Order of reaction question - HELP!

how do you do this question? its question (5) (c) (i) on this chemstry of materials ocr salters past paper.

http://http://www.ocr.org.uk/Data/publications/past_papers_2007_january/A_GCE_Chemistry_Salters_2849_01_January_2007_Question_Paper.pdf

i understand the concept of order of reaction, but these values in the table mean absolutely nothing to me. NO is second order, CO is zero and O2 is zero, but i dont understand why!

Thank you!

Reply 1

EierVonSatan

21

Experiment 1 -> 2 CO value doubles, everything else is constant - rate doesn't change => rate = k[2CO]⁰

Experiment 1 -> 3 CO value doubles (doesn't affect rate since zero order), NO doubles, O2 is constant - rate increases by 4rate = k[2NO]^x => x = 2 (since 2² = 4)

Experiment 3 -> 4 O2 value doubles, everything else constant - rate doesnt change => rate = k[2O2]⁰

:smile:

Reply 2

hodgey90

OP

8

EierVonSatan

Experiment 1 -> 2 CO value doubles, everything else is constant - rate doesn't change => rate = k[2CO]⁰

Experiment 1 -> 3 CO value doubles (doesn't affect rate since zero order), NO doubles, O2 is constant - rate increases by 4rate = k[2NO]^x => x = 2 (since 2² = 4)

Experiment 3 -> 4 O2 value doubles, everything else constant - rate doesnt change => rate = k[2O2]⁰

:smile:

oh i seeeeeee, that makes so much more sense! so you just see where the concentration of the reactants changes between the experiments, and then just look at what happens to the rate of reaction at the same time. if it stays the same, its zero order. if the rate doubles, first order. if the rate quadruples, its second order. right?

thank you

Reply 3

EierVonSatan

21

hodgey90

oh i seeeeeee, that makes so much more sense! so you just see where the concentration of the reactants changes between the experiments, and then just look at what happens to the rate of reaction at the same time. if it stays the same, its zero order. if the rate doubles, first order. if the rate quadruples, its second order. right?

thank you