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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 38 of 39
Marks: +1, -0
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea_a.
Solution:  
Given :T1=298 K,T2=308 K,: T_{1}=298\ \mathrm{K}, T_{2}=308\ \mathrm{K},
Ea=?,r1=rE_{a}=?, r_{1}=r
(( say )) then r2=2rr_{2}=2r
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]
Putting the values log⁡21=Ea2.303×8.314[1298−1308]\log \frac{2}{1} = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{1}{298} - \frac{1}{308} \right]
0.3010=Ea2.303×8.314[10298×308]0.3010 = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]
or,
Ea=0.3010×2.303×8.314×298×30810E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}
or, Ea=52.897 kJ mol−1E_{a}=52.897\ \mathrm{kJ}\ \mathrm{mol}^{-1}
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