Answer to Question #148188 in Physical Chemistry for Sd

Question

The rate constant of second order reaction is 5.7 x 10-5 dm3 mol-1
s
-1 at 298 K and 1.64 x
10-4 dm3 mol-1
s
-1 at 313 K. Calculate the activation energy and Arrhenius-pre
exponential factor.

Answer 1

lnK= lnA- E_a/RT

Ln5.7×10^-5=lnA-E_a/ (298×8.314)

2477.572 (Ln5.7×10^-5) = 2477.572- E_a

- E_a=-24211.97-2477.572lnA…. (i)

Ln1.64×10^-4= lnA-E_a/ (313×8.314)

2602.282(Ln1.64×10^-4) = 2602.282lnA- E_a

- E_a=-22744.82-2602.282 lnA…. (ii)

-24211.97-2477.572lnA= -22744.82-2602.282 lnA

LnA= 11.76

A=31.979

- E_a=-24211.97-2477.572lnA

- E_a=-24211.97-2477.572(11.76)

E_a=53348.21672J/mol