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

$53.6 k j m o l^{- 1}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$58.5 k j m o l^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$60.5 k j m o l^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$53.6 k j m o l^{- 1}$

From Arrhenius equation, $l o g \frac{k_{2}}{k_{1}} = \frac{- E_{α}}{2.303 R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
Given, $\frac{k_{2}}{k_{1}} = 2 T_{2} = 310 k$
$T_{1} = 300 k$
On putting values,
$\Rightarrow l o g 2 = \frac{- E_{α}}{2.303 \times 8.314} (\frac{1}{310} - \frac{1}{300})$
$\Rightarrow E_{α} = 53598.6 J / m o l = 53.6 k J / m o l$

Suggest Corrections

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:

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

300 K

310 K

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

log 2 = 0.301]

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

l o g_{10} 2 = 0.301)

27^{o} C

37^{o} C

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

Join BYJU'S Learning Program