Ch.13 - Chemical KineticsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Rate of Reaction
Average Rate of Reaction
Arrhenius Equation
Rate Law
Integrated Rate Law
Collision Theory
Additional Practice
Instantaneous Rate of Change
Energy Diagram
Michaelis-Menten Equation
Reaction Mechanism
Identifying Reaction Order

Rate Law represents an equation for a chemical reaction that connects the reaction rate with the concentrations or pressures of the reactants and the rate constant. 

Rate Law

When it comes to the rate of a reaction the Rate Law focuses on the reactant concentrations, while ignoring the product concentrations. 

Concept #1: Understanding the Rate Law. 


Hey, guys, in this brand new video, we’re going to take a look at the new topic, rate law.
Now up to this point, we should know what general rate is. General rate is just the change in the concentration or amount of a compound, whether be a product or a reactant, over time. Now we are talking about rate law. Rate law is very different than general rate. For one, rate law does not care about the amounts of the product. It only focuses on the reactants. So that’s the biggest thing you need to worry about here, cares only about the reactants.
Remember I told us earlier that sometimes our reactions go to completion, they have a solid arrow going towards products, but sometimes our chemical reactions may have double arrows, meaning they’re at equilibrium. That means our reaction goes in the forward and the reverse direction.
We’re going to say that although these chemical reactions can be reversible, meaning they can go either way, we will only look at the forward reactions so we’re only looking at the forward direction for now. And because we are looking at only the forward direction that means we’ll be ignoring the reverse direction. So we’re ignoring how it goes backwards.
And if we’re only looking at the forward direction, then the rate is only dependent on reactant concentrations because as we’re going forward what can affect our speed of our reaction is how much reactants do I have to begin with. How long is it going to take me to break down all of those reactants to make my products?
And not only the reactant concentration, but were also going to focus on a new variable, k. Now we’re going to say that rate law is equal to k. Then we’re going to have the concentration of just the reactants. So as you can see rate law only cares about the reactants the product is nowhere to be found within our equation.
Then we’re going to say NO is to the x, O2 is to the y, these are also new variables we have to familiarize ourselves with. Here we going to say k is called our rate constant, and then we’re going to say x and y are called our reaction orders. 

By focusing on only the reactant concentrations we only care about the forward direction of our chemical reaction. 

Concept #2: The Rate Determining Step. 


This is what we need to remember here. We're going to say unless the reaction, the reaction has to say this. Unless the reaction is classified as a slow step, and the slow step is the rate determining step, then x and y one must be calculated experimentally.
If we take a look at the balanced equation, we have 4 moles of NO, 1 mole of O2. Just because those are the coefficient does not mean that x here is 4 and y here is 1. The only time that would be true is if your professor tells you explicitly that it's a slow reaction. If they don't tell you it’s slow, you cannot do this. I never told us the reaction is slow, so we would have to find that x and y experimentally.
Later on we'll see what exactly does that mean, and what kind of work do we need to do to find x and y. But just remember, for now it has to say slow step for you to just use the coefficients. 

If the reaction says it’s a SLOW step or a ONE-STEP MECHANISM then we can simply look at the coefficients of the reactants to determine the reaction orders. 

Example #1: For the following reaction, use the given rate law to determine the best answer for the reaction with respect to each reactant and the overall order.

H2O2 (aq) + 3 I (aq) + 2 H+ (aq)  --->  I3 + 2 H2O (l)  


Rate = k [H2O2]2 [I ]


a) H2O2 is 1st order, I is 1st order, 2nd order overall.

b) H2O2 is 2nd order, I is 1st order, 3nd order overall.

c) H2O2 is 0th order, I is 1st order, H+ is 1st order, 3rd order overall.

d) H2O2 is 2nd order, I is 1st order, H+ is 0th order, 3rd order overall.

Example #2: Answer each of the following question based on the following chemical reaction:

Calculate the reaction order for reactant A. 

Practice: Calculate the reaction orders for Reactants B and C.

Example #3: Calculate the rate constant and the new rate for the given reaction if the initial concentration of [A] = 0.300 M, [B] = 0.150 M and [C] = 0.150 M.

The Rate Constant k

The Rate Constant k can have a direct impact on the rate of the reaction. In order to determine its units just utilize the equation below, where n is the overall order of the reaction: 

Example #4: A certain chemical reaction has the given rate law:

Rate = k [A]3 [B]2 [C]-1

What are the units of the rate constant for the given reaction?

a) M-2 s-1      b) M2 s-1       c) M-4 s-1      d) M-3 s-1      e) M3 s-1


Hey guys! In this new video, we're going to take a closer look at rate and look at the calculations involved. We’d figure out the units for the rate constant k.
Let's take a look at the first question. It says: A certain chemical reaction has the given rate law: Rate equals k A to the 3, B to the 2, and C to the negative 1. What are units of the rate constant for the given reaction? They're asking us to find the units for rate constant k. All you need to realize here is that when it comes to figuring out the units for k, just use the simple equation: k equals the molarity to the negative n plus 1, times time inverse.
Remember, we talked about the variable n earlier. We said n equals the overall order. How do we find out the overall order of a reaction? We just added up all of the reaction orders. Here 3 plus 2 plus negative 1. When we do that, we get n equals 4, so n here equals four. We're going to plug that in. M to the negative 4 plus 1, times here all the units of time are in seconds inverse so we'll just use times second inverse. Remember, time could be in minutes, in seconds, hours. It doesn't really matter. Then we're going to say here negative 4 plus 1 is negative 3. Here the correct answer would be m to the negative 3 times seconds to the negative 1. The correct answer would be D.
Remember, this is really saying M to the inverse three. What does that really mean? We could have also written the answer as this. Inverse means that it's really 1 over M to the third and then seconds is inverse too, so that's 1 over s. Those two are multiplying so it’d really be 1 over M to the third times seconds. Remember, this is just basic algebra. Inverse really means one over whatever that power is. If I gave you something to the inverse half, then that really means 1 over A to the half. Remember if you're the half power, that's another way of saying square root.
We’ll take a look at more of how do we manipulate inverse functions later on down this page.

