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Main content
Current time:0:00Total duration:10:39
AP.Chem:
TRA‑3 (EU)
,
TRA‑3.B (LO)
,
TRA‑3.B.2 (EK)
,
TRA‑3.B.3 (EK)
,
TRA‑3.B.4 (EK)
,
TRA‑3.B.5 (EK)

Video transcript

let's take a reaction where a + b gives us our products in a lowercase a and the lowercase b represent the coefficients for our balanced equation it makes sense that if we increase the concentration of a and B right a and B would be closer together in space and more likely to react therefore increasing the rate of our reaction and this is true for most reactions if you increase the concentration of your reactants you increase the rate of your reaction we can check this by doing some experiments so let's say we want to figure out what the effect of the concentration of a has on our rate of our reaction so we're going to hold the concentration of B constants we hold the concentration of B constant in our in our experiments we change the concentration of a and we see what effect that has on the rate of our reaction we're going to use the initial rate of the reaction and that's because as our reaction proceeds the concentration of products will increase and since reactions are reversible if we have some products present right that can affect the rate of our reaction and that's not our goal our goal is to figure out what the concentration what effect the concentration of our reactants has on our rate and so we use the initial rate where we have only reactants present and no products so in our first experiment let's say the concentration of a is 1 molar and the rate of our reaction the initial rate of our reaction is point zero one molar per second and our second experiment we increase the concentration of a to two molar we hold the concentration of B constant and we observe the rate of our reaction to increase two point zero two molar per second so we've increased the concentration of a by a factor of two and what happened to our rate our rate went from point zero one to point zero two so the rate increased by two as well alright let's compare our first experiment with our third experiment now we're going from a concentration of a of one to a concentration of a of three so we've increased the concentration of a by a factor of three and what happened to the rate the rate went from point zero one to point zero three so the rate increased by a factor of three alright to figure out the relationship you have to think to yourself to to what power x is equal to two obviously that would be two to the first two to the first is equal to two all right we could have done it for our other comparison as well three to what power X is equal to three obviously three to the first is equal to three so the rate the rate of our reaction is proportional to that's what this funny symbol means here the rate of our reaction is proportional to the concentration of a to the first power alright let's do the same thing for the concentration of B so we do some experiments where we change the concentration of B and we see what effect that has on our initial rate so for all of these we're going to hold the concentration of a constant therefore whatever we do to B is reflected in the rate of our reaction so in our first experiment the concentration of B is one molar and the rate is point zero one molar per second and then we change the contrary of B to two molar right we double the concentration of B while holding the concentration of a constant and we observe the initial rate of our reaction to be point zero four molar per second so we've increased we've increased the concentration of B naught a and let me change that we've increased the concentration of B by a factor of two we've gone from one molar to two molar and what happened to the rate the rate went from point zero one to point zero four so we've increased the rate by a factor of four let's compare our first experiment with our third experiment now we're going from a concentration of B of one molar to three molar so we've increased the concentration of B by a factor of three and what happens to the rate the rate goes from point zero one to point 9 so we've increased the rate by a factor of 9 so now we think to ourselves - to what power I'll make it why - to what power is equal to 4 obviously Y would be equal to 2 2 to the second power is equal to 4 or 3 to what power y is equal to 9 obviously 3 to the second power is equal to 9 so we've determined that the rate of our reaction it is proportional to the concentration of B to the second power alright now we can put those together we can put these together to write what's called a rate law ok so we know that the rate of our reaction is proportional to the concentration of a to the first power and we know that our rate is proportional to the concentration of B to the second power and then we put it in what we put in what's called a rate constant here K and this represents our rate law so let's go through these one by one here so capital R is the rate of our reaction right this is the rate of our reaction all right K is what's called the rate constant so this is the rate constant and there's a difference between the rate of our reaction and the rate constant if we change the concentration of our reactants we change the rate of our reaction but if we change the concentration of our reactants we don't change the rate constant right this is constant it does depend on the temperature though so we'll talk about that in later videos here we have that the reaction is concentration of a to the first power we say the reaction is first-order in a so we say that our reaction is first order first order in a and we found we found that it's second order in B right so we had a 2 here so this is second order second order in B and we can also talk about the overall order of our reaction so for first-order in a right we're first order in a and second order in B the overall order the overall order would be one plus two which is equal to three so the overall order of our reaction is three all right let's go back up here to the general reaction that we started with right so let's go back right back up to here we have we have this and let's write a general rate law so if this is your reaction your general rate law would be R is equal to your rate constant all right times the concentration of a to some power I'll make it x times the concentration of B to some power which I will make why and the reason why I'm showing you this is to show you that you can't just take your coefficients right you can't take your coefficients and stick them into here all right so it doesn't work that way you'd have to know the mechanism of your reaction so these orders have to be determined experimentally so you have to look at your experimental data here and the orders affect the units for your rate constant for example let's go back down to here and let's figure out the units for the rate constant for this example so the rate of our reaction right the rate of our reaction was in more per seconds right this is molar per second we're trying to find units for K the units for concentration are molar all right so this would be molar and then this B to the first power and this would be molar to the second power so we'd have molar to the second power all right so solving for K right you could just go ahead and cancel out one of these molars right here and solve for K so you would get this would be 1 over seconds now on the left so 1 over seconds right and divided by molar squared so 1 over seconds times molar squared or you can write this 1 over molar squared times seconds so those are be your units for K for this reaction right with an overall order of 3 but it can change right you can change depending on the order now let's look at this reaction we have only one reactant a turning into our products and if we look at the two experiments in our first experiment the concentration of a is 1 molar and the initial rate of reaction is point zero one molar per second if we double the concentration of a to two molar the rate stays the same it's still point zero one molar per second so even though the concentration of a is going from one molar to two molar right that's doubling the concentration or increasing the concentration of a by a factor of two the rate stays the same so you could say the rate it's the rate times one because it's the same rate so two all right so two to what power X two to what power X is equal to one obviously X would have to be equal to zero two to the zero power is equal to one so any number to the zero power is equal to one so this reaction is zero-order it's zero order in a and if we wanted to write our rate law we would write the rate of the reaction is equal to the rate constant K times the concentration of a we have only one reactant here and since this is zero order in a right we could just write the rate of the reaction is equal to the rate constant K and so if you wanted to know the units for the rate constant K well the rate is in molar per second and so those would also be your units for K K would be in molar per second so here's an example of how your units for K change depending on the overall order of your reaction
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