8.3.1: Concentrations and Reaction Rates

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As we have seen as the probability of collisions between reactant molecules increases, the rate of reaction increases. In order to get information about the reaction mechanism we need to know the exact relationship between concentrations and rates. This can be done using a number of different techniques and experimental set-ups. But before we do that, we need to go over a few more terms. Recall that the rate of the reaction is the change in concentration of reactant per unit time. If the time interval is measurable and real, the rate we get is called the average rate (over that time interval), as shown in the earlier graph. If we imagine that the time interval drops to 0, we get the instantaneous rate, which is the slope of the tangent to the concentration versus time curve at a given time (more calculus). The rate at the beginning of the reaction can be obtained by taking the tangent at the start of the reaction (t = 0). This initial rate is useful in many situations because as the reactants form products, these products can interfere with or inhibit the forward reaction. This is particularly true in biological systems, where a product may influence its own formation. For example, it can bind to a site on the enzyme that catalyzes the reaction. This type of interaction is common, and often inhibits the enzyme’s activity (a form of feedback regulation).

We can measure the initial rate for a reaction using different initial concentrations of reactants. Using an appropriate experimental design, we can figure out how the rate of the reaction varies with each reactant. For many common reactions, the relationship between the rate and the concentration is fairly straightforward. For example, in the reaction:

(CH₃)₃CBr + ^–OH + Na⁺ ⇄ (CH₃)₃COH + Br^– + Na⁺,

the rate is dependent only on the concentration of t-butyl bromide [(CH₃)₃CBr], not on the

concentration of the sodium ion [Na⁺] or the hydroxide ion [^–OH]. “But why only the t-butyl bromide?” you might well ask. We will get to that point shortly, because it gives us some very interesting and important insights into the reaction mechanism. First, let us delve into a bit more background.

Because the rate is directly proportional to the [(CH₃)₃CBr], we can write the relationship between rate and concentration as: rate ∝ [(CH₃)₃CBr], or we can put in a constant (k) to make the equation:

rate = k[(CH₃)₃CBr]

We could also write

–Δ[(CH₃)₃CBr] /Δt = k[(CH₃)₃CBr],

or if we let the time interval drop to zero,

–d[(CH₃)₃CBr] /dt = k[(CH₃)₃CBr].

In all these forms, the equation is known as the rate equation for the reaction. The rate equation must be experimentally determined. It is worth noting that you cannot write down the rate equation just by considering the reaction equation. (Obviously, in this case, ^–OH or Na⁺ do not appear in the rate equation.) The constant (k) is known as the rate constant and is completely different from the equilibrium constant (K_eq). The fact that they are both designated by k (one lower case and one upper case) is just one of those things we have to note and make sure not to confuse. A rate equation that only contains one concentration is called a first-order rate equation, and the units of the rate constant are 1/time.

Now, in contrast to the first-order reaction of methyl bromide and hydroxide, let us compare the reaction of methyl bromide with hydroxide:¹⁵⁷

CH₃Br + ^–OH + Na⁺⇄ CH₃OH + Br^– +Na⁺,

For all intents and purposes, this reaction appears to be exactly the same as the one discussed on the previous page. That is, the bromine that was bonded to a carbon has been replaced by the oxygen of hydroxide.¹⁵⁸ However, if we run the experiments, we find that the reaction rate depends on both the methyl bromide concentration [CH3Br] and on the hydroxide concentration [–OH]. The rate equation is equal to k [CH3Br] [–OH]. How can this be? Why the difference? Well, the first thing it tells us is that something different is going on at the molecular level; the mechanisms of these reactions are different.

Reactions that depend on the concentrations of two different reactants are called second- order reactions, and the units of k are different (you can figure out what they are by dimensional analysis). In general:

rate = k [A] first order
rate = k [A][B] second order (first order in A and first order in B)
rate = k [A]² second order (in A)
rate = k [A]²[B] third order (second order in A and first order in B).

There are a number of methods for determining the rate equation for a reaction. Here we will consider just two. One method is known as the method of initial rates. The initial rate of the reaction is determined for various different starting concentrations of reactants. Clearly, the experimental design is of paramount importance here. Let us say you are investigating our reaction A + B ⇄ 2AB. The rate may depend on [A] and/or [B]. Therefore, the initial concentrations of [A] and [B] must be carefully controlled. If [A] is changed in a reaction trial, then [B] must be held constant, and vice versa (you cannot change both concentrations at the same time because you would not know how each one affects the rate).

The method of initial rates requires running the experiment multiple times using different starting concentrations. By contrast, the graphical method involves determining the rate equation from only one run of the reaction. This method requires the collection of a set of concentration versus time data (the same data that you would collect to determine the rates). Ideally we would like to manipulate the data so that we can obtain a linear equation (y = mx + b). For example, if we have a set of [A] versus time data for a reaction, and we assume the reaction is first order in A, then we can write the rate equation as: -d[A]/dt =k[A].

Now, if we separate the variables [A] and t to get: -d[A]/[A] =kt. We can then integrate the equation over the time period t = 0 to t = t to arrive a

ln [A]_t = –kt + [A]₀.¹⁵⁹

You will notice that this equation has the form of a straight line; if we plot our data (ln [A] versus t) and if the reaction is first order in [A], then we should get a straight line, where the slope of the line is –k. We can do a similar analysis for a reaction that might be second order in [A]:

rate = k[A]².

In this case, we can manipulate the rate equation and integrate to give the equation:

1/[A]_t = kt + 1/[A]₀.

Therefore, plotting 1/[A] versus t would give a straight line, with a slope of k, the rate constant. This method of analysis quickly becomes too complex for reactions with more than one reactant (in other words, reactions with rates that depend on both [A] and [B]), but you can look forward to that in your later studies!

Order	Rate Law	Integrated Rate Law	Graph for Straight Line	Slope of Line
0	Rate = k	[A]_t =-kt + [A]₀	[A] vs. t	–k
1	Rate = k[A]	ln[A]_t =-kt + ln[A]₀	ln [A] vs. t	–k
2	Rate = k[A]²	1/[A]_t = kt + 1/[A]₀	1/[A] vs. t	k

The two approaches (multiple runs with different initial conditions and the graphical method finding the best line to fit the data) provide us with the rate law. The question is, what does the rate law tell us about the mechanism? We will return to this question at the end of this chapter.

Questions to Answer

It turns out that most simple reactions are first or second order. Can you think why?
Design an experiment to determine the rate equation for a reaction 2A + B ⇄ C. Using the method of initial rates and a first experimental run using 0.1-M concentrations of all the reactants, outline the other sets of conditions you would use to figure out what that rate equation is.
What is the minimum number of runs of the reaction that you would have to do?
How would you determine the rate for each of your sets of conditions?
Now imagine you have determined that this reaction 2A + B ⇄ C does not depend on [B]. Outline a graphical method you could use to determine the rate equation. What data would you have to collect? What would you do with it?

Questions for Later

Why do you think it is that we cannot just write the rate equation from the reaction equation?
Why do you think that the most common rate equations are second order?

References

¹⁵⁷ In fact, this reaction has a number of different products. For now we will concentrate on this one.

¹⁵⁸ We call these kinds of reactions substitution reactions because one group has been substituted for another. In fact, they are also nucleophilic substitution reactions, because the hydroxide is acting as a nucleophile here.

¹⁵⁹ It is not necessary to be able to follow this mathematical reasoning; it is included to show where the equation comes from.