The Rate Expression and Orders of Reaction
Simplistically one would imagine that doubling the concentration of a reactant would double the rate of a reaction. This seems logical; there will be double the number of collisions. However, although this is often the case it is not always true. Indeed, sometimes doubling the concentration of a reactant could have no affect on the rate and sometimes it will increase the rate by more than double, for example the rate could be quadrupled.
How can this be so? The answer lies in the fact that reactions usually take place by a series of steps and one of the steps (the slowest one (rate determining step RDS)) will control the rate. Only reactants involved in this step will affect the rate. For reactants that are not involved in the RDS their concentration will have no affect on the rate. The order with respect to this reactant will be zero and this reactant will not appear in the rate expression. If a reactant is involved once in the RDS then doubling its concentration will cause the rate to double. It is said to be first order with respect to this reactant. The rate expression would be rate ≈ [conc] or Rate = k [conc]. It could be the case that a reactant is involved twice in the RDS, in which case, doubling its concentration will cause the rate to go quadruple i.e. go up by 4. The rate expression would be: Rate ≈[conc]2 or Rate = k[conc]2. K = the rate constant and the powers are the orders for each of the reactants. The overall order is simply the addition of all of the powers in the rate expression.
For example for the rate expression: Rate = k [BrO3-] [Br-]
What is the order with respect to each reactant?
What is the overall order?
What units will k have in this rate expression. Note that the units of k will vary depending on the expression.
Remember that the rate expression can only be determined experimentally by following the rate and changing one reactant at a time. The stoichiometry of a reaction tells you nothing about the rate expression or possible mechanism.
Deducing a Mechanism from the Rate Expression
This can be quite complex. However, simplistically if a reactant has zero order then it is not in the RDS but must be in a faster step. If a reactant is first order then it is directly or indirectly involved once in the RDS. If a reactant is second order then it’s either directly or indirectly involved twice in the RDS. Indirectly means that it could be involved with producing an intermediate, which is then involved in the RDS.
It can be fun proposing mechanisms that fit with the rate expression. There may well be several possibilities.
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