Reaction Theory

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Reaction order really just boils down to what is the sum of the exponents in the rate law. So, here we have a rate law, the exponent is 0, and anything raised to 0 is just itself, right? So we can really say that for zeroth order, rate is equal to k. The units of k, and this is good to pay attention to cuz sometimes you can get a little bit of extra information out of a problem by looking at the units.

The units of k must be molarity per second, because that's all there is and that's the units of rate. An interesting example of a zeroth order rate law is saturated enzymes, right? There's a zeroth order because it doesn't actually matter how much more substrate you put in, right? You can have 10 times as much, a 100 times as much substrate.

If the enzyme is saturated, and the enzyme is the only real pathway to get to the product, then we can say that the reaction kinetics don't depend on the concentration anymore, they only depend on the enzyme. And the last thing we can take a look at is, what happens when you take a graph of concentration versus time. How might you get a straight line?

For zeroth order it is the concentration versus time, that's it, right? So zeroth order is pretty straightforward. Let's look at first order. So, here again, the superscript is what's most important, k times A to the 1. Well, one is first order, right? That's all there is, so we're gonna say this is just k times a.

The units of the rate laws, or sorry the units of the rate constant are gonna be 1 over seconds, right? Cuz one over seconds times molarity, which is what the brackets mean is molarity per second, right? Learning a little bit more about the units of k can sometimes give you extra information about a question or about what to expect in a certain situation.

Besides chemical reactions that you have calculated at a first order rate law for, you might wanna know that radioactive decay also follows first order rate laws. There's two ways that we can calculate concentrations from radioactive decays, or there's two ways we can use first order equations. One is that the amount at some time later equals the initial amount or A0 times e raised to the time constant times time.

We can also use if we manipulate this, that the natural log of A over the initial amount, A over A0. So this is really nice cuz you can just plug a fraction in here without actually knowing specific amounts =-k rate constant times time. Okay, the last little bit that is helpful for first order reactions is to know about the graphs or what kind of graph gives a straight line.

And that is the graph of the natural log of A versus time. And you can kinda get that out of the second question here, and it would be a negatively sloped line as well, just like the zeroth order. All right, so now we can move on to second order reactions. So here we get a little bit of quote unquote, creativity with one of the possible rate laws because we need the exponents to add up to two.

So if there's one reactant called A, then could just be two. There's a second reactant and could just be B raised to the 2, or it could be A raised to the 1 and B raised to the 1, right? Cuz one plus one for exponents would also give us two. Here again, we have an interesting situation for units, which is that we know we need molarity per second.

So the units of k must be well with molarity squared of some form. The units of k must be in the bottom we have seconds and we have to get rid of quote unquote, get rid of one molarity so much clarity like this. All right, so again, don't be too confused. If you see this in a passage and they're trying to write it out on one line. They might say that the units of k are something like molarity negative one, times the seconds, the negative one, this is the same thing as 1 over molarity times seconds.

There's no specific interesting reactions for second order like we have the saturated enzyme in zeroth order or nuclear decay first order. But we do wanna know about the graph of the straight, what graph cuz this is a straight line when we do second order reactions. And that is the graph of 1 over the concentration of whatever is raised the second power either A or B.

If it's A and B then it's gonna be a little trickier, but we're going to have a positively sloping line in this situation. Because, right, remember the first two are downward sloping. This is positively sloping because as A goes on, as A gets smaller 1 over A gets bigger, right? One over a smaller numbers is a larger number.

So as the reaction goes on, you get a larger and larger value for one over A. All right, so now we can talk about mixed order reactions. For mixed order, it really just boils down to the uniting theme cuz there's lots of different options is that the exponent is a non-integer, right? So, 1.5, 2.5 whatever it is, it could be k times A times or raised to the 0.5 and what can be k times A raised to the 0.5 and B raised to the 1, right?

This would be 1.5 for the reaction order. There is no hard and first rule for the units of k in this situation, you just have to remember that you always wanna get back to molarlity per second. So, whatever you have over here, right for a, the units of the k value would be 1 over. It's going to be, we'll need seconds in the bottom, so it's always there.

