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 $l o g_{10} 2 = 0.301)$

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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48.6 k J m o l^{- 1}

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Solution

The correct option is C $53.6 k J m o l^{- 1}$
$∵ \frac{r_{2}}{r_{1}} = \frac{k_{2}}{k_{1}} = 2$
Now $2.303 l o g_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$2.303 l o g_{10} 2 = \frac{E_{a}}{8.314} \times [\frac{310 - 300}{310 \times 300}]$
$∴ E_{a} = \frac{2.303 \times 0.301 \times 8.314 \times 310 \times 300}{10} = 53598.6 J$
$= 53.6 k J m o l^{- 1}$

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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:

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

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=8.314JK−1mol−1 and log102=0.301)

The correct option is C 53.6kJmol−1∵r2r1=k2k1=2Now 2.303log10k2k1=EaR[T2−T1T1T2]2.303log102=Ea8.314×[310−300310×300]∴Ea=2.303×0.301×8.314×310×30010=53598.6J=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:
$(R = 8.314 J K^{- 1} m o l^{- 1}$ and $l o g_{10} 2 = 0.301)$