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College Chemistry: Help and Review15 chapters | 207 lessons

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Lesson Transcript

Instructor:
*LaRita Williams*

LaRita holds a master's degree and is currently an adjunct professor of Chemistry.

In this lesson, we will give the equation for the second order integrated rate law. We'll discuss its graph and then set up and solve a simple example problem.

Today's technology makes it possible for almost anything to happen fast. Breaking news can be at your fingertips with the click of a button. Full meals can be microwaved in a just a couple of minutes. Cross-country travel might only take a few short hours. In a society obsessed with getting things done quickly, why not also be concerned with the speed of your chemical reactions? The investigation of kinetics allows us to do just that. **Kinetics** is the study of the rate at which chemical reactions occur.

In kinetics, reactions are classified according to their order. The **order of a reaction** is the experimentally determined exponent to which each reactant concentration must be raised in the differential rate law equation. If a reaction is second order with respect to a certain reactant, then that reactant is being raised to a power of 2 in its differential rate law equation.

For the generic reaction you can see below (Equation 1), where reactant A goes to products B and C, the differential rate law is given. Note, that in the differential rate law equation, the concentration of reactant A is raised to a power 2, indicating that this reaction is second order with respect to A.

The differential rate law models how the rate of a reaction depends on the concentrations of species in that reaction. On the other hand, the **integrated rate law** shows how the concentrations of species in a reaction depend on time.

An integration of the **second order** differential rate law that we looked at a moment ago gives us the second order integrated rate law equation, shown on your screen below:

In this equation, Asub0 represents the initial concentration of reactant A; *t* is the variable for time; *k* is the rate constant of the reaction, and A represents the concentration of reactant A at time, *t*. If the values of *k* and Asub0 are known, we can calculate the concentration of A at any given time, *t*, thus modeling how the concentration of that reactant is affected by time.

Look carefully, and you'll notice that the equation for the second order integrated rate law is in the form *y* = *m**x* +*b*, where *y* = 1/A; *m* = *k*; *x* = *t*; and *b* = 1/Asub0. We learned in other lessons that *y* = *m**x* +*b* is the equation for a line. So, if we plot our *x* and *y* coordinates from the equation (*t* and 1/A), we will produce the graph of a straight line. Moreover, the slope of that line (*m*) will be equal to the rate constant of our reaction (*k*).

Let's try an example problem.

The second order reaction given a moment ago in Equation 1 is set up with A at a 5.0 M concentration and is allowed to react for over 500 minutes. When the data from the first 500 minutes is plotted, the slope of the resulting line is found to be 8.0 x 10^-3 M^-1Min^-1. What will the concentration of A be at 525 minutes?

Since we know the reaction is second order, if we want to solve for the relationship between reactant concentration and a given time, we must use the second order integrated rate law equation. When we plug in our known/given values, the slope of the line = *k* = 8.0 x 10^-3 M^-1Min^-1, *t* = 525 minutes, and Asub0 = 5, we get the following set-up:

From there, we just have to work out the math and solve for A, which turns out to be 0.23 M.

To sum up, **kinetics** is the study of the rate at which chemical reactions occur and the **order of a reaction** is the experimentally determined exponent to which each reactant concentration must be raised in the differential rate law equation. A reaction is **second order** when its reactant (or sum of its reactants) is being raised to a power of 2 in the differential rate law equation. The integration of that second order differential rate law gives the second order integrated rate law.

The equation for the second order integrated rate law takes the form *y* = *m**x* +*b*, where *y* = 1/A; *m* = *k*; *x* = *t*; and *b* = 1/Asub0. Thus, the graph of the second order integrated rate law is a straight line when 1/A is plotted against time, *t*.

When the equation for the second order integrated rate law is plotted, the slope of the line (*m*) is equal to the rate constant of the reaction (*k*). Once *k* and Asub0 are known, we can calculate the concentration of A at any given time, *t*.

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College Chemistry: Help and Review15 chapters | 207 lessons

- Rate of a Chemical Reaction: Modifying Factors 8:44
- Rate Constant and Rate Laws 6:35
- Rate of a Chemical Reaction: Effect of Temperature 4:00
- Activation Energy and Catalysts 5:29
- Reaction Mechanisms and The Rate Determining Step 4:34
- Second Order Integrated Rate Law 5:06
- Go to Kinetics: Help and Review

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