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

When we include the variable of time to our Rate Law then we obtain the Integrated Rate Laws. 

The Integrated Rate Laws

Concept #1: Rate Law vs Integrated Rate Law


Hey guys. So here we're talking about the order of our reactions. So, we're going to say, if you were super observant you might have noticed that although we were talking about rate, the rate law didn't include a very important variable, it didn't include the variable of time, we never talked about how long it took our reactant to break down to form our products. So, here we're going to say the good thing is the integrated rate laws will help us to answer important question in kinetics, how long does it take for x moles per liters of a to be consumed. So, basically the integrated rate law topic to identify how long will it take for the concentration or molarity of our reactants to disappear to help form products, we're still dealing with the rate law, we are just dealing with the integrated rate law. So, we still are only concerned with the reactants, we don't care about how much product we're making over time. Now, from this point on ready to take a look at each of the integrated rate laws and talk about that.

Concept #2: First, Second and Zeroth Order Reactions


Okay. So, let's look at the integrated rate laws. So, we have 0 order, first order and second order. Notice that each of them have the same variables they all have At here this represents the final concentration of your reactants and therefore Ao represents the initial concentration of our reactants before any time has occurred, okay? So, that little 0 means 0 time is passed. So, we're starting with the initial amount and little t that certain amount of time has passed, so how much can you have left of our reactants, k here represents our rate constant and t equals our time, okay? And here, one important thing to realize that K and t have to agree in terms of units of time, meaning that if my K, let's say k equaled 0.15 molarities inverse times seconds inverse, because k is using seconds inverse that would mean that time is in seconds, if k had used minutes inverse then time would have to be in minutes. So, you look at the unit of K to figure out, what the unit of time will be, in seconds, minutes, years, days, weeks, whatever. Now, one thing we need to realize is that each one of these integrated rate laws is connected to a graph and because they're all connected to a graph they all follow the same equation, y equals Mx plus b, y equals Mx plus b y equals Mx plus b and remember any time you a plot, a plot is of the y versus x axis. So, y and x, y and x, realize here, let's take a look, we're going to say all of them have the same x axis time, that's because if you take a look all of them x is t time, that's why they all had that in common, next realize that we can figure out the order of a reaction but looking at its graph but how do you look at the graph and determine the order, you can tell by the y-axis. So, for a 0 order y is just the concentration of your reactant, so it's concentration of A, for first order it's ln of my reactant, that is why this is ln of my reactant and for second-order it's 1 over my reactant, that's why this is 1 over the reactant.

So, if they ever give you graph and they ask you to figure out the order, look at the y axis, the y axis is label, tells us the order of the reaction, also remember m is your slope. So, here your slope is negative k, that's why zeroth order is decreasing over time from the of the slope, it's a negative slope, here for first order it's also a negative k, that's why it's decreasing over time, second order is different, for second order k is positive, that's why it's increasing over time. So, you can also use your slope as a way of indicating, which one is second order because second order is the only one that has a slope that is positive, that's increasing over time. So, that's how each one of our equations connects to each one of these graphs and then finally each one of them has their own half-life equation. So, for 0 order half-life equals initial concentration divided by 2 times rate constant, for first-order half-life equals ln two over k, ln of 2 is just 0.693 when you punch it into your calculator, and then for second-order half-life equals 1 over k times initial concentration. Now here, one thing that's important about half life, when we're talking about something breaking down by half life processes that means that how much time it takes for it to lose half of its amount. Now, we're talking about radioactive processes, radioactive processes are always first order processes because for first-order half-life is constant, let's say it takes five minutes for it to lose half of its amount, it's always going to take five minutes to lose half, 0 order and second order they're more variable. So, when we talk about radioactive processes they're strictly first-order processes. So, these are the important things that you need to take in mind when you're talking about the integrated rate laws, and each one of the orders, as we delve deeper into this section of kinetics, we'll take a look at different types of questions and sometimes they may not be apparent what the order is, so we'll learn little tricks and tips on how to identify the order and which equation to use.

Example #1: The oxidation of ethane follows a first order mechanism, with a very high rate constant of 32 s-1, to form H2O and CO2 as products. If the initial [C2H6] is 4.12 M, what is the concentration after 1.12 x 10-3 minutes? 


