Question

The rate constant of a reaction is 2*10⁻³ min⁻¹ at 300 K temperature.By increase in temperature by 20 K, its value becomes three time; then calculate the energy of activation of the reaction.What will be its rate constant at 310 K temperature? Calculate the given example.

Answer 1

Explanation:

oh no I don't know this answer sorry

Answer 2

Answer:

using arrhenious equation

         k = - Ea / RT

         In k1/k2 = Ea/R(1/T1 - 1/T2)...........T1 = 300K & T2 = 320K

Putting values  

       In(5x10^-4)/(3x5x10^-4) = Ea/1.987(1/320 - 1/300)

Ea= 10478.12 cal/mol

rate constant at 37 degree C = 310 K

       In k  = - 10478.12/1.987x310

               = -17.01

            k = 4.09x10^-8

hope this helps u <3