The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate *E*_{a}.

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#### Solution

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} = 2*k*

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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