Hey guys! Let's take a look at the following integrated rate law questions in the following video. Here it says the oxidation of ethane follows a first-order mechanism, with a very high rate constant remember rate constant, remember rate constant is lowercase k, of 32 seconds inverse to form H2O and CO2 as products. If the initial concentration of ethane is 4.12 molar, what is the concentration after 1.12 times 10 to the negative 3 minutes?
What we need to realize here is we're asked to find the concentration. We're looking for our final concentration. Here, this is our initial concentration so this is Ao. Here they give me minutes which is time. What we need to remember here is that k and t have to both agree in terms of the units used for time. Time here is in minutes but our rate constant k here is in seconds inverse. Remember, time must always agree with k. Here we have to change minutes into seconds so that those seconds will match the seconds inverse of our rate constant k. We're going to say here for ever one minute that we have, we have 60 seconds. Minutes cancel out so I have that number times 60 which gives me 0.0672 seconds. Now my time t and my rate constant k are both in the same units of time.
We're dealing with first order so that means we're going to use the first order integrated rate law. ln of At equals negative kt plus ln of Ao. Here ln At equals, k is negative 32 seconds in verse. Remember seconds inverse just means that it's over seconds, times 0.0672 seconds. You can see why you want units to be the same because here these seconds cancel out with these seconds. Plus ln of 4.12. Now we have ln At equals, so what we're going to do here is going to multiply these two together. Remember the negative sign in front, plus ln of 4.12. Here ln At equals negative 0.7345. That is what ln of At is. But remember, we don't want ln of At. We just want At, our final concentration by itself.
Just approach it this way. We want to get rid of ln so we divide both sides by ln. We're not really dividing that number by ln because if you did that in your calculator, it will give you the answer of error. When I say divide by ln, I'm really saying we're taking the inverse of the natural log because natural log means ln. We're taking the inverse of the natural log. What I'm really saying here is anytime you divide a number by ln, what that really means is it becomes e to that number. Again, when you divide by ln, you're not really dividing by ln. It becomes e to that number. It does not change the sign at all. Here this was negative. Here it’s still negative.
We got rid of that ln here for At, so now At is by itself. When you punch that into your calculator, e to the negative 0.7345, you can get for your final concentration 0.4797. Since initial concentration was in molar, final concentration will be in molar. That would be our answer to this question and that's the approach we need to take in order to get the correct answer. Remember, first order this is our integrated rate law for first order. Remember, when it comes to our rate constant k and our time t, they both must agree in terms of units of time. Here k was in seconds inverse but time was in minutes, so change time t into seconds as well. It's easier to manipulate time T than to manipulate rate constant. Don't ever try to manipulate the k constant because it's not as easy as you think it would be.

Example #2: Iodine-123 is used to study thyroid gland function. This radioactive isotope breaks down in a first order process with a half-life of 8.50 hours at 800 K. How long will it take for the concentration of iodine-123 to be 74.1% complete? 

Practice: At 25oC, 2 NOBr (g) ----> 2 NO (g) + Br2 (g). The rate of the reaction is found to be: rate = k [NOBr]2. The constant at 25oC is 7.80 x 10-4 M-1 s-1. If 0.550 moles of HBr (g) is placed in a 5.0 L container, how long will take for the concentration to reach 0.063 moles of HBr (g)?

Practice: In a typical chemical reaction, nitrogen trioxide, NO3, reacts to produce nitrogen dioxide, NO2, and oxygen gas, O.     

                                                     2 NO3 (g) 2 NO2 (g)  +  2 O (g)

A plot of [NO3] vs. time is linear and the slope is equal to 0.183. If the initial concentration of NO3 is 0.930 M, how long will it take for the final concentration to reach 0.400 M?

Example #3: Part A: Given the following graph for a second order reaction:

a)  Calculate the frequency factor. 


