# Specific Case of the Rate law

My textbook says that for a chemical reaction with two reactants, $\ce{A}$ and $\ce{B}$ with equation $\ce{aA + bB -> Products}$, where $a$ and $b$ are coefficients, the general rate law is:

$\text{Rate} = k[\ce{A}]^m[\ce{B}]^n$

where $m$ and $n$ are the reaction orders for $\ce{A}$ and $\ce{B}$ respectively. Then it says that $m=a$ and $n=b$ if the reaction between $\ce{A}$ and $\ce{B}$ occurs in a single step and with a single activated complex. Why is this true?

According to collision theory, for a chemical reaction to occur, the reacting molecules must collide with each other.

For a bimolecular reaction (as given by you), collision occurs in the single step and thus becomes the rate determining step, as there are no other steps involved.

If you consider reactions like bromide ions reacting with bromate ions in the presence of acid, it involves 12 reacting particles. If all the 12 reacting particles collide simultaneously you can write rate law expression considering it to be elementary, but those events are extremely rare. In reality such reactions are found to be complex i.e reactions which take place through a sequence of two or more consecutive steps. Here, there are many steps involved, and one should consider slowest step to write rate law expression. But, why? That will become another SE question. In case of reactions taking place in single step, as there are no other steps left to compare with, whether it is slower or not, so you need to consider the theoritical equation.

One more thing to be noted is that elementary reactions are associated with single transition state and not single activated complex. For more info read this page: Activated complex

• This does not tell why the relationship between a and m is that one is equal to the other. Why is it that if you have 2 particles of A and 1 particle of B reacting (mole ratio) then the rate law has the reaction order being 2 in A and 1 in B? – user11629 Mar 6 '14 at 21:57
• In elementary reactions, 2 particles of A and 1 particle of B collide simultaneously to form products. So, rate of disappearance of A must be twice as that of B. Suggesting that order of reaction being 2 in A and 1 in B. – Immortal Player Mar 7 '14 at 11:57
• Twice is not quadratic, which is what order 2 means, though. – user11629 Mar 7 '14 at 18:12
• Your question is not clear. – Immortal Player Mar 7 '14 at 18:15
• Basically I'm saying, why is it what you said 2 comments ago implies that order of reaction is quadratic. How does a directly proportional relationship in concentration lead to a quadratically proportional relationship in order? – user11629 Mar 7 '14 at 18:32

First some definitions:-
Elementary reaction= The reaction that takes place in a single step i.e one transition/activated complex according to the transition theory or one single collision event according to the Collision theory of kinetics.

Molecularity= Number of independent species involve in the collision event in case of an elementary reaction. These all collide together with sufficient energy (threshold) and proper orientation to convert themselves into products.

Now, let us assume the collision model for gaseous reaction where there are two species $A$ and $B$ which react together in a reaction according to :- $$\ce{aA + bB ->}\text{ products}$$

This reaction will occur whenever $a$ molecules of $A$ collide with $b$ molecules of $B$ in proper orientation. Suppose out of all collisions of $a$ molecules of $A$ with $b$ molecules of $B$, a fraction $P$ occur in proper orientation. The probability of collision of these specified number of reactants is proportional to the $a^{th}$ power of $[A]$ and $b^{th}$ power of $[B]$. This is because the concentration of the reactant gives crudely the probability of finding that reactant in a unit volume. For collision probability, you multiply the probability of finding one reactant to the probability of finding another reactant. Since there are $a$ molecules of $A$, therefore the probability of finding $A$ will be multiplied $a$ times and similarly for $B$. Thus the resultant rate based on collision probabilities will be :- $$k=PZ[A]^a[B]^b$$ Where Z is an appropriate constant (since the probablities are only proportional to the concentrations and not equal to them, there is some appropriate constant for each proportionality which finally multiplies to give $Z$). This is true only for elementary reactions adhering to the Collision theory-Mechanism of reaction.

• I understand this, except for the notion that probability of collision is proportional to the ath power of A. Could you please elaborate? – user11629 Mar 7 '14 at 18:30