And I guess we only really have, right? We have a square root of molarity. So this would have to be actually also, not just one but molarity to the 0.5. For the second one, well, we have 1.5. So it's going to be molarity to the 1.5 and times some value of k, which is something over seconds, and that has to come back to molarity per second.

Well, I have, quote unquote too much molarity in the top. So this is going to be seconds, little bit of room down here. Times molarity to the 0.5, and 1 in the top. So that would be a way to cancel out, 0.5 and 0.5 and then you get back to the units. So just keep in mind where you want to be, and you can always calculate the units for what you need.

So these mixed order rate laws are a little bit tricky but the good news is that on the MCAT you're basically guaranteed to not see one unless you're given the rate law or you're told that it's mixed order reaction. You wouldn't be able to calculate it if you went from the data just because you don't have a calculator. And so a lot of these calculations would be kind of complicated or at least the time input would be a bit longer than you would be expected to put in if you had to somehow calculate square roots or 1.5 exponents by hand.

So for the most part you won't deal with these or if you do deal with them, you'll know exactly what you're working with cuz you'll be told. So let's go ahead and practice this reaction order information that we now know. Certain detoxifying enzymes are being investigated as potential cost effective solutions for patient care in areas with venomous creatures.

The following data are observed. What order kinetics does the enzyme being investigated appear to follow? A zeroth order, B first order, C second order, or D not enough information to determine. Go ahead and pause the video, give it a shot and than come on back. So hopefully you're able to look at this determine that this is a three fold multiplication, right?

From 2.5 to 7.5, and this is also roughly a three fold increase, okay? Experimental data will always have some amount of variability. There's some extra information here that might have given you some pause, by the way, three to three cuz it's a one to one ratio it first order. The extra information here is that, notice this is milliliters of enzyme. This is milli molar here and so what we have is a volume and a concentration which will give us an amount of enzyme, which in whatever reaction volume, the experiment most of us will give a specific concentration.

So this is constant for the concentration added so you'll get the same relative change. Same thing here, right normally rate is in molarity per second here. We have millimolar and it's per 5 minutes, this is still a rate because it's molarity per time. It's millimolar for five minutes, but we can work that out to be molarity per second if we wanted to, but we do have the comparison between concentration and time, or concentration and rate, rather.

Let's try one more, an opportunisitc strain of P aureus catalyzes a reaction to cleave disulfide bonds found between the heavy chains of IgG anitbodies. In order to develop an inhibitor for this reaction, studies are focused on identifying the rate limiting step. Given that a graph of the natural log of cytoplasmic extract versus time produces a straight line, which of the following could be the rate limiting step of the reaction.

A, 2 R-SH groups + 2 chloride ions produces a disulfide bridge plus 2HCl. B, disulfide bridge plus chloride ion produces RSCL plus RS minus. C, A disulfide bridge produces to RS groups, D not enough information to determine. Go ahead and pause the video, work on it and I'll see for the answer afterwards.

So there is a lot of information here and what you wanna do is forget what's most important. Number one, what are we looking for, rate limiting step. Number two, a graph of the natural log of extract which just means a natural log of the reactants versus time gives us a straight line. What that means, if it's the natural log that means it's a first order reaction.

And we know that the rate limiting step, is always what's going to determine the rate law. So this must be a first order, this must be a step where there is one reactant, okay? So in A, I see that there's 1, 2, 3, 4 components, this would be a fourth order reaction.

I see in B, there's two components, this should be a second order reaction. And C this is just one this would be a first order reactions as possible. So just to make sure that you see what I mean, let me just read out what the each rate law would be if this was the rate limiting step. The first one would be rate equals k times RSH raised to the 2 times Cl squared.

And that would be fourth power rate equals k times RSSR times Cl- that would be a second order reaction. And here we have the rate is equal to k times RSSR raised to the 1 which means this is a first order reaction. So a lot of information but if you know what you're looking for you can get to where you need to be with your rate laws.

So here we have the summary, make sure you've captured all these in your notes and you understand all the ideas. You're good to go on your reaction orders.

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