# Determining Rate Equation, Rate Law Constant & Reaction Order from Experimental Data

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• 0:04 Rate Law
• 2:00 Reaction Order
• 2:24 Rate Constant
• 3:09 Some Practice
• 4:57 Lesson Summary
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Lesson Transcript
Instructor: Julie Zundel

Julie has taught high school Zoology, Biology, Physical Science and Chem Tech. She has a Bachelor of Science in Biology and a Master of Education.

Is your head spinning from rate laws, reaction orders and experimental data? Don't worry, this lesson will help you become a pro at all of these things by walking through two comprehensive problems that address all of these topics.

## Rate Law

Buckle your seatbelt, we are about to do a whole lot of chemistry! Let's get started right away with rate laws, sometimes called rate equations, which are equations that relate the concentrations of reactants with the reaction speed. Remember, when you look at a chemical reaction, the reactants are on the left side of the arrow and the products are on the right side.

Take a look at this rate law equation:

Where:

• A and B are the concentrations of compounds or molecules
• x and y are the reaction orders
• r is the rate
• k is the rate constant

Consider the reaction where NO and H2 are the reactants and N2 and H2 O are the products.

2NO(g) + 2H2 (g) â†’ N2 (g) + H2 O (g)

Look at the experimental data table that relates the concentrations of reactants (note: M is molarity) with the speed of the reaction (note: M/s is molarity per second).

Experiment Concentration of NO (M) Concentration of H2 (M) Initial Rate (M/s)
1 0.0050 0.0020 1.25 x 10-5
2 0.0100 0.0020 5.00 x 10-5
3 0.0100 0.0040 1.00 x 10-4

We can use this data to help us figure out x and y and write a rate law equation for this reaction. In order to find x, we need to look at the data and determine what happens to NO when H2 remains the same (Experiments 1 and 2). Notice that NO doubles from Experiment 1 to Experiment 2.

How does this doubling impact the rate? The rate increases by a factor of four. In order to figure out x, we can relate the doubling and the increase by a factor of four in the following way: 2x = 4. Doing some math, we know that x = 2.

Next, we need to figure out y. Looking the table, you can see that Experiments 2 and 3 hold NO constant, while H2 is doubled. You can also see that when H2 is doubled, the rate is doubled. We can relate these two factors in the following way to determine y: 2y = 2. So we know y = 1.

Using our rate law equation, we can plug in A, B, x and y.

## Reaction Order

It's safe to say we are cruising on through this! Are you ready for more? Now that we have determined the rate law, we can determine the reaction order, which tells us if and how the concentration of reactants impacts the rate.

In order to determine the reaction order, add the exponents x and y together. In our previous example, we would add 2 + 1 = 3. This tells us that we have a third order reaction.

## Rate Constant

We have one more piece to bring everything together. Remember k from earlier? This is the rate constant, which relates the concentration of reactants to the rate of a reaction. We can use our rate law and our experimental data to determine k for our equation.

We can plug data from our experiment and then solve for k. We can use any experiment, but let's use the data from Experiment 2.

Plug in the molarity for NO and H2 and the rate from the data table. Now solve for k, which gives you 250.

The units are a little tricky, but if you do it right, you should end up with 1/ M2 s.

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