Ncert Exercises & Solutions

4:8The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.

It is given that T $_{1}$ = 298 K
$∴$ T $_{2}$ = (298 + 10) K
= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k $_{1}$ = k and that of k $_{2}$ = 2k
Also, R = 8.314 J K $^{- 1}$ mol $^{- 1}$

Now, substituting these values in the equation:

$\log \frac{k_{2}}{k_{1}} = \frac{E}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

we get :

$\log \frac{2 K}{k} = \frac{E_{a}}{2.303 \times 8.314} [\frac{10}{298 \times 308}]$
$\Rightarrow \log 2 = \frac{E_{a}}{2.303 \times 8.314} [\frac{10}{298 \times 308}]$
$\Rightarrow E_{a} = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log 2}{10}$
$= 52897.78 J m o l^{- 1}$
$52.9 k j m o l^{- 1}$

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

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