The integrated rate law is very useful when determining the reaction order and rate constant, particularly given that it can utilize all the data collected from a single experimental trial. However, thus far we have only considered example reactions involving one reactant, whereas most reactions involve two or more reactants. How can we use integrated rate laws to find the reaction order and rate constant for those reactions?
Let’s consider a generic reaction:
Flooding refers to running a reaction that involves two or more reactants with a large excess of all but one reactant. For example, we can run the reaction above with a large excess of reactant B so that reactant A is the limiting reactant by a significant amount. In that case, [B]0 >> [A]0, and the concentration of B would effectively remain constant during the course of the reaction. For example, if [B]0 = 0.10 M and [A]0 = 0.0010M, then at a time t when all the reactant A has reacted, [B]t = (0.10 – 0.0010) M = 0.10 M (0.099 M if extend the significant figures), which is essentially no change in the concentration of B.
Under this condition of [B]t ≈ [B]0 = constant, the rate law becomes:
where kobs, the rate constant we observe during the flooded experiment, is kobs = k[B]0n. This new rate equation allows us to use the integrated rate laws we have just discussed to determine the reaction order with respect to [A], and also to determine kobs. The order of the reaction, m, is called a pseudo order because it is obtained under flooding conditions and is not necessarily the overall order of the reaction. If m = 1, we say the reaction is pseudo-first-order and kobs is called a pseudo-first-order rate constant.
Here is an example experiment:
For trial 1, flood the reaction mixture with a large excess of reactant B ([B]01) and measure [A]t as the reaction progresses. Then plot ln[A]t vs t. If the resulting graph is linear, then the reaction is first-order with respect to [A]. (In this case, the flooded reaction a pseudo-first-order reaction.) The slope of this graph is –kobs1.
For trial 2, flood the reaction mixture with a different large excess of B ([B]02 ≠ [B]01). The plot of ln[A]t vs t has a different slope corresponding to –kobs2.
The ratio of the two kobs allows us to determine n, the reaction order with respect to [B]:
In the above equation, all the variables aside from n are known or have been experimentally determined.
Finally, the actual rate constant for the reaction, k, can be determined from the relationship kobs = k[B]0n. The data from all trials should be averaged to get the best value of k.
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