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)

$48.6 k J m o l^{- 1}$

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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 D $53.6 k J m o l^{- 1}$
As per Arrhenius equation:

$l o g \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.3 R} [\frac{T_{2} - T_{1}}{T_{2} T_{1}}]$

$2.3 l o g 2 = \frac{E_{a}}{8.314} [\frac{10}{300 \times 310}]$

$∴ E_{a} = 53.6 k J m o l^{- 1}$

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

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:

Q. The rate of a reaction doubles when its temperature changes from

300 K

310 K

. Activation energy of such a reaction will be:

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

and

log 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 particular reaction doubles when temperature changes from

27^{o} C

37^{o} C

. 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 300 K to 310 K. Activation energy of such a reaction will be: (R=8.314JK−1 mol−1 and log2=0.301) (IIT-JEE-2013)

The correct option is D 53.6kJmol−1As per Arrhenius equation: logK2K1=Ea2.3R[T2−T1T2T1] 2.3log2=Ea8.314[10300×310] ∴Ea=53.6kJmol−1

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)

The correct option is D $53.6 k J m o l^{- 1}$
As per Arrhenius equation:

$l o g \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.3 R} [\frac{T_{2} - T_{1}}{T_{2} T_{1}}]$

$2.3 l o g 2 = \frac{E_{a}}{8.314} [\frac{10}{300 \times 310}]$

$∴ E_{a} = 53.6 k J m o l^{- 1}$