## Section 3: Chemical Kinetics and Rate Laws

What is Chemical Kinetics?
Reaction Rate
Practice Problem
Basics
Why Use Rate Laws?
1st, 2nd, 3rd Order

#### What is Chemical Kinetics?

Chemical kinetics is the study of how reactions take place and what the steps are that a reaction goes through. In chemical reactions, it is important to understand what is reacting and how it is reacting. You also must understand the spontaneity of reactions and what factors play a role in the rate of a reaction. This is what the field of chemical kinetics is all about.

#### Reaction Rates

In chemical kinetics, scientists use a term called reaction rate to describe how fast or slow a reaction occurs. Reaction rate is defined as the change in concentration of a reactant or product per unit of time or mathematically ...

Rate = the change in [A] / the change in time

(Please not that if the reaction rate is negative then the substance is being used up.)

#### Practice Problem

If you were given the reaction, H2 + Cl2 -----> 2HCl, and you were given the below information:

Time (s)[H2] mol/L[Cl2] mol/L
01.52
25.751

What is the reaction rate for hydrogen gas? and what is the reaction rate for chlorine gas?

Rate (H2) = change in H2 / change in time = .75 - 1.5 / 25 - 0 = -.03 mol / L * s Rate (Cl2) = change in Cl2 / change in time = 1 - 2 / 25 - 0 = .04 mol / L * s

In rate law reactions, the coefficients can be used to shad some light on the rate of reaction and can also be used as a check step. For example, in the reaction above, H2 + Cl2 = 2HCl, we can tell by using the coefficients that H2 and Cl2 should be used up at the same rate and HCl should be produced twice as fast as H2 and Cl2 are used up. (Note the practice problem above, is ficticious, that is why this rule doesn't work for that problem.)

#### Basics

In the previous section, we discussed rate laws, and defined the rate law as:

Rate = change in [A] / change in time

but now we are going to also define it as:

Rate = k [A]n
where k is the proportionality constant and n is the order

K, the proportionality constant, can also be called the rate constant. Both, the rate constant and the order must be determined experimentally.

#### Why Use Rate Laws?

We use rate laws because rate laws can be helpful in determining the steps that a reaction goes through. Determining what these steps are is the ultimate goal of the chemists in this field.

#### 1st, 2nd, 3rd Order

To determine order, a table of experimentally measured concentrations and times must be presented. Below is an example of the data that must be presented to you to solve the question of what order the reaction is.

Reaction:
BrO3- + 5Br- + 6H+ -----> 3 Br2 + 3H2O

Experiment[BrO3-][Br-][H+]Initial Rate
1.1.1.18 * 10 -4
2 .2.1.11.6 * 10 -3
3 .2.2.13.2 * 10 -3
4.1.1.23.2 * 10 -3

To solve this problem, the first step is to find 2 concentrations in BrO3- that are the same and 2 concentrations in Br- that are the same for the same experiment number.

In experiment 1 and experiment 4, BrO3- has a concentration of .1 and Br- has a concentration of .1. Now look at the concentrations of H+ and decide how it changes. The concentration doubles. Then you have to compare it to the quotient of the initial rates. So:

2n = 3.2 * 10-3 / 8 * 10-4
n = 2, so the order for BrO3- is second

Next, we need to find 2 experiments where the concentrations of BrO3- in each experiment are the same and the concentrations of H+ in each experiment are the same.

In experiment 2 and experiment 3, BrO3- has a concentration of .2 and H+ has a concentration of .1. Now look at the concentration of Br- and decide how it changes. The concentration doubles. Then you have to compare it to the quotient of the initial rates. So:

2p = 3.2 * 10 -3 / 1.6 * 10 -3
p = 1, so the order of Br- is 1

Finally, we must do this same procedure again for BrO3-. We must find 2 experiments where the concentrations of Br- in each experiment are the same and teh concentrations of H+ in each experiment are the same.

In experiment 1 and experiment 2, Br- has a a concentration of .1 and H+ has a concentration of .1. Now look at the concentrations of H+ and decide how it changes. The BrO3- concentration doubles. Then you have to compare it to the quotient of the initial rates. So:

2q = 1.6 * 10 -3 / 8 * 10 -4
q = 1, so the order of BrO3- is 1.

Finally, the general formula of the rate law for the reaction of BrO3- + 5Br- + 6H+ -----> 3 Br2 + 3H2O is:

Rate = k [BrO3-]1 [Br-]1 [H+]2