If we double the concentration of NO alone in the reaction: 2NO+O2—2NO2, the reaction rate increases 4 times.

If we double the concentration of both NO and O2, the reaction rate increases 8 times. What is the kinetic law of the reaction?

A) v = K[NO]²[O2]

B) v = K[NO]²[O2]²

C) v = K[NO]²

D) v = K[NO][O2]/[NO2]²

E) v = K[NO][O2]

Hi!

The answer for this question is A).

I thought the answer would be simply v = k[A]^(stochio number) * [B]^(stochio number) which gives A). Is this correct?

I’m confused, how do we find the exponents by looking at the changes of reaction speed?

Thanks!

hey

What you are saying is correct and conventional, and this is what we learned at school. However, everything gets complicated once you dig deeper so this case is not an exception

Basically, “stochio number formula” does not always work and in this type of questions you should kinda work it out by yourself. How? The exercise should give you a clue:

- initially rate=k[NO]^x*[O2]^y ;

if [NO] doubled becoming [2*NO], rate would increase 4 times (4r=k[2*NO]^x*[O2]^y) thus it’s clear that x=2 (which is “by accident” same as stochio number:)

2)second hint the exercise give will be used to find O2 stochio number y (which is also “by accident” same as stochio number), 8r=[2*NO}^2*[2*O2]^y thus y should be 1

good luck

1 Like

Thank you for your answer it’s much clearer now!