Question

What would happen to the rate of a reaction with rate law `rate = k[NO]^2[H_2]` if the concentration of `H_2` were halved?

Answer 1

If we assume that `Delta[NO] = 0`, then the rate is halved. Try making up numbers and performing the calculation.

Let `["NO"] = "0.2 M"` and `["H"_2] = "0.5 M"`, and let `k = "2"` `1/("M"^2*"s")`.

`color(green)(r(t)) = 2*0.2^2*0.5`

`color(green)"= 0.04 M/s"`

Okay, now do the same thing when `[H_2] = "0.25 M"`. Can you predict what you get without writing out any actual math?

What does that tell you about the order of `H_2` in the reaction? What, then, is the order of `NO` in the reaction? How do you know?