Second Order Reaction Chemistry Problems - Half Life, Units of K, Integrated Rate Law Derivation PROFESSOR: In this video, we're going to focus on second order reactions and how to solve problems associated with it. So in this problem we have the initial concentration of a reactant and it's a second order reaction, it's 0.95 moles per liter, we're given the rate constant k, and what will be the concentration after 3 minutes? So the integrated rate law expression for a second order reaction is this, 1 over A final, is equal to positive kt, plus 1 over A initial. So our goal is to calculate the final concentration after 3 minutes. K is 0.0045. And now notice that k has the unit seconds in it, which means we need to convert the time from minutes to seconds. 1 minute is equal to 60 seconds, so 3 minutes is 3 times 60, which is 180 seconds. And then we have 1 over A initial, where A initial is 0.95. So 0.0045 times 180, that's 0.81, plus 1 divided by 0.95, this is equal to 1.8626. So once you get to this part you could basically flip both fractions. If you raise both sides to the negative 1, 1 over A final simply becomes A final. And on the right side 1.8626 to the minus 1, is simply 1 over 1.8626 and that's 0.537. So that's it for part A. Now let's move on to part B. How long will it take the concentration to be reduced to 0.25? So now we need to calculate the time and so we're going to use the same formula. So this time the final concentration is 0.25. k is still 0.0045. We need to calculate t and A initial is still 0.95. So 1 divided by 0.25 is 4. And 1 divided by 0.95, that's 1.05263. So 4 minus 1.05263, that's 2.94737 and that's equal to 0.0045t. So t is 2.94737 divided by 0.0045 and that should give you 655 seconds if you round it to the nearest whole number. Now if you want to you can convert that to minutes, just divide it by 60. So that's going to be around 10.9 minutes. And so that's how you can calculate how long it's going to take for the concentration to change from 0.95 to 0.25. Now let's work on the second problem. So the concentration of a reactant in a second order reaction changed from 0.75 to 0.45 moles per liter in 5 minutes. Calculate the value, the rate constant. So let's rewrite the integrated rate law expression for a second order reaction. So the final concentration is 0.45, our goal is to calculate the value of k, t is 5 minutes and the initial concentration is 0.75. So if you type in 1 divided by 0.45 minus 1, divided by 0.75, that's going to give you 0.8889 and that's equal to 5k. So take 0.8889 or 0.8 repeating, and divide it by 5. And so k is 0.1777 repeating or 78 if you want to end it. So that's the value of the rate constant k. Now what are the units of k? So for a second order reaction let's write the rate law expression, rate is equal to k times A squared. So let's isolate k, so k is going to be the rate divided by A squared and the rate of a reaction is molarity per unit time, in this case, the time is going to be in minutes. So it's molarity per minute, which you can write that as molarity to the first power, times minutes to the negative 1 power. And the concentration of A is in molarity but it's A squared, so it's molarity squared, which is molarity times molarity. So I wrote it like this so you can see that these are going to cancel. And as I move this from the bottom to the top, the 1 is going to change from positive to negative. So on top is going to be m to the negative 1 minutes to the minus 1. And so that's how you could determine the units of k for a second order reaction. So it's always going to be m to the minus 1 and some unit of time, which could be seconds to the minus 1, minutes to the minus 1, or hours to minus 1. And if you want to you can replace m with moles per liter. So m to the first power is moles to the first power, liters to the negative 1. So now that we have the units of k, what is the half life of this reaction? Now the formula to calculate the half life for a second order reaction is it's 1 over k, times A initial. So we have k, that's 0.17778, and the initial concentration is 0.75 in this example. So if you multiply 0.17778 by 0.75, that's going to give you 0.13 repeating. So the final answer is about 7.5 minutes. So that's the half life of this reaction, it takes 7.5 minutes for the reactant to lose half of its value. Now what we're going to do is talk about how we can convert the rate law expression into an integrated rate law expression. So for a second order reaction, the rate law expression is rate is equal to k times A squared. And the rate is the change in the concentration divided by the change in time and for a reactant we need to put a negative sign. So here we have the differential rate law and now let's replace the change in concentration of A, divided by the change in time with dA over dt. Now we need to integrate by separation of variables. So we got to get A on one side and t on the other side. So what I'm going to do first is multiply both sides by dt. And so these will cancel on the left. So on the left all I have is negative dA, on the right I have k times A squared, times dt. Now I'm going to divide both sides by negative A squared. So on the left the negative signs will cancel, so I have 1 divided by A squared, times dA, and on the right, I have a negative sign, so it's going to be negative k, times dt. Now 1 over A squared is the same as A raised to the minus 2. So at this point we can now take the antiderivative of both sides. To integrate A to the negative 2, add 1 to the exponent, negative 2 plus 1 is negative 1, and then divide by that result. On the right side, the antiderivative of negative kdt is simply negative kt, and then we need to add a constant C. So A to the negative 1 is like 1 over A but we have a negative sign in front, so it's a negative 1 over A and that's equal to negative kt plus C. Now we need to find out what C is equal to and in order to do that, we need to replace t with 0. Right now this represents the final concentration because it's dependent on time. It becomes the initial concentration when t is 0. So when t is 0 we have negative k times 0, so that whole thing becomes 0. So negative 1 over A is equal to C and because t was 0 this represents the initial concentration. So C is technically negative 1 over A initial. So let's go ahead and replace C with this expression. So right now we have negative 1 over A final is equal to negative kt, minus 1 over A initial. So now let's multiply everything by negative 1. So that's going to change all of the negative signs into a positive sign. So 1 over A final is equal to positive kt, plus 1 over A initial. And so that's how you can derive the integrated rate law expression for a second order reaction.