Are there some smart chemists here

Use the following data to determine the rate law for the reaction

2NO(g)+H2(g) => N2O(g)+H2O(g)

expt# [NO]o [H2]o Initial rate
# 1 .021M .065M 1.46M min^(-1)
#2 .021M .260M 1.46M min^(-1)
#3 .042M .065M 5.48M min^(-1)


Help and explain please!!!!!!!!!!!

Background information on rates:
Rate= k[NO]^m[H2]^n
m and n can either be an order of 0,1,2, or 3. These numbers work by powers of 2. This is how it looks:
0th order= the Initial rates divided by each other is 1( the rates are the same number)
1st order= the initial rates divided by each other is 2
2nd order= the initial rates divided by each other is 4
3rd order= the initial rates divided by each other is 8

What you are looking for in your rpoblem are the powers m and n.

Lets find m first.
To solve for m look for when H2 is constant. This means the 2 experiments when [H2] is the same number, which in this case is .065M, or exp# 1 and 3. Now just take the Initial rate laws at those 2 experiment numbers which are 1.46 and 5.48 and divide them and you get 5.48/1.46 which is about 4. From what I showed at the top of this page a 4 means that this is 2nd order so m=2

Now to solve for n just do the same thing:
Look for when NO is constant which is exp# 1 and 2, or .021M. So once again divide the Initial rate laws by each other and you get 1.46/1.46=1. From the information at the top of the page a 1 means that this is 0th order so n=0

Now just plug those numbers into the rate equation, Rate= k[NO]^m[H2]^n
Rate= k[NO]^2[H2]^0
Since anything to the power of 0 is 1, [H2]=1 so it isn't in the final rate law so the final answer is
Rate= k[NO]^2

I hope this helps.