IIT JEE past question Kinetics 7

The time required ofr 10% completion of a first order reation at 298 K is equal to that required for its 25% completion at 308 K.

If pre exponential factor for the reaction is 3.56 x 10⁹ s^-1, calculate its rate constant at 318 K and also the energy of activation.

1 Answers

k1.t= 2.303log10/9

k2.t = 2.303log4/3

k1/k2 = log10/9 / log4/3 ---------#1

again, k1/k2=e^{-Ea/R[1/T1 -1/T2]} -------------- #2

comparing 1 and 2 and putting the temperatures we get Ea.

now from the arrhenius eqn,

k=Ae^-Ea/RT, we get the rate constant at 318K.

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