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

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

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Solution

It is given that T1 = 298 K

T2 = (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 k1 = k and that of k2 = 2k

Also, R = 8.314 J K−1 mol−1

Now, substituting these values in the equation:

`log  k_2/k_1 = E_a/(2.303R) [(T_2-T_1)/(T_1T_2)]`

We get

`log  "2k"/k = E_a/(2.303xx8.314)[10/(298xx308)]`

`=>log 2 = E_a/(2.303xx8.314)[10/(298xx308)]`

`=> (E_a = 2.303xx8.314xx298xx308xxlog2)/10`

= 52897.78 J mol−1

= 52.9 kJ mol−1

Concept: Rate of a Chemical Reaction
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APPEARS IN

NCERT Class 12 Chemistry
Chapter 4 Chemical Kinetics
Q 8 | Page 116

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