In this problem we're asked to come up with a rate expression. And it says to do it in terms of the disappearance of reactants and the appearance of products. So let's look at what we're going to do in part A. A rate expression is a way of determining a rate of reaction. You can determine it by observing any one of the reactants and products. Now the reactants are disappearing. And if we monitor the changing concentration of ammonia for example, that concentration is decreasing with time. But once the rate of the reaction will be a positive number, so we put a minus sign there and we will divide by delta t. To make this equal to the rate of the reaction, we also must divide by the coefficient of 4. We can do this for each substance in the reaction, so since oxygen is also a reactant, we will do a minus sign, change in concentration of O2, with respect to time, without forgetting to put in the five on the denominator. For the products, we do not need the negative sign. But we still need the coefficient, so we will divide it by four delta T. And for water, it will be the change in concentration of H2O divided by 6 delta T. So I could monitor any one of those substances and determine the rate of the reaction. Now somebody went and observed the reaction and noticed that oxygen was disappearing at a rate of 0.335 M/s. So that would be the change in concentration of O2 with respect to time. Now it's disappearing. So its change of concentration is negative. When we say it's disappearing at this rate, they're actually giving me the positive of that. So this is what we know. What we want to know is the rate of formation of NO. So we're trying to determine this value. We can come up to a rate expression and notice that we have those two pieces in the middle part of this rate expression. We also see that the change in O2 is observed. And I'll put kind of a dash circle around it. We see it right here. Okay we see it observed right there, and this is the value that they gave for us. So I'm going to write down this expression, minus the change in concentration of O2, divided by 5 delta T is equal to the change in concentration of no divided by 4 delta t. I will put in place of this part, the numerical value that they gave me, which is 0.355 molarity per second. Over 5 is equal to the changing concentration of no divided by 4 delta t. We don't want the 4. Look up here what we're trying to define. Just how NO is changing with time. So I'm want to multiply both sides by the 4. And that will leave me with a value of, let's see, 0.284 molarity per second for the change in concentration of NO with respect to time, so the rate of change of NO, how fast is it appearing? Now let's look at the numbers and do some comparison. We see that NO is appearing at a slower rate than O2 is disappearing. See that number there? See this number here? Let me say that again. We see that NO is appearing more slowly than the O2 is disappearing. If we look up here at the balanced equation and we see these coefficients that should make sense to us. For every 5 02 that we consume we are only going to make 4 NO, so give will be disappearing more rapidly than NO will be appearing, so our number makes sense to us.