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๐Ÿงชย ap chem

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๐Ÿ‘Ÿย Unit 5

ย ย โ€ขย ย โฑ๏ธ3 min read

5.4 Writing Rate Laws

Dylan Black

dylan black

โฑ๏ธ June 2, 2020


๐Ÿ”„ Refresher on Rate Laws

To learn how to write out rate laws, let's refresh ourselves on what exactly a rate law is. Given a reaction A --> B, the rate law for this reaction is R = k[A]^n, where R is the reaction rate, k is the rate constant, and n is the order of the reaction in A. The rate law shows us that rate is directly proportional to the concentration of the reactants, or in this case just reactant.

Rate laws are simply what they look like: mathematical expressions to find the rate of a reaction. However, given any reaction, is there a way to find exactly what the rate law is? Let's find out!

๐Ÿงช Rate Laws Found Experimentally

Unfortunately, the only true way that rate laws can be found are through an experiment. It's a common mistake of chemistry students to look at the stoichiometric coefficients of the reactants and use those as reaction orders, but you CANNOT DO THAT! The reaction 2A --> B is not necessarily second order, it could be, but we won't know until we run an experiment. The experiment run to find a rate law is quite simple. All it involves is running multiple trials of a reaction with different concentrations of each reactant, typically doubling one and then seeing how the rate reacts to this change in concentration. One important stipulation is that these reactions MUST and I mean MUST be run at the same temperature. If the temperature fluctuates, the rate can change dramatically, giving incorrect results. This is because k, the rate constant, is temperature dependent. Once you find the orders of a reaction, you can plug in values from an experiment to find k.


How Does Concentration Affect Rate?

You might be wondering, how does doubling the concentration change the rate? Well, let's take a look at a general example. Let's say at 1M, the rate of a reaction is 1 mol/Ls and at 2M, the same reaction at the same temperature is 4 mol/Ls. This means that by doubling the concentration, the rate quadrupled. This is a quadratic effect. Essentially, if R = k[A]^n and doubling [A] leads to a quadrupling in rate, n must be two. We can see this through some simple algebra:

1 = k[A]^n and 4 = k(2[A])^n ==> 4 = k(4[A]^2) ==> 1 = k[A]^2

As a general rule of thumb: if doubling [] leads to rate doubling, it's first order, quadrupling rate means second order, octupling rate means 3rd order, etc., though it's rare to see more than 2nd order reactions.

โœ๏ธ Example Problem

Find the rate law for the reaction 2NO + 2H2 --> N2 + 2H2O given the reaction data below at 1280C.

Finding Order of NO

Between experiments 1 and 2, [NO] doubles and [H2] remains constant, so we can use these to find the order of the reaction in NO. When [NO] is doubled, rate increases by a factor of 5.0/1.25 = 4 (we can cancel out the 10^-5 here). Therefore, the reaction is second order in NO.

Finding Order of H2

Similarly, between experiments 2 and 3, [NO] remains constant and [H2] doubles. Therefore, we can do the same as before: [H2] doubles, and rate increases by a factor of 1 * 10^-4/5 * 10^-5 = 2. Thus, the reaction is first order in H2.

Finding k and putting it all together

We now know the rate law is in the form R = k[NO]^2[H2]. Now, by plugging in values you can find k and write out the full rate law. Give it a try! The first two steps really are the hard parts.

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โš›๏ธย  Unit 1: Atomic Structure and Properties

๐ŸŒ€ย  Unit 3: Intermolecular Forces and Properties

๐Ÿงชย  Unit 4: Chemical Reactions

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