Question

A certain reaction id 50% complete in 20 minutes at 300 K and the same reaction
is again 50% complete in 5 minutes at 350 K. Calculate the activation energy if is
a first order reaction. Given R = 8.314 JK-1 mol-1, log4 = 0.602.
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Answer 1

Answer:

Explanation:

Given A certain reaction id 50% complete in 20 minutes at 300 K and the same reaction

is again 50% complete in 5 minutes at 350 K. Calculate the activation energy if is

a first order reaction. Given R = 8.314 JK-1 mol-1, log4 = 0.602.

For first order reaction T ½ = 0.693 / K

We know that T1 = 300 K

So K1 = 0.693 / T ½  

         = 0.693 / 20 x 60

        = 0.693 / 1200

      = 0.0005775

     = 5.775 x 10^-4 /s

Now at T2 = 350 K

K2 = 0.693 / 5 x 60

      = 0.693 / 300

     = 2.31 x 10^-3 / s

We know that

Log (K2/K1) = Ea[(1/T1) – 1/T2)] / R

Log (2.31 x 10^-3) / (5.775 x 10^-4) x R = Ea (1/300 – 1/350)

Log 4 x R = Ea(1/2100)

0.602 x R = Ea(1/2100)

0.602 x 8.314 = Ea (1/2100)

Ea = 10.51 KJ/mol

So activation energy will be 10.51 KJ/mol