[BLANK_AUDIO]. Okay, so now, we've introduced the concept of order reaction, let's just consider some simple orders, and we'll start with a zero order reaction. So by this, we mean that the reaction rate is independent of reactant concentrations. So let's see that in action. And let's take a simple process, which we have considered before, where we've got a moles of A plus b moles of B going to give us products. And we've already shown that in that particular case, that the rate of the process is equal to minus what is being used up, one of the stoichiometry, 1 over a, times the rate of change of a with respect to time. Okay, so if this is a zero order process, it's going to depend upon the concentration of [A] raised to the power 0, times the concentration of B raised to the power 0, times some constant. In this case, we'll call the constant k prime. It doesn't really matter what we call it. Okay, well anything raised to the power of 0 is equal to 1. And consequently, this whole thing immediately simplifies down just into k prime. In other words, it is a constant. [BLANK_AUDIO]. So what we've shown here is that the rate of reaction, the change in concentration with time, is going to be a constant. Okay, we've actually got two numbers here which we can play with a bit. We've got the stoichiometry a and constant k prime, and we can just absorb the stoichiometry a into the constant simply by multiplying both sides of this equation by a. So if we do that, we're going to get minus d[A] by dt is equal to another constant, let's call it k0, where k0 is going to be equal to k prime multiplied by the stoichiometry a. So here's our very simple rate expression for a zero order reaction. You can also see the units of k in this process, because it's just going to be equal to the units of d[A] by dt which is a concentration divided by a time. So the units in this case will be concentration, which will be moles per decimeter cubed, divided by time. So that will be per second. Moles per decimeter cubed per second. So here's our rate expression for a zero order reaction. Now we're going to need to do something with this, because it's not in a very pleasant form to actually be meaningful to us. So the first thing we really need to do is to integrate this expression. So if we want to integrate this, the first thing we need to do is to separate the variables, the two variables that we have of concentration and the time. So we take those onto opposite sides of the equation. So, on the left hand side we'll just leave the d[A] and take the minus sign onto the other side. And we'll take the other variable, dt, to the other side. Now we're in a position to integrate this expression, so we can integrate d[A]. And we can integrate the other side. We don't need to integrate a constant. So you can put the integral sign just in front of the dt. And in order to have this solved without having constants of integration, going to put some limits. So, we start at time equal zero and if at time equal zero, and the concentration of [A] is A0. And at some time later, time t, the concentration is just going to be the concentration [A]. We can now solve this. So on the left hand side we have the integral d[A], which is just [A] itself. And we put in the upper limit for [A] and subtract the lower limit for [A]. You have the solution to the left hand side. The solution to the other side, the integral dt is just t. I'm putting in the upper limit is t, minus 0, so the integral of this expression is just t. So the right hand side just simplifies to minus k0 times t. Or we would sometimes more normally just rearrange this expression in order to give us the concentration at any particular time t by taking the [A]0 onto the other side and then to have [A]0 minus k0 times time. So now we have an expression which relates the concentration of [A] at any given time t. We need to know the rate constant and also the initial concentration of [A]. And that is called zero order integrated rate expression. Okay so for a zero order expression, we have shown that the rate of reaction is a constant given by minus d[A] by dt equals k0. Consequently, if we plot the rate as a function of the concentration of [A], that rate is going to be independent of the concentration of [A]. And it will actually have a constant value throughout. If we integrate that expression, we've shown that the concentration of [A] itself changes as a function of time. When t is equal to 0, it's going to be equal to [A]0. And then, as the reaction proceeds, because it's a minus sign, concentration is going to drop. This is the equation of a straight line. So, if we plot the concentration [A], as a function of time t, it is going to give us a straight line with a negative slope. Slope is going to be equal to minus k0. So we have a way to determine the rate constant. And then the intercept here at this point is going to be equal to the initial concentration of [A] naught. Now zero order reactions are actually not very common. But here are a few examples of processes, for example, gasses on metals. So the reaction of ammonia which is catalyzed over a tungsten catalyst to give you nitrogen and hydrogen, is a zero order reaction. Some solution reactions, for example, the iodination of methoxyethane in an aqueous solution is a zero order reaction. And then finally, other catalyzed reactions, such as enzyme reactions, are often zero order. [BLANK_AUDIO]