Rate law for reaction of NO and H2

[i_love_science]

Homework Statement

Determine the rate law for this reaction given the data below.

2 NO + 2 H2 -> N2 + 2H2O

Reaction 1:
Initial [NO] = 0.20
Initial [H2] = 0.15
Initial rate = 2.0 * 10^(-4)

Reaction 2:
Initial [NO] = 0.40
Initial [H2] = 0.15
Initial rate = 8.0 * 10^(-4)

Reaction 3:
Initial [NO] = 0.40
Initial [H2] = 0.30
Initial rate = 1.6 * 10^(-3)

I think the rate law is rate = k[NO][SUP]2[/SUP][H2], since when you double the NO concentration, you quadruple the rate.

The solution is rate = k[NO][H2]. Is this wrong? Thanks.

Relevant Equations

rate law

My solution and question are in the homework statement due to some formatting issues. Thanks.

 

[mjc123]

Unless there's an error in the question, you are right. There could easily be a misprint in the solution.