Hey guys! In this new video, we're going to continue with our discussion about the integrated rate laws but now tie in plots and the Arrhenius equation. If we take a look at the first question, it says: Given the following graph for second order reaction. They want us to calculate the frequency factor. We know the frequency factor is dealing with the variable A, capital A. Since we're dealing with a graph here, we have to use the graphical form of the Arrhenius equation because it's the Arrhenius equation that deals with the variable A, the frequency factor.
We did this earlier how we manipulate our Arrhenius equation to its graphic form. It's going to be ln k equals negative Ea over R times 1 over t plus ln A. Remember, this is our y equals mx plus b. What are we trying to do? We're trying to solve for our frequency factor, so our A. A is right here. But there seems to be a lot of things mixed in which makes it difficult to just isolate that one variable. All we're going to do to solve for this question is find the y-intercept. Remember, when we're looking for the y intercept, all that means is we're going to make x equal to 0 which is going to help get rid of a lot of variables.
Here we’re going to say if x is 0, that means that 1 over t will be 0. If 1 over t is equal to 0 and it's multiplying times negative Ea over R, that whole thing will become 0. Now our equation becomes ln k equals ln A. When x is equal to 0, according to our chart here when x equals 0, y equals 0.569. Then that means our y is this number, this ln k. Our equation simply becomes 0.569 equals ln A. Remember, we don't want ln A. We just want A. The last thing we have to do is divide both sides by ln. Just remember, anytime we divide something by ln, it becomes e to whatever it was. We're dividing 0.569 by ln so it becomes e to that number. That will equal just A. When you punch that into your calculator, we’ll find that the frequency factor is equal to 1.7665.
This type of question required us to manipulate the Arrhenius equation so it fits towards a plotted graph. But remember, there's a graphical form of the Arrhenius equation which we learned how to do from earlier videos.

Example #4: Part B: Calculate the energy of activation in (J/mol). 


Knowing that, we can now do Part B. Part B says calculate the energy of activation in joules over moles. Remember, the energy of activation is Ea. To solve this question, we're just going to say Ea is involved here. It's negative Ea over R. We're going to say that this fraction in a sense is also the same thing as a slope, as M. If they're both the same exact thing, they're both equal to each other. Remember, M is your slope. What's your slope? Your slope is the change in Y over the change in X. If we take a look here at this graph, we have one point here and one point here.
Here, this first point up here, X is equal to 0, Y is equal to 0.569. In this second point here, X what you're going to say X2 is equal to 0.0047 where Y2 is equal to 0. Slope is equal to Y2 minus Y1 over X2 minus X1. Let's just plug in what we know. Y2 is 0, Y1 is 0.569. X2 is equal to 0.0047, X1 is equal to 0. When we work this out in our calculator, it’s going to give you back -121.064. That's what our slope is equal to. Remember, R here has a constant value. This is our constant when it comes to energy. Remember, we said that this number would be 8.314 joules over moles times K. That's what R is equal to when we’re dealing with energy.
We have all variables known except for activation energy. All we have to do now is just solve for it. Multiply both sides by 8.314. Negative Ea equals negative 1006.52 joules over moles. We know it's in joules over moles because R had joules over moles. Then we're going to divide out than the negative because we don't want negative Ea. We just want Ea, activation energy. Then we're going to say activation energy then becomes positive 1006.52 joules over moles.
Don't worry. These two questions were a bit of a challenge. I did that intentionally just to show you how we can manipulate the Arrhenius equation towards a potential graph. Even though you may not see something like this in the lecture, you most definitely would see some variation of it in lab. It's always good to remember the techniques we've used here.
Now that we've done this one, I want you guys to attempt to do the practice one left on the bottom of the page. What you should realize here is I'm asking us to calculate the rate constant. Basically before we can do that, we have to figure out what is the order of our reaction. I give us three graphs. Only one of them gives you a linear, line, a straight line. The one with the straight line gives you the identity of the reaction. It gives you if its first order, zeroth order or second order. Since this is the only one giving us a straight line, from this plot of y versus x, you should tell me what the order of the reaction is. From that, you need to know which integrated rate law to use. From that, you can solve for your rate constant, K. First tell me what's the order of the reaction based on this graph with the straight line and then from that information, you can find out what K is. Good luck guys!

Practice: The three plots were done based on a chemical reaction. What is the rate constant of the reaction if it takes 21.2 minutes for the reaction to be 38.0% complete? 


Concept #3: Half-life is defined as the time it takes for half of the amount of a substance to decay in a certain amount of time.

Concentration and Integrated Rate Laws

Concept #4: For zeroth order reactions the concentration of reactants decreases in a uniform manner over time. 

Concept #5: For first order reactions the concentration of reactants decreases in a logarithmic manner.

Concept #6: For second order reactions the largest drop in concentration occurs initially followed by a decreasing loss of reactant.