Question

NO reacts with O₂ in the gas phase according to the following chemical equation:

2NO(g) +O₂(g) --------> 2NO₂(g)

It is observed that, when the concentration of O₂ is reduced to 1/4 of its initial value, the rate of the reaction is also reduced to 1/4 of its initial value. When the concentration of NO is multiplied by 9.56, the rate of the reaction increases by a factor of 91.4.


(a) Write the rate expression for this reaction, and give the units of the rate constant k, assuming concentration is expressed as mol L^-1 and time is in seconds.

rate =
k units =

(b) If [NO] were multiplied by 4.58 and [O₂] by 3.30, what change in the rate would be observed?

The rate would increase by a factor of ?   

Answer 1

a)
when the concentration of O2 is reduced to 1/4 of its initial value, the rate of the reaction is also reduced to 1/4 of its initial value
so, order of O2 is 1

When the concentration of NO is multiplied by 9.56, the rate of the reaction increases by a factor of 91.4
since 9.56^2 = 91.4
so, order of NO is 2

rate law is:
rate = k[O2][NO]^2

rate = k[O2][NO]^2
put units:
M/s = k * M * M^2
k = M-2.s-1

rate = k[O2][NO]^2
unit of K = M-2.s-1

b)
Given Rate law is:
rate = k[O2][NO]^2
rate new / rate old = ([O2]new/[O2]old)*([NO]new/[NO]old)^2
here:
[O2]new/[O2]old = 3.3
[NO]new/[NO]old = 4.58
putting values
rate new / rate old = (3.3)*(4.58)^2
rate new / rate old = 69.2221
Answer: 69.2 times