D15.5 Flooding Method: Pseudo-Order Reaction
The integrated rate law can be 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:
where:
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 we 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]:
![\dfrac{k_{obs_2}}{k_{obs_1}} = \dfrac{k([B]_{0_2})^n}{k([B]_{0_1})^n} = \left(\dfrac{[B]_{0_2}}{[B]_{0_1}}\right)^n](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-52bd48545236f716bd5b605f658e8dc6_l3.png "Rendered by QuickLaTeX.com")
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 laboratory trials should be averaged to get the best value of k.
Activity: Determining a Rate Constant Using Flooding Method
The three graphs below represent data collected at 25 °C for the reaction of methyl acetate with strong base (OH−). The products of the reaction are acetate ion and methanol. In your notebook, write an equation for the reaction.
![Three graphs are shown. Left: [ methyl acetate ] (M) vertical, time (s) horizontal; graph curves from upper left to lower right. Middle: l n [ methyl acetate ] vs time (s); linear with negative slope. Right: 1 / [ methyl acetate ] vs time (s); curves up.](https://wisc.pb.unizin.org/app/uploads/sites/727/2021/06/methyl_acetate_flooding.png)
The initial concentrations of methyl acetate and hydroxide ion were 0.0098 M and 1.00 M, respectively. The linear graph obeys the equation
y = -0.014x – 4.654
Use this information to determine the order of the reaction with respect to methyl acetate.
When the reaction was repeated with [OH−] = 2.00 M, the initial rate was twice the initial rate when [OH−] = 1.00 M. Determine the rate law for the reaction and the rate constant.
Write in your notebook, then left-click here for an explanation.
The equation for the reaction is:
CH3COOCH3(aq) + OH−(aq) ⟶ CH3COO−(aq) + CH3OH(aq)
Assume that the rate law has the form:
rate = k[CH3COOCH3]m[OH−]n
The concentration of hydroxide ions is more than 100 times the concentration of methyl acetate, so the concentration of hydroxide ions would change by less than 1% by the time the reaction is complete. Thus, the reaction is flooded with hydroxide ions and kobs = k[OH−]n can be obtained from the data.
The linear graph plots ln[methyl acetate] vs time. Thus, the data imply that the integrated rate law is
ln[methyl acetate]t = −kobst + ln[methyl acetate]0
which is an integrated first-order equation. Thus, m = 1 and the reaction is first-order in methyl acetate.
The slope of the graph is −kobs, so
−kobs = −0.014 s−1 and kobs = 0.014 s−1
Because the initial rate doubles when the concentration of hydroxide ions doubles, the reaction must be first-order in hydroxide ions. Thus, n = 1. Solve for k in the equation kobs = k[OH−]n and substitute data from the experiment where [OH−] = 1.00 M to calculate the rate constant.
![k = \dfrac{k_{obs}}{[\text{OH}^-]^1} = \dfrac{0.014 \;\text{s}^{-1}}{1.00 \;\text{M}} = 0.014 \frac{1}{\text{M} \cdot \text{s}}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-22b24794a5acca9e219d88ece0aeff13_l3.png "Rendered by QuickLaTeX.com")
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