Rate of reaction!! please help me

http://www.xtremepapers.me/IB/Chemistry/Higher/2008%20May/Chemistry%20HL%20paper%202%20TZ2.pdf

when you open the link there is a question 1.

i am so confused about how to get the initial rate of reaction for experiment 5??
please help me please

ok.. let's start here..
for the reaction A ---. B + C..

rate = change in concentration of A per time.. agreed?

and in math lingo.. where [ ] means concentration
(1)...rate = - d[A] / dt

where.. d[A]/dt means change in concentration of A per time and it's - because it's decreasing.

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in addition.. rate is proportional to [A].. the more A you have.. the faster the rate. ie.. the higher the concentration.. the faster the rate. and if we put in a proportionality constant...k.. we get
rate = k x [A]

and sometimes the mechanism of reactions require 2 atoms of A to collide.. sometimes just one will decay on it's own.. so we put an exponent on that [A]
(2).. rate = k x [A]^n

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combining those to we get...
(3).. rate = -d[A] / dt = k x [A]^n

where "n" is called the "order" of the reaction.. n=0. zero order.. n=1.. first order.. n=2 2nd order..

and..that "n" must be determined experimentally. we change the concentration and monitor the rate of reaction. And that is what this table shows.

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for this particular problem.. you have two reactants... and we modify that equation (2) for multiple reactants, A&B like this.
rate = k x [A]^n x [B]^m

and in this case
rate = k x [H2]^n x [NO]^m

so let's determine n and m before we do anything else.

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from experiment 1 to 2 [NO] is constant.. concentration NO is constant.. so we can do this..
rate 1 = k x [H2 #1]^n x [NO]^m
rate 2 = k x [H2 #2]^n x [NO]^m

rearranging.
rate 1 / [H2 #1]^n = k x [NO]^m
rate 2 / [H2 #2]^n = k x [NO]^m