 ## General Chemistry

Learn the toughest concepts covered in Chemistry with step-by-step video tutorials and practice problems by world-class tutors

15. Chemical Kinetics

# Rate Law

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.

Understanding Rate Law
1
concept

## Rate Law Concept 1 1m
Play a video:
Now the rate law is just an expression that relates the rate of a reaction to its reaction change and reacting concentrations, rate constant and reaction orders. Now your rate constant is a proportionality constant. That links the rate to these reactant concentrations and your reaction orders. Well, these are just the exponent component for the given concentrations. Now they're determined mathematically with a chart which which has given values or from a series of steps called a reaction mechanism here, we're not going to go into detail and reaction mechanisms just yet. Because pretty abstract Later on, we'll go in greater detail on what reaction mechanisms are and how we can use them to figure out our reaction orders Now. Also, what's important is that rate law ignores products because notice I only talked about the concentration of the reactant. I never mentioned concentrations of products because when it comes to rate law, we ignore products altogether. Now our rate law has its rate law expression which is rate equals K. Which is your rate constant times A. To the X. B. To the Y. Now A and B represent your reaction concentration or reacting concentrations here. If your chemical equation had more than two reactant and then we continue, we have sea to the sea X and Y. These are your exponents or reaction orders. So just remember your rate law is trying to connect the ideas of changes in concentration, foreign reactant rate constant and reaction orders in order to find the overall rate of any given chemical reaction
2
example

## Rate Law Example 1 2m
Play a video:
the chemical reaction has a rate law of K. Eight of the three B. To the one C. To the zero. By what factor would the rate increase if the concentration of A. Were tripled, the concentration of B. Was cut by half and the concentration of C. Was increased by half and the rate constant K. Was kept constant. Alright so here we're going to say rate equals K. eight of the three B. to the one c. to the zero. They're telling me that our rick constant K. Staying the same. It's not changing so it's not going to affect any change in rate so we can ignore it for now and here we're talking about tripling, cutting in half, increasing by half each one of these reactant concentrations to make it easy for us. They don't give us their initial amounts so just make them all one molar. Okay? Just to make it easy because proportionately it wouldn't matter. I can make them all three moller doing what it tells me to do, I still get the same answer. So here we're going to assume that they're all starting at one molar. And here I'm going to triple a. It's going to go from one molar to three molar And it's still to the three B. Is getting cut by half. So you start at one moller, you cut it by half so now you only have 10.5 left Still to the one And see we're still gonna do. See but it's not gonna really matter, we're increasing it by half so that it becomes 1.5 C really doesn't matter because any number to the zeroth power is equal to one so I can make this concentration inside for C a million. It won't matter because a million to zero is still equal to one. So really all that is important to figure out. The changing rate is A. And B. So three to the three is 27 Times . is 13.5. This means that my rate would be expected to increase by 13.5 fold. This is how much faster my rate would go because increasing the concentration of my reactant would cause an increase in the rate. Here we're seeing it mathematically being shown to us. So here option C would be the correct answer.
3
concept

## Rate Law Concept 2 45s
Play a video:
Now the units for the rate constant K can be determined by first calculating the overall order. Now our overall order uses the variable end. It's a numerical value calculated from the addition of all reaction orders. And here the formula to figure out the units for your rate constant K is K equals M two and plus one times time inverse. So remember Kesari constant, capital M is your polarity, and and again is your overall order. So adding up all the orders of your rate law helps you to figure out and take that and plug it into this formula and you'll know what the units for K will be.
4
example

## Rate Law Example 2 1m
Play a video:
what is the overall order? And the units for the rate constant K. For the following chemical reactions from below that has a rate equal to K N 02 to the two cl two to the one. Alright, so remember to and one are your reaction orders to find your overall order? You just add them together. So this would be two plus one. So your reaction would be third order overall. Now for K we'd say remember K is M to the negative end plus one times time inverse. Here, time could be in units of seconds, minutes, gears, whatever. So this would be adding to the negative three plus one. Here, let's just say seconds are the units of time that we're looking at? And so this would be m to the negative two times seconds inverse. So whatever the actual value of K would be, these would be the units that follow it. So it would be some value em to the negative two times s to the negative one. Okay, so we'd be third order overall. And these would be the units for our rate constant K.
5
concept

