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

$[R = 8.314 J K^{- 1} m o l^{- 1}$ and $log 2 = 0.301]$

53.6 k J m o l^{- 1}

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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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Solution

The correct option is A $53.6 k J m o l^{- 1}$
Let,
$K_{1} =$ Rate of reaction at 300 K temperature
$K_{2} =$ Rate of reaction at 310 K temperature
$Δ E_{a c t} =$ Activation energy

$log (\frac{K_{2}}{K_{1}}) = \frac{- Δ E_{a c t}}{2.303 K} (\frac{1}{310} - \frac{1}{300})$

$log (2) = \frac{- Δ E_{a c t}}{2.303 \times 8.314} (\frac{- 10}{300 \times 310})$ $(∵ K_{2} = 2 K_{1})$

$∴ Δ E_{a c t} = 53.59 K J m o l^{- 1}$

Hence, the correct option is $A$

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Similar questions

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:

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

and

l o g_{10} 2 = 0.301)

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

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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=8.314 JK−1mol−1 and log2=0.301]

The correct option is A 53.6 kJ mol−1Let,K1=Rate of reaction at 300 K temperatureK2= Rate of reaction at 310 K temperatureΔEact= Activation energylog(K2K1)=−ΔEact2.303K(1310−1300)log(2)=−ΔEact2.303×8.314(−10300×310) (∵K2=2K1)∴ΔEact=53.59KJmol−1Hence, the correct option is A

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

$[R = 8.314 J K^{- 1} m o l^{- 1}$ and $log 2 = 0.301]$