# Reaction Rate Problems

###### Explanation

In class, you may often be asked to solve different types of** reaction rate problems**. When solving reaction rate problems, it is important to remember the reaction rate laws and the basics of balancing equations.

###### Transcript

Alright so when you're union of kinetics and you have rate law problems here's some practice problems that you might come across that are a little bit more difficult so if you have just wanted to talk of general rate laws you might want to look at the actual videos that we have from before this. So here's one practice problem that I want to talk about that talks about the concepts within it. Okay let's consider a solution where we have 0.1 molar of H+ ions and a 0.1 molar solution of thioacetamide at 25 degrees Celsius okay. So the rate law looks like this, it's second order rate, second order reaction and they're both in the first order. Okay at each question we're going to ask does the rate increase, decrease or remain the same and does k our rate constant increase, decrease or remain the same?

Okay so let's say we were to add water to the solution, how is that going to affect the rate? Well it's not going to affect k at all, k is never going to change except in one scenario and that's when temperature changes, so the rate law constant is actually going to be the same it's not going to be affected but what will be affected is the rate. And why would the rate be affected, because we're diluting these guys, we're lowering the concentration and we add water we're lowering the concentration of the thioacetamide and the hydrogen ion. So the rate is actually going to be slower, which makes sense because according to our collision theory the concentration plays a major part in how fast the reaction goes.

Number 2, the reaction is heated to 75 degrees Celsius, well we're adding heat to this reaction so it should proceed faster but if you notice and so the reaction, the rate does increase in this case if we increase temperature but how is that shown here? Well k is temperature dependent so even though concentrations remain the same k will change when temperature increases. So yeah k will increase as well, it's the only time k will change. What if we add sodium hydroxide to this solution? Well why would that make a difference? Because nothing in here as sodium hydroxide it may dilute the system but the concentration it's actually going to be quite similar. So the main thing that's actually going to affect is actually sodium hydroxide is going to have a reaction. The reaction it'll, this is the base it will react with this acid and so will go down which will decrease the rate of reaction. So this concentration is going to go down, this will remain the same, the rate is going to lower as well. So the k will not change at all because again k is only going to change when temperature changes.

Okay, so that's one type of a problem that you might see, another type of a problem that you might see or dealing with this type of finding the actual rate law using data. And we've talked that in the rate law, when we're dealing with rate laws and I talked to you about rate laws. But let's talk about how this problem might be a slight bit different, so you're supposed to figure out the rate when comparing 2 different trials. So let's look at trial b because these 2 are the same and we're going to look at how a is affected. So we're going to say okay the concentration of a and the second one is 8, the concentration of a and the first one is 2 and how is that going to affect my superscript? Because don't forget the backload of the rate law is going to be rate equals k or rate constant times the concentration of a to some certain power to the concentration of b to some certain power. And don't forget these coefficients that you have in this reaction are not going to necessarily be the coefficients or the powers in your rate law.

Let's actually, so right now it's time to find the powers so as A changes what's going to happen to rate, rate goes from it goes 4.0 times 10 to the negative 2 and the break from the first one is 2.5 times 10 to the negative third and so I find this to be 4 to the m equals 8, no I'm sorry let's actually figure this out 16. Okay so what 4 to the what power is going to equal 16, well we know 4 squared is equal 16 so m is going to equal 2. So alright, so I'm going to change this to 2, fantastic so now I need to do b. Well b is changed, there's no time when a is constant, so this is actually what makes it more difficult, there's no time when a is constant but b there is a time when b is changing. So what do we do now if a, if we can't compare? Well we already know the superscript, we already know the order of a so we can use that to figure out the order of b. So let's go over here and say okay I know that, let's just compare 2 and 3.

And so we're going to say okay 16.0 over 8 this is a new problem and this is going to be, we know this is squared because it's sorted already times 3.0 over 1.5 to the n because we don't know that is going to equal 3.2 times 10 to the negative 1 over 4.0 times 10 to the negative 2. Okay so this 16 over 8 is 2, 2 squared is 4 times 3 over 1.5 is 2 to the nth power we don't know, equals this and let's just write that to make sure we have the right answer. 3.2 so I get negative 1 divided 4.0 I get negative 2 and this equals 8. So we're going to divide both sides by 4, 2 to the n equals 8 over 4 which is 2 so n is going to equal 1. So now I can change this to 1. So that is how you figure out, find the rate law if something, if one of your, in your trials one of your reactants is not changing. That might something you might get in a more advance class like in honors class or something and this is how you go about doing it. So now we have our rate law and I would typically have to find our k value I'm not going to sit there and find our k value or the plugging charge problem but if you're to find, we're then going to find the units for k and the units for k are pretty easy because if you remember k is equal to the units for k I'm going to write units, is equal to 1 over the molarity times the order of the reaction minus 1 times the unit of time in this case it's seconds, so it's times seconds. So the unit for k in this case is k equals 1 over m the order in this case is third order 2+1 is 3 so the order minus 1 is 2 times seconds, this is the unit for this k.

Okay, so the next question might say, if we have, which one is the correct mechanism? So the mechanism just because this follows the rate law actually the mechanism is the slowest step in the reaction and actually has to be the one that follows the rate law. Meaning that there has to be a 2 in front of the a and a 1 in front of the b in the slow step of the mechanism in the rate step. And if you want to learn more about mechanism, actually there's a video on mechanism describing everything about them that you would want to know. So here, which one is the correct mechanism? So here is the slow step, the slow step is a rate determining step. We would say from here, we would say our rate is equal to k times ay because our slow step tells us that however we can't have y in our rate law. y cannot be in our rate law because it's not in our reaction, so we have to look for where y was formed. y was formed over here so we can, we're going to substitute y for b because y was made with just b. So I can substitute y for b and say the rate in, if this was the mechanism, this would be the rate.

However, we said the rate was this using our data, this cannot be right, so it must be this one. But let's prove it, rate in this cases is this slow step so we're going to say okay rate equals k times a times y. Again y is in our rate law I can't in our rate law because it's not in our reaction. So I'm going to take y, y was formed here and I'm going to change it to a times b so then I can say okay I'm going to rewrite this and say k equals a squared, b equals rate does this match our rate law? Rate equals k, a squared b yes this is right so this must be the correct mechanism. So these are the more difficult problems that you're going to come across when dealing with kinetics and rate laws, so just use the information that you know and actually try it Mathematically you can actually always figure something out, so rate law can be pretty difficult and as we didn't go there's actually even more integrated rate laws and things like that half life activation energy that you might deal with in Chemistry so but this just your basic honors rate law problem.