## Rate Law Concept 3 1m
Play a video:
6
example

## Rate Law Example 3 7m
Play a video:
the initial rates of reaction to N. O gas plus cl two gas giving us two N O C L gas are given below Here we have experiments +12 and three. And each experiment has attached to it reacting concentrations and initial rate. Now here it says determine the new rate if given new initial concentrations for NONCL two. Now, remember our sequence for solving is we first have to determine our reaction orders. Once we do that, then we can figure out our rate constant K. And then finally we can determine our new rate. Alright, so for step one, we're gonna choose a reactant and look at two experiments where its concentration changes, but the other reactions stay the same. These other reactant we're going to ignore. So you're gonna ignore the reactant whose concentrations remain the same here, we're gonna look for the reaction order of N O. So let's make it X. Cl two would have y you can make the many variables you want abc, whatever. Here, I just do X and Y. We have to look for two experiments where was changing but cl two is staying the same. That occurs with experiments two and 3 with experiments two and three. We cno is changing and the concentrations of cl two remain the same. Once we figured that out, we go to step two where we create a pair of ratios for the reactor that sets react rates equal to concentrations of the reactors. This is important. You place the larger rate value on top of the smaller rate value to get whole numbers when solving. And here you're going to raise the reacting concentrations to an unknown power or variable for the reaction order and solve. All right. So if we take a look here, experiments two and three we can see that rate three. Its rate initial rate three. Because it's connected to experiment three, it's larger than initial rate too. So that would mean that we're gonna have rape three over rape two equals the concentrations of N. O. To the X. Now we're gonna plug in the number. So 18.2 divided by 9.08 equals 0.500 divided by 0.0 to 50. Still to the X 18.2 divided by 9.08 is approximately two And then .0500 divided by .0250 is two and it's still 50 x. Now you say to to what number gives you to the answer would be one. So the reaction order for N. O. Is one. Now we're gonna do the same thing for step three. We're going to repeat the process for any remaining reactant until all reaction orders are determined. Right? So now we have to look for cl two. We look for two experiments where cl two is changing but N. O. Is staying the same. We see that happen with experiments one and 2. Now here initial rate one is larger than initial rate too. So if we come down here we're gonna say rate one over rate two equals C. L. Two divided by C. L. Two to the Y. So let's see that was 18.2 18.2 divided by 9.8. And then here we're gonna plug in the concentrations for cl two. So here for cl two it's 20.510 and this is 0.255 to the wat. This also comes out to two equals 2 to the Y. To to what number gives us to. The answer is one. So it's first order for both of our react it's now step forward if necessary. Here it is necessary because we need to find our new rate to find our new rate we have to figure out our react our rate constant. So here we're gonna say if necessary to solve for the rate constant. K. Plug in the reacting concentration and reacting orders into the rate law. Alright so here we're going to say that our rate We're gonna say rate one. You can use any of the rates but just to keep it simple. Always go with rate one. So rate one equals Kay And we're gonna stay here n. O to the one Cl 2 to the one. Okay and then we're gonna start plugging the numbers. So 18.2 we're looking for Kay we don't know what K. Is. Come back up here. Let's use the concentration for. So since we're using rate one we have to use the concentrations of experiment one. So that's 10.0 to 50 and 500.510 To the one in .0510 to the one. Alright so here when we multiply these together that's gonna give me point .001275. Still multiplying with K. rate one again is 18.2 divide both sides by 0.1275. And we'll have our K. Here K equals 14117.647. Now here we could technically give the units for cake because remember to find the units of K. It's M to the negative end plus one times time in verse. And here's our overall order which is two and it comes from adding up your reaction orders together. So that's negative two plus one. Here let's just use seconds in verse. Since our initial rates are polarities per second. So that would be M to the negative one times seconds inverse. Okay finally we can figure out our new rate and again if necessary we're gonna say to solve for the new rate plug in the k. The reaction orders and the additional reaction concentrations given. So here we're going to say that our new rate equals K. And we're gonna plug in the new concentrations given to us in the very beginning. So let's come up here. So here we have 0.730 for R N O. And then we have 00.510 for our cl two. So those are the new concentrations we're gonna plug into here, So let's come down here. So this is .0730 actually. And that's again for N. O. To the one in C. 02 to the one. So new rate. So okay, we got is 14117.647 N O. Is 00.730 to the one cl two is 20.510 to the one when we plug all that in. That gives us 52.6 polarities per second for a new rate. So we can see that if they're asking us for the new rate, it is quite an ordeal. You have to go through a lot of things to get to your final answer. Remember the sequence of solving is important. First figure out reaction orders before you can figure out your rate constant K. Before you can figure out your final new rate
7
Problem

Given the following chemical reaction, A → B. If the concentration of A is doubled the rate increases by a factor of 2.83, what is the order of the reaction with respect to A.

8
Problem

The data below were collected for the following reaction:CH3Cl (g) + 3 Cl2 (g) → CCl4 (g) + 3 HCl (g) Calculate the value and units for the rate constant k.

A. 1.25 M–0.5•s–1

B. 5.6 M–1•s–1

C. 25 M3/2•s–1

D. 7.9 M–5/2•s–1 