[BLANK_AUDIO]. So we've developed an equation now for the rate of reaction for a simple process. R going to P. But now let's just consider. The effect of stoichiometry on the reaction. So lets consider a different reaction where reactants R go to 2 P that is one molecule of R goes to two molecules of P. Now, in this particular case the product P is appearing in moles twice the rate that the react, reactant R is disappearing. So, we're going to have to modify our reaction expression. So, in this case we can say therefore the rate of the reaction is still going to be equal to minus the change in the concentration of r, d r by d t. Buut in terms of the products p, it's going to be positive but we're going to have to multiply it by one half times dp by dt. So let's have a look at a more general process, for example, where we have A moles of molecule A combining with B moles of molecule B, giving us C moles of molecule C and D moles of molecule D. In this case we can write out a rate expression using any one of the four components, they're all changing as a function of time. So if we consider either of the reactants then we're going to need a minus sign because they're disappearing. And if we consider it in terms of molecule A we're going to have to multiply by 1 over the stoichiometry 1 over little a times the rate of change, the concentration over A with time. And that's equal to minus 1 over b times d B by d t. If we want to do it in terms of the products, they're all going to be positive. So you're going to have 1 over c, times the concentration of c, with respect to time. Or 1 over the concentration, the moles of d times the change in concentration of d with time. Any one of these 4 expression would be perfectly valid way to define the rate and they are all equal to,to one another. So, the trick here is to always multiply by one over the stoichiometry of a particular component using a minus sign if it's being used up. And keeping it positive if the molecule is being formed. So now we have developed the equation for the rate of reaction in terms of the concentration, but of course, as our reaction proceeds, the concentration will, in fact, change with time. So in order to express this, we introduce something that's called the order of a reaction. And this describes this relationship between the rate and the concentration of species, whether they be reactants or products. So for instance if we had a simple process where we've got a molecules of a combining with b molecules of b, to give us products. Then if we vary the concentration of both A and of B and we make a note of how the rate changes, we will be able to develop an expression for the rate in terms of both the concentration of A and also the concentration of B. And the concentration will be raised to some power which is the order of the reaction to the respect A and B also raised to some power which should be the order in respect to B. So, in this case the order is x with respect to A and Y with respect to B. And there will need to be a constant, a small k, which is called the rate constant, multiplying the beginning of this equation. So in this expression here, k is the rate constant. So now, we can define an order reaction. We're going to have an order of x with respect to the concentration of a, and we're going to have an order of Y with respect to the concentration of B and also be an overall order for the reaction which is the sum of x plus y. So, we got the orders x plus y. Remember, you also have the stoichiometries of the reaction a and b. And we should just make a note now of these stoichiometries at a and b are not necessarily the same as the orders of the reaction x and y, although in some cases these actually can be equal to one another. Okay so lets consider some examples. For instance if we have two molecules of 2 N2O5(g) phase, we are going to use four molecules of NO2 and one molecule of O2. And if we measure how the reaction varies in terms of the reactants N2O5, you find that the rate depends upon the concentration of N2O5 to the power 1, which we don't have to write, times the rate constant. In other words, this particular action is first order. So in this particular case the power here, the power of 1 does not match this stoichiometry over here. However, if we go to a different reaction; where we take 2 molecules of n o, combining with 1 of oxygen to give us 2 NO2. In this particular case, the rate depends upon the concentration of NO, here, squared, times the concentration of oxygen, over here, to the power one. So in this particular case these stoichiometries match up with the order, so you've got 2 matching up with 2 here and 1 matching up with 1 here. So in this particular case, this reaction is second order with respect to n o. But it is first order with respect to oxygen. And overall this going to be 1 plus 2 is going to be 3rd order. Now consider another reaction, let's consider 2 so 2 combining with oxygen to give us 2 so 3, this reaction is catalyzed by adding Platinum to the reaction and if we look at the rate of this reaction we find that it depends upon the concentration of SO2 this component to the power one. But the concentration of S O 3, one of the products, to the power minus a half. So this tells us something else, it tells us the orders of the reaction do not even need to be integers. In this case, we have uh,1 over the square root of the concentration of S O three. Later on in the course, we will consider how these, rates can actually tell us something about the mechanism of the reaction. So ultimately the order is determined by experiment and you need to monitor the rate as a function of the concentrations in the reaction. [BLANK_AUDIO]