Question

Ozone gas, O3 decomposes into O2 in presence of NO and gaseous O atoms The following data was obtained Determine the experimental rate law for the reaction given the following data and the value for the rate constant

O 3= 2.3 x 10^-4 atm NO = 1.6 x 10^-4 O = 1.5 x 10^-4 rate1 = 2.6 x 10-3

O3 = 2.3 x 10^-4 NO = .8 x 10^-4 O = 1.5 x 10^-6 rate 2 = 1.3 x 10^-3

O 3 = 4.6 x a0^-4 NO = 1.6 x 10^-4 O = 1.5 x 10^-6 rate 3 = 5.2 x 10^-3

O3 = 4.6 x 10^-4 NO = 1.6 x 10^-4 O = 3.0 x 10^-6 rate 4 = 5.2 x 10^-3

Answer 1

Let the rate law of the reaction be

rate = K [03]^a [N0]^b [0]^c

here

k is rate constant

a is order with respect to 03

b is order with respect to NO

c is w.r.t 0

now

conisder condition 1

rate1 = K[03]^a [N0]^b [0]^c

2.6 x 10-3 = K [2.3 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c

now

consider condition 2

rate2 = K[03]^a [N0]^b [0]^c

1.3 x 10-3 = K [2.3 x 10-4]^a [0.8 x 10-4]^b [1.5 x 10-4]^c

now

conisder condition 3

rate3 = K[03]^a [N0]^b [0]^c

5.2 x 10-3 = K [4.6 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c

now

conisder condition 4

rate4 = K[03]^a [N0]^b [0]^c

5.2 x 10-3 = K [4.6 x 10-4]^a [1.6 x 10-4]^b [3 x 10-4]^c

from the above obtained equations

now divide condition 1 by condition 3

2.6 x 10-3 / 5.2 x 10-3 = K [2.3 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c / K [4.6 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c

on solving we get

a =1

now divide condition1/ condition 2

we get

2.6 x 10-3 / 1.3 x 10-3 = K [2.3 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c / 1.3 x 10-3 = K [2.3 x 10-4]^a [0.8 x 10-4]^b [1.5 x 10-4]^c

solving we get

b=1

now divide condition 3 / condition 4

5.2 x 10-3 / 5.2 x 10-3 = K [4.6 x 10-4]^a [1.6 x 10-4]^b [1.5 x 10-4]^c / 5.2 x 10-3 = K [4.6 x 10-4]^a [1.6 x 10-4]^b [3 x 10-4]^c

on solving we get

c=0

so the rate law is given by

rate = K [03]^1 [N0]^1 [0]^0

so the rate law is

rate = K [03] [N0]

now

consider condition 1

2.6 x 10-3 = K [ 2.3 x 10-4 ] [1.6 x 10-4 ]

K = 7.065 x 10^4

so

the rate constant = 7.065 x 10^4