JEE Main · 2013 · 07 Apr 2013 · Offline · Chemical Kinetics

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:
(

R = 8.314 JK^{- 1} mol^{- 1}

and

lo g 2 = 0.301

)

Answer: (D)

Solution

Use the two-temperature Arrhenius relation,

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

. The rate doubles, so

k_{2} / k_{1} = 2

, with

T_{1} = 300 K

and

T_{2} = 310 K

. Switching to base-ten logs,

2.303 lo g 2 = \frac{E _{a}}{8.314} (\frac{1}{300} - \frac{1}{310})

, and putting

lo g 2 = 0.301

and solving gives

E_{a} \approx 53.6 kJ/mol

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option