Question

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O3(g)+NO(g)⟶O2(g)+NO2(g) O3(g)+NO(g)⟶O2(g)+NO2(g) The rate law for this reaction is rate of reaction=k[O3][NO] rate of reaction=k[O3][NO] Given that k=4.09×106 M−1⋅s−1k=4.09×106 M−1⋅s−1 at a certain temperature, calculate the initial reaction rate when [O3][O3] and [NO][NO] remain essentially constant at the values [O3]0=5.84×10−6 M[O3]0=5.84×10−6 M and [NO]0=8.65×10−5 M,[NO]0=8.65×10−5 M, owing to continuous production from separate sources.

Answer 1

Initial rate of reaction is
`2.07* 10^(-3)M.s^(-1)`.

Step-by-step explanation:

It is a second order reaction.

Initial rate of reaction =
`k[O_(3)]_(0)[NO]_(0)` , where k is rate constant,
`[O_(3)]_(0)` is the initial concentration of
`O_(3)` and
`[NO]_(0)` is the initial concentration of NO.

Here, k =
`4.09* 10^(6)M^(-1).s^(-1)`,
`[O_(3)]_(0)=5.84* 10^(-6)M` and
`[NO]_(0)=8.65* 10^(-5)M`

So, initial rate of reaction =
`(4.09* 10^(6)M^(-1).s^(-1))* (5.84* 10^(-6)M)* (8.65* 10^(-5)M)`

=
`2.07* 10^(-3)M.s^(-1)`

So, initial rate of reaction is
`2.07* 10^(-3)M.s^(-1)`