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

60.5 k J m o l^{- 1}

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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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Solution

The correct option is B $53.6 k J m o l^{- 1}$
$0.3010 = \frac{E a}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$\Rightarrow 0.3010 = \frac{E a}{2.303 \times 8.314 \times 10^{- 3}} (\frac{310 - 300}{310 \times 300})$
$E a = 53.6 k J m o l^{- 1}$

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Similar questions

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 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

310 K

. Activation energy of such a reaction will be:

(R = 8.314 J K^{- 1} {m o l}^{- 1} a n d log 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:

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

and

l o g_{10} 2 = 0.301)

The rate of reaction doubles when its temperature changes from 300K to 310 K .activation energy of such a reaction will be (R=8.314 kj/ mol and WG2= 0.031 )

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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:

The correct option is B 53.6 kJ mol−10.3010=Ea2.303R[T2−T1T1T2] ⇒0.3010=Ea2.303×8.314×10−3(310−300310×300) Ea=53.6 kJ mol−1

The correct option is B $53.6 k J m o l^{- 1}$
$0.3010 = \frac{E a}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$\Rightarrow 0.3010 = \frac{E a}{2.303 \times 8.314 \times 10^{- 3}} (\frac{310 - 300}{310 \times 300})$
$E a = 53.6 k J m o l^{- 1}$