Example #5: The reaction of 3 A + B ---> 2 C + D, was found to be: Rate = k [A]2 [B]3. How much would the rate increase by if A were tripled while B were increased by half?

a) 0.50              b) 30.38            c) 0.75              d) 20.25            e) 1.125

Practice: If the rate law for the following reaction is found to be the following, what are the units for the rate constant, K?

Cl2 (g) + HCCl3 (g) → HCl (g) + CCl4 (g)

a)          b)           c)           d)           e) 

Reaction Mechanisms

Concept #3: A   Reaction Mechanism   involves a series of elementary reactions that give the overall equation. 


Hey guys. In this new video we're going to take a look at reaction pathways. Now, we're going to say that a reaction mechanism is a sequence of single steps that add up to give our overall chemical reaction, what we should realize is that sometimes you're given one chemical reaction but in actuality it took several steps and those several steps added up together give us the overall reaction, which represents that one equation. So, for example, here we have A plus 2B gives us E. Now, this one single reaction is actually coming from this reaction mechanism. Now, this reaction mechanism, we say it's made up of three equations, these equations are called elementary steps. So, we have three elementary steps and together they're called the reaction mechanism, them working together helped to create our overall reaction right here, now, how does this work? Well, we're going to say here we have a car C as a product but we also have a C as a reactant, they look the same but they're different one is a reactant one is a product, when we see that we get to cancel them out then we're going to say, we also have a d as a product and a d as a reactant again, when this happens they get to cancel out, everything else left behind gets to come down. Now, both of these B's our reactants since they're the same type of compound we can add them together because they're both reactants. So, again, if they're both reactants or both products you get to add them up, if one's a product and one's a reactant they cancel out, so this is going to be A plus 2B gives us E. So, that's how we're able to get this overall reaction here, we dealt with three elementary steps, which together form our reaction mechanism. Now, knowing this we can look at this example. So example one says, the elementary reaction two moles of NOBr gas gives us two moles of NO plus Br2, is an example of a blank reaction. Now, we're going to say here, elementary steps reaction mechanisms, they're highly focused on our reactants. So, they kind of follow rate constant, rate law, and we're going to say, when it comes to the rate law the rate law only cares about reactants, we've said this before. So, we're going to figure out the molecularity of this and to do that we look at the moles of reactants, here we have two moles of NOBr and because we have two moles of NOBr, we're going to say that this reaction is bimolecular, because bi means two, bimolecular is basically saying we have two moles of reactants in an elementary step, unimolecular we would mean we have one mole of reactant, dimolecular doesn't exist, that was a trick question, tetra means we have four moles of reactants in our elementary steps and if you wanted three it would be termolecular, termolecular would mean three, let's say we want this reaction in an opposite way, let's say it was really 2NO plus Br2 gives us 2NOBr and let's say that this was our new elementary step, here we'd say we have a total of three moles of reactants, you would add up all the moles of reactants that comes out to three. So, this example up here would be termolecular. So, just remember molecularity only looks at the reactives just like our rate law, going to example two, here we're saying answer the following questions for the following reaction. So, here we have a fast step and a slow step reaction and don't worry, you won't be asked to figure out which one is fast and which one is slow, you'll be given that information by your professor. So, first they're going to say identify the intermediate or intermediates, what we should realize here, intermediates are compounds that appear first as products. So, appear first as products then in the following reaction. So, in the next reaction, reaction line, they appear as reactants, okay? And because one of the products is a reactant we expect them to cancel out. So, in this reaction mechanism above, we would say that the intermediate is BrO, it first appears as a product then in the very following reaction it appears as a reactant. So, it's no longer there. Now, catalyst is basically the opposite, we're going to say a catalyst appears as one of the first reactants, then as one of the last products. So, who would be a catalyst, we'd say oxygen appears as one of the first reactants then appears as one of the last products. So, we 'd say oxygen represents our catalyst.

Now, if there was a reaction in between these two O it'd still be the catalyst, because remember it appears as one of the first reactants and then as one of the last products in the last reaction given, these guys here, they help speed up the reaction but they themselves are not part of the reaction. So, we can cancel them out too because one's a reactant and one is a product. Now, finally they're asking us, what is the overall reaction. So, all we do here is we write down what didn't get cancelled out. So, bring down everything. So, we have these two Br's as reactants, come down, you add them up plus O2 plus O gives me Br2O plus O2. So, I basically wrote down all the reactants and all the products that didn't get cancelled out because they're not intermediates or catalyst. So, that's how we'd handle these types of reaction mechanism questions. So, remember a reaction mechanism basically numerous steps reactions combine and cancel out together to give us our overall reaction, each of these individual reactions is called an elementary step, to find the molecularity of it just look the moles of reactants and just remember the difference between an intermediate and a catalyst, they're similar but they're quite different from one another.