 In this video, we are going to look at zero-order reactions. We are going to understand what a zero-order reaction means. So the first thing we need to look at is what do we mean by a zero-order reaction? So what is a zero-order reaction? And because we are interested in studying the kinetics of chemical reactions. So we are going to look at how the rate of the reaction, rate of reaction varies with different parameters for zero-order reactions. We are going to look at how the rate of the reaction varies with say concentration and with time for zero-order reactions. So let's start. So what is a zero-order reaction? Well, if I have any chemical reaction, say a giving me products, then we know that from the rate low, we can write the rate of the reaction to be equal to some constant k into concentration of the reactant raised to some power x. Now this x is an experimentally determined quantity. It's an experimentally determined quantity. You can't just look at the reaction and say what the x is. You have to figure it out experimentally. And this experimentally determined quantity is called the order of the reaction. Now if it turns out, if it turns out that for this particular reaction, if the rate of the reaction can be written as k into a to the power zero, that is, if experimentally x is found to be equal to zero, then the order of this reaction is zero and so we will call it a zero-order reaction. So what is a zero-order reaction? A zero-order reaction is nothing but a reaction in which the order with respect to the reactant has been experimentally found out to be equal to zero. So therefore zero-order reactions are those reactions in which the rate of the reaction can be written as concentration of the reactant raised to the power zero. Let's now move forward and see how the rate of reaction changes with concentration and time for these zero-order reactions. Let's start by looking at how the rate of the reaction changes with the concentration of the reactants. So let's say we have our hypothetical chemical reaction A giving me products. And if this was a zero-order reaction, then I can write the rate of the reaction to be equal to the rate constant k multiplied by the concentration of the reactant raised to the power zero, right? So what does this mean? So say if this reaction is happening in a container and let's say that the concentration of A out here, the concentration of A out here is 10 molar, then at this instant the rate of the reaction can be written as the rate constant k multiplied by the concentration of the reactant which is 10 molar and because it's a zero-order reaction, so raised to the power zero. Now 10 to the power zero is going to be equal to one, right? So the rate of the reaction will simply be equal to the rate constant k. Now instead of 10 molar, if the concentration of A was something different, if the concentration of A was say 20 molar, then even out here at this instant, the rate of the reaction will be equal to the rate constant k multiplied by the concentration of the reactant which is 20 to the power zero. Now even out here, 20 to the power zero is going to be equal to one. So the rate of the reaction will simply be equal to the rate constant k. So as you can see, whatever be the concentration of the reactant, for a zero-order reaction, this term will always come out to be equal to one. So the rate of the reaction will be a constant. It will be equal to the rate constant k. So whatever be the concentration of the reactants, the rate of zero-order reaction will always be equal to k. So if I plot a graph of how the rate of the reaction changes with the concentration of the reactant. So if I have to plot a graph of rate of reaction versus concentration, whatever be the concentration of A, the rate of the reaction will always be equal to this constant value. So this will be a straight line at k. So for zero-order reaction, the rate of the reaction will always be equal to the rate constant k, whatever be the concentration of the reactant. So the rate of the reaction is a constant independent of the concentration of the reactant, of the concentration of the reactant. Right? Let's now look at how the rate of reaction changes with time for zero-order reactions. So how does rate of reaction evolve with time for a zero-order reaction? Let me elaborate this a little bit further. Let us say we have a reaction A giving me products and let's say that we start this reaction at say time t equal to zero with 10 molar of A. So initially at time t equal to zero, I had 10 molar of A. Now as this reaction keeps progressing, as time evolves, some of these reactants will get converted into the products, right? So after some time say t1, the concentration of A will no longer be 10 molar. It will drop down to say 8 molar. So after some time t1, the concentration of A let's say drops down to 8 molar. See with more passage of time as the reaction keeps progressing, the concentration of A will drop down further and say it drops down to say 5 molar at say time t equal to t3. So at time t equal to t3, the concentration of A drops down to 5 molar. So now if I want to plot a graph of how the rate of reaction changes throughout the course of this reaction. How the rate of reaction changes with time? What would that graph look like? You can pause the video and see if you can come up with the answer. Well, with time the concentration of the reactant keeps gradually decreasing, right? However, since this is a zero-der reaction, since the rate