What is the significance of a reactant being zero, first, or second order when calculating the rate of a reaction?

In simple terms, we need to know whether a reactant is first, second or zero order if we want to understand what effect changing the concentration of that reactant has on the rate of the reaction.

Zero order= change in conc. has no effect on rate

1st Order= change in conc. causes same change to rate e.g. conc. x2 = rate x2

2nd Order= change in conc. causes the same change of rate to the power 2 e.g. conc x2 = rate x4

If I was to give an example for the reaction:

2X(g) + Y(g) --> A(g) + B(g)

The effect the concentration of each reactant has on the rate of the reaction is displayed through the rate equation, which includes the reaction constant (k).


K is constant at a constant temperature.

We can see from the rate equation that the only reactant that has an effect on the rate of the reaction is X, as it is the only one mentioned. This means the rest of the reactants are zero order and have no effect.

What we can also derive from the rate equation is that X is second order in relation to rate, as it has a next to it's concentration. This means that when the concentration of X is doubled, the rate of the reaction increases by a factor of 22 = 4.

If the rate equation was rate= k[X], when the concentration of X was doubled, the rate of the reaction would be doubled (if [X] was tripled, rate would be tripled etc). This is known as a first order relationship.

Naomi  B. A Level Chemistry tutor, A Level Physical Education tutor

2 months ago

Answered by Naomi , who has applied to tutor A Level Chemistry with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist


Ryan M. GCSE Biology tutor, A Level Biology tutor, GCSE Chemistry tut...
View profile
£20 /hr

Ryan M.

Degree: Medicine (Bachelors) - Birmingham University

Subjects offered: Chemistry, Science+ 4 more

English Literature
-Medical School Preparation-

“Hi! Thank you for visiting my profile. My name is Ryan and I am a third year Medical student at the University of Birmingham. My devotion to learning motivates me to provide the most positive, enjoyable and useful tuition I possibly c...”

MyTutor guarantee

Oliver W. A Level Chemistry tutor, GCSE Chemistry tutor, A Level Math...
View profile
£22 /hr

Oliver W.

Degree: Physical Natural Sciences (Chemistry) (Masters) - Cambridge University

Subjects offered: Chemistry, Maths


“Second year chemist, from the University of Cambridge, with several years experience tutoring in Chemistry and Core Mathematics.”

PremiumOsama H. Uni Admissions Test -Medical School Preparation- tutor, Ment...
View profile
£26 /hr

Osama H.

Degree: Medicine (MBBS/BSc) (Bachelors) - Imperial College London University

Subjects offered: Chemistry, Geography+ 4 more

.BMAT (BioMedical Admissions)
-Personal Statements-

“Hi I'm Osama, a medical student at Imperial College London. I'm happy to help with core subjects at GCSE, and Biology and Chemistry and A-level. I am also happy to help you get that coveted place at medical school!”

About the author

Naomi  B. A Level Chemistry tutor, A Level Physical Education tutor
View profile

Naomi B.

Currently unavailable: until 26/09/2016

Degree: Sport and Materials Science (Joint Honours) (Bachelors) - Birmingham University

Subjects offered: Chemistry, Physical Education

Physical Education

“Hello! I'd love to use this space to tell you a little more about myself. I am studying a Joint Honours degree in Sports Science and Materials Engineering at Birmingham University, where I play hockey for the 1st Team. I love that my ...”

MyTutor guarantee

You may also like...

Other A Level Chemistry questions

How does the reactivity of group 2 elements change down the group, and what is the cause of this trend?

Explain the trend in ionisation energy down group 2? (3)

What's the difference between an electrophile and a nucleophile?

How do mass spectrometers work?

View A Level Chemistry tutors


We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss