 Let's try solving a numerical. The decomposition of N2O5 in CCl4 at 318 Kelvin has been studied by monitoring the concentration of N2O5 in the solution. Okay. Initially, the concentration of N2O5 is 2.33 moles per liter. Moles per liter as we know is molarity. So initially, the concentration of N2O5 is given to be 2.33 molar. And after 184 minutes, it is reduced to 2.08 molar. Right? So after 184 minutes, it's 2.08 molar. The reaction takes place according to the equation. So this is an equation that's given. And we are asked to calculate the rate of production of N2O5 during this period and the average rate of this reaction in terms of hours. Okay. So how do we solve this problem? Let's see. So what we have out here is N2O5 that decomposes into NO2 and O2 according to this balanced chemical reaction. Right? Now, initially at time t equal to 0, the concentration of N2O5 is given to be 2.33 molar. So the concentration of N2O5 initially is 2.33 molar. Let's assume that no NO2 and O2 is formed as of yet. So the concentration of NO2 and O2 is 0 molar. Now after 184 minutes, after 184 minutes, the concentration of N2O5 drops down to 2.08 molar. Right? So the concentration of N2O5 drops down to 2.08 molar after 184 minutes. Now this is all the information that's given to us in the equation. And we are asked to calculate the rate of production of NO2 during this time period. So how do we do that? Let's start by first asking ourselves, what do we mean by rate of production of NO2? Now rate as we know is the change in concentration that happens per unit time. So if we can figure out the change in concentration of NO2 that's happening in this time period, we can easily figure out the rate of production of NO2, right? So how do we do that? How do we figure out the change in concentration of NO2 that's happening out here? You can pause the video and see if you can come up with the answer. Well, we can think of the amount of NO2 that has formed by first thinking about the amount of N2O5 that has reacted out here, right? So in 184 minutes, the concentration of N2O5 dropped down from 233 to 2.08 molar. So the amount of N2O5 that has reacted out here is 2.33 minus of 2.08. So this will come out to be 0.25 molar. So in 184 minutes, the amount of N2O5 that has reacted is 0.25 molar. Now if you look into this equation, we can see that 2 molar of N2O5 will give me 4 molar of NO2. So the amount of NO2 formed will be twice the amount of N2O5 that has reacted. So in 184 minutes, since 0.25 molar of N2O5 has reacted, so the amount of NO2 that will be formed is 0.5 molar, right? Now looking at this information, we can say that the rate of formation of NO2 is the change in concentration of NO2 and changes final minus of initial. So this is 0.5 minus of 0 by the change in time which is 184 minus of 0. So this is 0.5 molar in 184 minutes. And if you do the math, 0.5 divided by 184, this will come out to be 0.00271. So it's 2.7 into 10 to the power minus of 3. So this is equal to 2.7 into 10 to the power minus of 3 molar per minute. So the rate of production of NO2 is 2.7 into 10 to the power minus of 3 molar per minute. Now one thing that I'd like to add out here is that initially we had considered the concentration of NO2 and O2 to be 0 molar. However, it's not a necessary condition. If you look into this question, there is no information that is given regarding NO2 and O2. So maybe there was some NO2 and O2 in the container. Initially, maybe I don't know. So let's see what happens if we had say x molar of NO2 and y molar of O2 initially present in the container. In fact, I don't need O2 because I don't need to calculate anything about O2 in this question. So let's forget about O2. What if we had x molar of NO2 initially present in the container? What do you think would be the rate of production of NO2 in this scenario? Do you think it will change or will it remain the same? You can pause the video if you want some time to make up your mind. So in 184 minutes, 0.25 molar of N2O5 reacted. So this 0.25 molar will give rise to 0.5 molar of NO2. But because x molar of NO2 was present from before, so the concentration of NO2 at 184 minutes should be x plus 0.5 molar. Right? Now even out here, if we try to calculate the change in concentration of NO2, change again is final minus of initial. The final concentration this time will be x plus 0.5 molar minus the initial concentration which is x molar. So the change in concentration will still be 0.5 molar and the rate of production of NO2 will still be 2.7 into 10 to the power of minus of 3 molar per minute. So when it comes to calculating the rate of disappearance of the reactants or the rate of appearance of the products, what we are really concerned with is the change in concentration that's happening rather than the exact values of the initial and final concentration. So to summarize, if we look into this chemical equation, 2 molar of N2O5 on complete reaction will give rise to 4 molar of NO2. So 0.25 molar of N2O5 on complete reaction in 184 minutes is bound to increase the concentration of NO2 by 0.5 molar. So the rate of formation of NO2 in this time interval has to be 2.7 into 10 to the power of minus of 3 molar per minute. So now keeping all these things in mind, instead of solving this question in this way, we can also use a more direct approach. So once again these are all the information that's given to us in the question and we need to figure out the rate of production of NO2 during this time interval. So now if you look at this question closely, we will realize that we are given the change in concentration of N2O5 that's happening over this time interval. In other words, we know the rate of disappearance of N2O5, we can calculate it using this data and we are asked to calculate the rate of production of NO2. Now because rate is the change in concentration that's happening per unit time, so we can always relate these rates of appearances and disappearances just by looking at our balanced chemical reaction. We should be able to find out a relationship between the rate of production of NO2 and the rate of disappearance of N2O5 using this balanced chemical reaction. You can pause the video and take some time and see if we can find out any such relationship. Now looking at this equation, we can say that 2 molar of N2O5 on complete reaction gives rise to 4 molar of NO2. So if the concentration of N2O5 decreases by 2 molar, then the concentration of NO2 in the same time interval is bound to increase by 4 molar. So if the concentration of N2O5 decreases by 1 molar, the concentration of NO2 is bound to increase by 2 molar in the same time. So the rate of formation of NO2 will always be 2 times the rate of disappearance of N2O5. So the rate of production of NO2 will always be 2 times the rate of disappearance of N2O5. So all we now need to calculate is the rate of disappearance of N2O5 and the rate of disappearance of N2O5, the rate of disappearance of N2O5 will be the change in concentration of N2O5 that's happening over the change in time. And change in concentration of N2O5 change is always final minus of initial. So it will be 2.08 minus of 2.33 molar which will come out to be minus of 0.25 molar and the change in time is 184 minutes. So this value will come out to be a negative quantity. But because we report the rate of disappearances as a positive value, so we put a minus sign out here. So this will come out to be plus of 0.25 divided by 184 minutes. So now the rate of production of NO2 will be 2 times this value. So this is going to be 2 times of 0.25 divided by 184. So this is going to be 0.5 divided by 184 molar per minute. So like before, this will come out to be equal to 2.7 into 10 to the minus of 3 molar per minute. So to summarize, we can always relate the rate of disappearance of the reactants and the rate of appearance of the products from the balanced chemical reaction. And we can always figure them out if one or the other is given to us in the question. Let us now try to solve the second part of the question. We need to calculate the average rate of this reaction. So we need to calculate the average rate of this reaction in terms of hours. So what is the rate of a reaction? Well out here for this reaction, we had calculated that the rate of production of NO2 in 184 minutes was 2.7 into 10 to the minus of 3 molar per minute. Now what we had actually calculated was the average rate in 184 minutes because as we have seen in our videos, chemical reactions generally do not happen at a constant speed. But instead, the rate of conversion of the reactants into the products continuously keeps changing throughout the course of the reaction. So what we have out here is actually the average rate of production of NO2 in 184 minutes. Now we had also seen that we could relate the rate of production of NO2 as 2 times the rate of disappearance of N2O5. So the rate of disappearance of N2O5 will be half times the rate of production of NO2. So I can write the average rate of disappearance of N2O5 as half of this value which will come out to be 1.35 into 10 to the minus of 3 molar per minute. So the rate of production of NO2 is 2.7 into 10 to the minus of 3 molar per minute. The rate of disappearance of N2O5 is 1.35 into 10 to the minus of 3 molar per minute. But what exactly is the rate of the reaction? Well, as we have seen from our videos, for any balanced chemical reaction, the rate of disappearance of the reactant divided by the stoichiometric coefficient and the rate of appearances of the products divided by their stoichiometric coefficient, these values are always going to come out to be equal. And this value by definition is called the overall rate of the reaction. For example, in this reaction, if you look at N2O5 and O2, by the time 2 molar N2O5 disappears, 1 molar of O2 would have formed. So by the time 1 molar of O2 forms, 2 molar of N2O5 disappears. So the rate of disappearance of N2O5 will be 2 times the rate of appearance of O2. So I can write this as rate of N2O5 by 2 will be equal to rate of O2 by 1. So from this, we come up with this relationship. And similarly, if we compare N2O5 and NO2. So by the time 2 molar of N2O5 disappears, 4 molar of NO2 would have formed. So by the time 2 molar of N2O5 disappears, 4 molar of NO2 would have formed. So by the time 1 molar of N2O5 disappears, 4 by 2 molar of NO2 would have formed. So the rate of appearance of NO2 is 4 by 2 times the rate of disappearance of N2O5. And this is nothing but 2 times the rate of N2O5. But just to build up this relationship, let us just keep it this way. So if I like rearrange, then rate of NO2 by 4 will be equal to rate of N2O5 by 2. So this time, I'll come up with this relationship, right? So for any chemical reaction, the rate of disappearances of the reactants and the products divided by their respective stoichiometric coefficients will always be equal. And chemists all over the world have agreed to call this value as the rate of the reaction. So the rate of reaction can be written as rate of N2O5 by 2 or rate of NO2 by 4 or rate of O2 by 1. All of this will have the same value. So let me write the rate of reaction as rate of NO2 by 4. So let me write it as rate of NO2 by 4. Now rate of NO2 is 2.7 into 10 to the power minus of 3 molar per minute. So the answer is going to be 2.73 into 10 to the power minus of 3 molar per minute divided by 4, right? However, if we look into the question, we need to find out the rate of this reaction in terms of hours and not in terms of minutes. So we first need to figure out the rate of production of NO2 in terms of hours and not minutes. Now in 1 hour there are 60 minutes and so if in 1 minute, if in 1 minute 2.7 into 10 to the power minus of 3 molar of NO2 is produced, then in 60 minutes the amount of NO2 that is going to be produced will be 2.7 into 10 to the power minus of 3 multiplied by 60. So this will come out to be equal to 2.7 into 60. So this will be 162 into 10 to the power minus of 3 molar. So I can write it as 1.62 into 10 to the power minus of 1 molar, right? So the rate of production of NO2 in terms of hours can be written as 1.62 into 10 to the power minus of 1 molar per hour divided by 4, right? So the rate of the reaction in this case will be 1.62 divided by 4. So this will be equal to 0.405 into 10 to the power minus of 1 molar per hour. So I can write this as 4.05 into 10 to the power minus of 2 molar per hour.