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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 30 of 39
Marks: +1, -0
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:  
Given r2=4r1,r_{2}=4 r_{1},
T1=293,T2=313 K,Ea=?T_{1}=293, T_{2}=313\ \mathrm{K}, E_{a}=?
We know, r2r1=k2k1=41\frac{r_{2}}{r_{1}}=\frac{k_{2}}{k_{1}}=\frac{4}{1}
Using Arrhenius equation,
log⁡k2k1=Ea2.303R(1T1−1T2)\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)
log⁡4=Ea2.303×8.314(1293−1313)\log 4 = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{313} \right)
or, Ea=log⁡4×2.303×8.314×293×31320E_{a} = \log 4 \times \frac{2.303 \times 8.314 \times 293 \times 313}{20}
or, Ea=52.864 kJE_{a}=52.864\ \mathrm{kJ}
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