Question

In the reaction \[\ce{2NO + O2 -> 2NO2}\], the rate law is rate = k[NO][O₂]².

1. How will the rate of reaction change if [NO] concentration is doubled and [O₂] concentration is halved at the same time?
2. Write the order of reaction if [NO] concentration is in large excess.

Answer 1

i. Let’s analyse the rate law:

rate = k[NO][O₂]²

If [NO] concentration is doubled, the rate will increase by a factor of 2.

If [O₂] concentration is halved, the rate will decrease by a factor of `(1/2)^2 = 1/4`.

Combining these effects, the overall change in rate will be:

`(2) xx (1/4) = 1/2`

So, the rate of reaction will decrease by half.

ii. If [NO] concentration is in large excess, its concentration will not significantly affect the rate of reaction. In this case, the rate law can be simplified to:

rate ≈ k[O₂]²

Since the rate depends on the square of [O₂] concentration, the order of reaction with respect to O₂ is 2.

Therefore, the overall order of reaction is 2.