will be : $(R = 8.314 J K^{- 1} m o 1^{- 1}$ and $log 2 = 0.301)$

48.6 k J m o l^{- 1}

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58.5 k J m o l^{- 1}

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60.5 k J m o l^{- 1}

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53.6 k J m o l^{- 1}

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Solution

The correct option is C $53.6 k J m o l^{- 1}$
As per Arrhenius equation:
$ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{2.303 . R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
$\frac{k_{2}}{k_{1}} = 2$ ; $T_{2} = 310 K$ , $T_{1} = 300 K$
$l o g (2) = \frac{- E_{a}}{2.303 \times 8.314} (\frac{1}{310} - \frac{1}{300})$
$E_{a} = 53598 J = 53.6 k J / m o l$

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The rate of a reaction doubles when its temperature changes form 300 K to 310 K. Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

$(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

(IIT-JEE-2013)

Q. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

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$(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

(IIT-JEE-2013)

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. Calculate the energy of activation for such reaction.

[R = 8.314 J K^{- 1} m o l e]

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The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be :(R=8.314JK−1mo1−1 and log2=0.301)

The correct option is C 53.6 kJmol−1As per Arrhenius equation:lnk2k1=−Ea2.303.R(1T2−1T1)k2k1=2 ; T2=310K, T1=300Klog(2)=−Ea2.303×8.314(1310−1300)Ea=53598 J=53.6 kJ/mol

The rate of a reaction doubles when its temperature changes from $300 K$ to $310 K$ . Activation energy of such a reaction will be : $(R = 8.314 J K^{- 1} m o 1^{- 1}$ and $log 2 = 0.301)$