of the reaction is k into concentration of A to the power zero, so whatever be the concentration of the reactant, the rate of the reaction will always be equal to the rate constant k, right? So it doesn't matter if the concentration of A is 10 molar or 8 molar or 5 molar, the rate of the reaction will always be equal to the rate constant k. So now if we check it with respect to time, then initially the rate of the reaction was equal to k and now after some time t1, it will still remain k and it will forever remain k. So over time in zero-order reactions, the rate of the reaction doesn't change and it keeps constant throughout the course of the reaction. So with time over the course of a zero-order reaction, the rate of the reaction doesn't change and it is constant throughout the course of the reaction. Now if you have stuck around till now, I'd like to add that zero-order reactions actually do not make much sense. In fact, to be honest, zero-order reactions kind of go against everything that we have learned till now. Let me show you how. Now we have always maintained that a chemical reaction happens when the reactant molecules collide against each other. Now if the reactant molecules are few and far between, then the probability of these collisions will be lower. So when the concentration of the reactant molecules is low, the rate of the reaction should be low and if the concentration of the reactant molecules is high, then the probability of these reactant molecules colliding against each other will drastically increase. And so if the concentration of air increases, then the rate of the reaction should increase. However, zero-order reaction totally does not fit into this picture. For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactant. So it doesn't matter if I increase or decrease the concentration of the reactant, the rate of the reaction won't change. Wow. Are such reactions even possible? Let's see. Let's take a look at this reaction, the decomposition of N2O into N2NO2. Now if we carry out this reaction in the gaseous phase, so if we like simply take a container and fill it with N2O molecules, then these N2O molecules will collide against each other and some of the old bonds will get broken down and some new bonds can get formed. So some of these collisions will lead to the formation of N2O2. Now if we think about it, if I increase the concentration of N2O out here, the overall number of collisions between the reactant molecules will increase. So the probability of successful collisions will also increase. So the rate of the reaction should increase, right? So if I increase the concentration of N2O, the rate of the reaction should increase. Now experimentally, the rate of this reaction has been found out to be equal to the rate constant K multiplied by the concentration of N2O to the power 2. Now this is nothing contradictory, even this shows that if I increase the concentration of N2O, then the rate of the reaction will increase and if I decrease the concentration of N2O, the rate of the reaction will decrease. Nothing surprising out here, right? However, this is a very slow reaction and instead of carrying out this reaction in this way, it is generally carried out in the presence of a catalyst like nickel. So instead of simply taking it in a container and letting the collisions happen, if we do this reaction in the presence of nickel, then experimentally, the rate of the reaction is found out to be equal to the rate constant K multiplied by the concentration of N2O to the power 0. So this becomes now a zero-order reaction. So what exactly is happening out here? Let's see. Now in the presence of nickel, the N2O molecules get adsorbed at this nickel surface. In other words, they use some of their electrons to form bonds with this nickel metal. So because they use some of their electrons to make new bonds, so their old bonds gets weaker and therefore these N2O molecules that are adsorbed at the nickel surface, these N2O molecules on collision can break the bonds much more readily, leading to the formation of N2O2. So this is how nickel catalyzes the decomposition of N2O. However, having said that, as you can see the rate of this reaction is actually limited by the surface area of nickel that's available for adsorption. Now generally, the number of gaseous molecules that we have is much much greater compared to the available adsorption sites. So once these adsorption sites are filled, which happens pretty quickly, these unabsorbed N2O molecules have to wait for some of these reactions to happen and only when these sites get freed up, only then they can get adsorbed. So in this scenario, if I keep increasing the concentration of N2O, this is not going to lead to a consequential increase in the rate of the reaction, right? So if I increase the concentration of N2O, there is going to be no change in the rate of reaction under these circumstances. So that is why under these conditions, the rate of the reaction is zero-order with respect to N2O. Now a quick note that I'd like to add out here is that as this reaction keeps happening, as more and more N2O molecules convert into N2O2, then ultimately it might so happen that the number of N2O molecules are not in excess compared to the adsorption sites. So in such a scenario, the reaction will no longer be zero-order and will revert to some other order. So what I'm trying to say is that in reality, we actually cannot have a zero-order reaction that's like a 100% zero-order all the time.