Practice problems for these concepts can be found at:

Thus far, only cases in which instantaneous data are used in the rate expression have been shown. These expressions allow us to answer questions concerning the speed of the reaction at a particular moment, but not questions about how long it might take to use up a certain reactant, etc. If changes in the concentration of reactants or products over time are taken into account, as in the **integrated rate laws**, these questions can be answered. Consider the following reaction:

Assuming that this reaction is first order, then the rate of reaction can be expressed as the change in concentration of reactant *A* with time:

and also as the rate law:

Setting these terms equal to each other gives:

and integrating over time gives:

where ln is the natural logarithm, [*A*]_{0} is the concentration of reactant A at time = 0, and [*A*]_{t} is the concentration of reactant *A* at some time t.

If the reaction is second order in *A*, then the following equation can be derived using the same procedure:

Consider the following problem: Hydrogen iodide, HI, decomposes through a second–order process to the elements. The rate constant is 2.40 × 10^{–21}/M s at 25°C. How long will it take for the concentration of HI to drop from 0.200 M to 0.190 M at 25°C?

Answer:

1.10 × 10^{20} s. In this problem, *k* = 2.40 × 10^{–21}/M s, [*A*]_{0} = 0.200 M, and [*A*]_{1} = 0.190 M. You can simply insert the values and solve for t, or you first can rearrange the equation to give *t* = [1/[*A*]_{t} – 1/[*A*]_{0}]/*k*. You will get the same answer in either case. If you get a negative answer, you interchanged [*A*]_{t} and [*A*]_{0}. A common mistake is to use the first–order equation instead of the second–order equation. The problem will always give you the information needed to determine whether the first–order or second–order equation is required.

The order of reaction can be determined graphically through the use of the integrated rate law. If a plot of the ln[*A*] versus time yields a straight line, then the reaction is first order with respect to reactant *A*. If a plot of versus time yields a straight line, then the reaction is second order with respect to reactant *A*.

The reaction **half–life**, *t*_{1/2}, is the amount of time that it takes for a reactant concentration to decrease to one–half its initial concentration. For a first–order reaction, the half–life is a constant, independent of reactant concentration, and can be shown to have the following mathematical relationship:

For second–order reactions, the half–life does depend on the reactant concentration and can be calculated using the following formula:

This means that as a second–order reaction proceeds, the half–life increases.

Radioactive decay is a first–order process, and the half–lives of the radioisotopes are well documented.

Consider the following problem: The rate constant for the radioactive decay of thorium–232 is 5.0 × 10^{–11}/year. Determine the half–life of thorium–232.

Answer: 1.4 × 10^{10} yr.

This is a radioactive decay process. Radioactive decay follows first–order kinetics. The solution to the problem simply requires the substitution of the k–value into the appropriate equation:

*t*_{l/2} = 0.693/*k* = 0.693/5.0 × 10^{–11} yr^{–1} = 1.386 × 10^{10}yr

which rounds (correct significant figures) to the answer reported.

Consider another case: Hydrogen iodide, HI, decomposes through a second–order process to the elements. The rate constant is 2.40° × 10^{–21}/M s at 25°C. What is the halflife for this decomposition for a 0.200 M of HI at 25°C?

Answer: 2.08° × 10^{21} s.

The problem specifies that this is a second–order process. Thus, you must simply enter the appropriate values into the second–order half–life equation:

*t*_{1/2} = 1/*k*[*A*]_{0} = 1/(2.40 × 10^{–21}/M s)(0.200 M) = 2.08333 × 10^{21} seconds

which rounds to the answer reported.

If you are unsure about your work in either of these problems, just follow your units. You are asked for time, so your answer must have time units only and no other units.

Practice problems for these concepts can be found at:

View Full Article

From 5 Steps to a 5 AP Chemistry. Copyright © 2010 by The McGraw-Hill Companies. All Rights Reserved.