Zero Order Reaction Chemistry Problems - Half Life, Graph, Slope, Units of K, & Integrated Rate Law INSTRUCTOR: In this video, we're going to go over some problems associated with a zero-order reaction. So let's look at this problem. The initial concentration of a reactant is 0.75 moles per liter, which is molarity, and is used up in a zero-order reaction. The rate constant K is 0.015M per minute. What will be the concentration of the reactant after 15 minutes? So let's write the integrated rate law expression for a zero-order reaction. So it's the final concentration, which is equal to negative Kt, plus the initial concentration. Our goal is to look for the final concentration after 15 minutes. We're given the value of the rate constant which is positive 0.015 molarity per minute, and the time is 15 minutes. The initial concentration is 0.75 moles per liter, or molarity. So notice how the unit minutes will cancel giving us the units of molarity. So it's going to be a negative 0.015 times 15 plus 0.75, and so you should get 0.525M for the final concentration. So this is the answer to part A. Now, let's move on to part B. How long will it take the concentration to be reduced to 0.06M? So that's the new final concentration, and how long means that we're looking for the value of the time. So let's replace the final concentration with 0.06, and K is still the same, 0.015. But this time, we need to calculate the value of t, and the initial concentration is still 0.75. So first, we need to subtract both sides by 0.75. 0.06 minus 0.75 is negative 0.69. Now, to get t by itself, we need to take a negative 0.69 and divide it by negative 0.015. So the value of t is 46 minutes. So that's how long it's going to take for the concentration to be reduced from 0.75M to 0.06M. So that's it for the first problem. Number two, the initial concentration of a reactant is 1.5M. It takes 40 seconds for the concentration to be reduced to 0.3M in a zero-order reaction. What is the value of the rate constant? So let's start with the integrated rate law expression for a zero-order reaction. So this time, we're looking for the value of K. The final concentration after 40 seconds is 0.3, and the time that it takes to go from the initial to the final concentration is 40 seconds, and the initial concentration is 1.5. So let's subtract both sides by 1.5. 0.3 minus 1.5 is negative 1.2. So now, we need to divide both sides by negative 40 to get the value of K. So K is negative 1.2 divided by negative 40. So K is 0.03, and the units of K is going to be molarity per second, since we have seconds here, and it's a zero-order reaction. For a zero-order reaction, the units of K is the same as the units for the rate of reaction. Calculate the half-life of this reaction at t equals 0. So how can we get the answer to part B? For a zero-order reaction, the half-life is based on the initial concentration. It's the initial concentration divided by 2K. In this example, the initial concentration is 1.5, and K is 0.03. So 2 times 0.03 is 0.06, and 1.5 divided by 0.06 is 25. So the half-life is 25 seconds. So what does this mean? The half-life tells us the time it takes for the initial concentration to change by one half of its original value. So let's say, if we go back to the integrated rate law expression, notice what's going to happen if we replace t with 25. So K is 0.03. Let's multiply that by 25%, and the initial concentration is 1.5. Negative 0.03 times 25 plus 1.5 is 0.75. So it takes 25 seconds for the initial concentration to change from 1.5 to half of its value which is 0.75, and that's what the half-life tells us. It tells us the time it takes for the concentration to change by one half, and in the case of the reactants, it's decreasing by half of its value. Number three, which of the following graphs will produce a straight line in a zero-order reaction? Is it A, B, C, D, or E? Now, the graph ln A versus t, that's for a first-order reaction. If you have a first-order reaction, it's going to produce a straight line. So that's not the answer we're looking for. Now, A versus t will produce a straight line for a zero-order reaction, so B is the answer. And 1 over A versus t, that's for a second-order reaction, and this just doesn't apply, and it's not going to be E. So make sure you know that a first-order reaction will produce a straight line, if you plot ln A versus t. And if you plot 1 over A versus t for a second-order reaction, it will produce a straight line, but the one we're looking for is A versus t. So that's going to produce the straight line for a zero-order reaction. Number four, the data table below shows the concentration of a reactant with respect to time for a zero-order reaction. What is the value of the rate constant K? Now, what you need to know is that the rate constant K is equal to the slope of the line, if you plot A versus t for zero-order reaction. If you plot this on a graph, you should get a straight line. So A's on the y-axis, time is on the x-axis, and it's decreasing, which means that the slope is negative. And K has to be a positive number, so K is equal to the negative slope, or slope equals negative K. So let's calculate the slope. The slope is the change in y divided by the change in x. So the y values are our concentration values, and the x