Lesson 12. 3 The Integrated Rate Law
Suggested Reading- Zumdahl Chapter 12 Sections 12.4 & 12.5
- What is the integrated rate law and how is it different from the differential rate law?
- Analyze data from rate experiments.
- Determine rate laws and rate constants.
- Calculate the half-life of a first or second order reaction.
The rate laws we have looked at so far express rate as a function of reactant concentration. These types of rate laws are known as differential rates, although they are usually just referred to as the rate law. The integrated rate law, which is based on the differential rate, gives a mathematical expression of how a reactant concentration changes over a period of time. Thus, the integrated rate law can be used to predict concentrations at different times. Using the integrated rate law we can answer questions such as "How long does it take for this reaction to be 50% complete?" Also, as you will see in lab, plotting concentration vs. time data provides an alternative to the initial rate method of determining rate laws (I suggest you pay close attention to this part of the lesson, as this will be one of the more complex labs we do). Calculus is used to transform the differential rate law into the integrated rate law, but since we will only work with the final equation the derivations will not be shown.
Concentration-Time Equations
The Integrated First-Order Rate Law
Let us first look at the first-order law. Let A be a substance that reacts to give products according to the equation
aA → products
where a is the stoichiometric coefficient of reactant A. Suppose that this reaction has a first-order rate law
Using calculus, you get the following equation, which is called the integrated first-order rate law.
where [A]t is the concentration of reactant A at time t, and [A]0 is the initial concentration. The ratio of[A]t/[A]0is fraction of reactant remaining at time t. Lets look at an example of how to use this equation.
Example: Using the concentration-time equation for a first-order reaction
The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.80 x 10-4/s at 45 ∘C.
a) If the initial concentration is 1.65 x 10-2 M, what is the concentration after 825 s?
b) How long would it take for the initial concentration to decrease to1.00 x 10-2M?
Solution:
a) This is basically plug and chug type work. Substituting gives
To solve for [N2O5], you take the antilogarithm of both sides (10x key). This removes the log from the left and gives antilog(-0.172), or 10-0.172 on the right, which equals 0.673.
b) You substitute into the concentration-time equation and solve.
The Integrated Second-Order Rate Law
Again, consider the reaction
aA → products
but suppose it has the second order rate law
Like before, using calculus, you obtain the following relationship between concentration and time for second order reactions.
This equation is applied in the same way as shown in the example for the integrated first-order rate law, but is used when the reaction is second order.
Half-Life Reaction
As a reaction proceeds, the concentration of a reactant decreases, because it is being consumed. The half-life, t1/2, of a reaction is the time it takes for the reactant concentration to decrease to one-half of its initial value. The concept of half-life is used to characterize a radioactive nucleus, whose decay is a first-order process. Carbon dating is a well known example.
For the general first-order rate law
the half-life is related to the rate constant, but is independent of the concentration of A. The equation derived from calculus is
This is the equation for half-life for first-order reactions. For second order reactions the equation is
In second-order reactions, the half-life does depend on the initial concentration and becomes larger as time goes on.
Example: Half-life of a first order reaction
Sulfuryl chloride, SO2Cl2, is a colorless, corrosive liquid whose vapor decomposes in a first-order reaction to sulfur dioxide and chlorine,SO2Cl2(g)→SO2(g)+Cl2(g). At 320 deg C, the rate constant is 2.20 x 10^-5/s.
a) What is the half-life of theSO2Cl2vapor at this temperature?
b) How long will it take for 50% of theSO2Cl2to decompose? How long for 75%?
Solution:
a)
b) The half-life is the time required for 50% of the vapor to decompose. From above we see this is 3.15 x 104 s or 8.75 hours. After another half-life, one half of the remaining vapor decomposes. The total decomposed at this point is 75%. The time required for two half-lives is 2 x 8.75 hours = 17.5 hours.
Graphing Kinetic Data
In the previous lesson you saw that the order of a reaction can be determined by comparing initial rates for several experiments in which different initial concentrations are used (initial-rate method). It is also possible to determine the order of a reaction by graphing the data for a particular experiment. The experimental data are plotted in several different ways, first assuming a first-order reaction, then a second-order, and so on. The order is determined from the graph that best fits the data, which is the one with the straightest line.
Lets take a closer look. You have learned that each reaction order has its own, differential rate law, integrated rate law, and half-life equation. Similarly, each reaction order has its own plot. The table 12.6 on page 548 of your textbook shows the relationships.
What this means is that for zero-order reactions, the plot of [A] vs t gives a straight line. For first-order reactions the plot of log [A] vs t gives a straight line, and for second-order reactions it is the plot of 1/[A] vs t. Thus, for a set of experimental data, you can determine the reaction order by creating these three plots. If your plot of [A] vs t give a straight line, your reaction is zero-order. If your plot of log [A] vs t results in a straight line your reaction is first-order, and if 1/[A] vs t is a straight line it is second order. Of the three plots, only one should result in a straight line. Get the picture? The textbooks tend to complicate this topic, so be careful.
In order to apply this in the lab, you must collect concentration versus time data. There are several methods, but we will use spectrophotometry. You will learn more about this method next semester. It is super fun!
Below are typical experimental results for a zero-order reaction:
A first-order reaction:
A second-order reaction
This work can be time consuming, but it is not hard. Unless, of course something goes wrong and none of your graphs results in a straight line! Take careful measurements!
HOMEWORK: Practice exercises 14.9, 14.10
Book questions page 278 questions 1, 3-6, 16 & 18