Here's is what I understand so far and the following is referenced directly from UNSW CHEM1061 lecture notes by Professor Pall Thordarson:

  • order of the reaction SHOULD be equal to the number of molecules of that reactant involved in each collision ( however this is not always the case)

  • it is only possible to predict the order of a reaction is it occurs in a single step (elementary reactions)

  • reaction order is experimentally derived from rate equation

so I understand all of the above but what I fail to fully comprehend is that actual DEFINITION of 'reaction order' what is it in it's entirety, what does it's value represent if it only sometimes correlates with molecularity (no. of molecules which react in an elementary reaction) ? If someone can answer this question could they please provide a qualitative answer that explains the value conceptually? Or is it just a value to relate concentration to reaction rate? that helps us work out more information?


1 Answer 1


Very Shortly, Rate of reaction is directly proportional to product of the active masses of reactants raised to some power which is Order of the Reaction.

Consider an elementry chemical reaction
$$\ce{2A(g) + B(s) -> C }$$
As reaction is elementry, stoichiometric coefficient and order will be same. The order of the reaction will be 2 (as the other component is solid, therefore will not affect concentrations). The rate law for reaction will be given as
$${R = K[A]^2}$$
where K is reaction constant. [A] is concentration of A.

If the order of the reaction would be 0, then Rate would be equal to reaction constant itself as the rate law would be given as
$${R = K[A]^0}$$

This explains the maximum use of Order of the reaction and also what order of reaction is.
Edit: for more wide explaination on this, see wikipedia. (Same explaination is provided but a widely expanded.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.