values are our time values in this example. So let's just pick two numbers. Let's use the first and the last result. So the final concentration is 0.16, and the initial concentration is 0.8. The change in time is 40 seconds. So 0.16 minus 0.8, that's negative 0.64 divided by 40, and so it's going to be negative 0.016 molarity per second. So that's the slope which means that K is positive 0.016 molarity per second. So remember, the slope is equal to negative K. K is always a positive number. Now, what we're going to do at this point is derive the formula for the integrative rate law of a zero-order reaction from the differential rate law. So for a zero-order reaction, we know that rate is equal to K times A to the 0 power, and A to the 0 power is 1. So the rate equals K. Now, the rate is the change in concentration divided by the change in time, and so that's going to be the change in the concentration of A divided by delta t, and for a reactant, the change is going to be negative. So we want the rate to be positive, so two negatives will give us a positive result. So this is known as the differential rate law expression, which you could simply call it the rate law expression. Now, I don't know if you've taken calculus, but the change in calculus, you might see it as dA over dt. So the rate of change of A with respect to t is dA over dt. Now, I'm going to multiply both sides by dt. So it's going to be negative dA is equal to K times dt. Now, for those of you who haven't taken calculus, and if you wish to understand how this works, it might be good to review basic integration. You could look that up on YouTube. Just learn some basic integration techniques and also integration by separation of variables. Now, just review some basics, let's say if I'm integrating a constant with respect to x, it's simply going to be 5x. If I'm integrating another constant, like 4, with respect to y, it's simply 4y. Or 3 with respect to r, it's going to be 3r. So if you're integrating the constant, just add the letter to it. Now, let's say, if I'm integrating 8x to the 1st power with respect to x. What we need to do is add a 1 to the exponent and then divide by 2. So it's 8x squared over 2 plus some constant C. So it's going to be 4x. So let's say, if I want to integrate 12x squared dx. Add a 1 to the exponent, so it's going to be 12x cubed divided by 3, and 12 divided by 3 is 4. So that's a simple way to integrate certain functions. So I'm going to integrate both sides. So I have the integration of negative 1 times dA. If there's no number here, just put a 1 in front of it, and that's going to equal the integration of 1K dt. Now, let's multiply both sides by negative 1. So this is going to be positive, and I'm going to transfer the negative sign to the right side. Now, to integrate this, I just need to add an A to the left side. The integration of 1dA is just going to be A, and K is a constant. So I'm just going to add t to that side. So it's going to be negative Kt, and anytime you have an indefinite integral, you always need to add some constant C. Now, on the left, we have the final concentration. We need to find the value of C. So when you plug in t, let's say, if you replace t with 0, then A will be equal to C. Now, the initial concentration is the concentration when t is 0, and when t is 0, notice that A is equal to C. So therefore, that's the initial concentration, which means C is A0. So I'm going to replace C with A0, since it represents the initial concentration. It's equal to A when t is 0. So thus, I have this equation. The final concentration is equal to negative Kt plus the initial concentration. So for this type of derivation, if you're dealing with a first-order reaction, and a second-order reaction, whenever you get this C part, just replace it with the initial concentration. Because whenever t is 0, it's going to equal A, which means that it is the initial concentration. So that's how you can derive the integrated rate law for a zero-order reaction starting with the differential rate law. Now, the last thing I want to mention in this video are the units for the rate constant K for a zero-order reaction. So it's always going to be molarity to the first power and some unit of time to the negative 1 power. So let's say, if the rate of the reaction has the units molarity per minute, it turns out K is going to have the same units, molarity per minute, or times minutes to minus 1. And keep in mind, you could replace M with moles per liter. So you could say that K is moles to the 1st power times liters to the negative 1 power times minutes to minus 1 power. Now, if you're given the units for the rate of the reaction in terms of molarity per hour, then K is going to be the same thing. It's going to be molarity per hour, which you can write it as M to the 1 times hours to the negative 1. Or that can be moles times liters to the minus 1 times hours to the negative 1, which you can write that as moles per liter per hour. So the units of K is going to be the same as the units for the rate of the reaction, and you can see this based on a rate law expression. For a zero-order reaction, rate equals K. So therefore, they have to have the same units for a zero-order reaction. So it's always going to be this-- M the first power, and time to the negative 1.