Question

Kinetics

For H2 +I2 ->2HI

10³k (1/⁰Cs)      0.54               2.5                   14                    25                    64

T (K)               599                  629                  666                  683                  700

Find E_a and A

Answer 1

from the data take first two rate constant values and first two temperatures.

when two tempareture are given the Arrehenius equation

K1-------------- rate constant at T1 temperature (low T) = 0.54

K2-------------- rate constant at T2 temperature (high T) = 2.5

T1= 599K

T2=629K

log(k2/k1) =( Ea/2.303R)[1/T1-1/T2]

     log (2.5/0.54) =( Ea/2.303x 8.314 x10^-3)[1/599-1/629]

Ea= 160 kJ/mol

another equation

K= A . e ^-Ea/RT

take first rate constant and temperature

0.54 = A . e ^-160/8.314 x10^-3 x599

A= 4.87 x10¹³

A= 4.87 x10¹³ rate constant units