 So we have this reaction where A and B react to form C and all of these are gases. So we want to know that at this point when they are in equilibrium, what if we introduce a few moles of inert gas? What will happen to the equilibrium? Will it shift towards the reactants or the products or will it stay the same? But before we begin, there are two things to note here. First, when we say inert gas, we are not just referring to gases like helium, argon and other noble gases. The word inert here is implying that the gas that is introduced is not reacting with any of these gases. So if we have some Z moles of a gas G, such that this gas does not react with any of these gases, we can say that this gas is inert for this reaction. So when we say inert gas, it could be any of these, but it can also be some other gas which is not reacting with A, B and C. So that's the first point. Next, when we're talking about this, we are assuming that the temperature is not changing. So we are at equilibrium and the temperature is fixed. And at this point, we want to think about what happens when an inert gas is introduced. So for that, we need to think of two variables, the pressure and the volume which can affect the equilibrium. So to look at them individually, let's take two cases. One, when we have the pressure constant and second, when the volume is constant. So first, let's take the case where an inert gas is introduced at equilibrium keeping the pressure constant. So first, let's look at the constant pressure case. And we'll go through this case by looking at this example. So let's say we have two gases, A and B, which combine to give a third gas, C. And we're assuming that the partial pressures of each gas is equal to 1 atm. And so the total pressure will be equal to the sum of all the partial pressures, which is 3 atm. So from our definition, we know that the equilibrium constant will be equal to the partial pressure of C divided by the partial pressure of B times the partial pressure of A. And because all of our stoichiometric coefficients here are 1, all of these would be raised to 1, but that does not make any difference. So we're just going to write it as this. And now let's say we introduce an inert gas here. Since we're assuming constant pressure conditions, the total pressure is going to be the same, which is 3 atm. But because we're adding one more gas and the total pressure has to be 3 atm, this would mean that each of these partial pressures will have to reduce. And we know this because if we look at how the partial pressure is defined, let's say for gas A, the partial pressure of A will be equal to mole fraction of A times the total pressure. And we know that mole fraction is defined as the number of moles of A divided by the total number of moles. So when we add an inert gas, the total number of moles goes up and the number of moles of A don't change. So the mole fraction of A decreases. And since it is a constant pressure condition, we know that the total pressure will not change. And so the partial pressure of A must reduce where an inert gas is added. So here if we make an assumption that each of these partial pressures has reduced by the same amount, which was also equal to the partial pressure of the inert gas, all of them would be equal to 3 by 4 atm. And we're just making this assumption to make our life easier. But you can verify that this assumption is valid because if you sum up the partial pressures of all four of them, which is 4 times 3 by 4, the total pressure will be still 3 atm. So after introducing this inert gas, the partial pressures of A, B and C have reduced and we know their value. So at this point, what if we calculate the reaction quotient and we know that the reaction quotient will be equal to the partial pressure of C which is 3 by 4 divided by the partial pressure of B which is also 3 by 4 times the partial pressure of A which is also 3 by 4. So since you can cancel out both of these, the reaction quotient after introduction of an inert gas is 4 by 3. And if we go back to our original equilibrium constant, the value of K was 1 because all of these partial pressures were equal to 1. So now after introduction of the inert gas, what we notice is the reaction quotient is greater than the equilibrium rate constant. And we know that when this happens, the equilibrium will shift in the reverse direction. So for this reaction, it will shift towards the reactants. So we'll have more amount of A and B being formed. And because this is a type of a formation reaction where we have one mole on the product side and two moles on the reactant side. So we can say that the equilibrium is shifting in the direction of increasing number of moles. And this was for a formation type reaction. So what happens if you have a dissociation type reaction where let's say A dissociates to give B and C. So for that case, let's carry out a very similar calculation and see how the equilibrium shifts. In this case, we have this dissociation type of reaction where A dissociates to give B and C. And again, because we know that all of them are gases, the rate constant is going to be the partial pressure of C times the partial pressure of B divided by the partial pressure of A. And the rest of the conditions are same as before. We're assuming that the partial pressures of A, B and C are 1 atm. So the total pressure is 3 atm. And when we introduce an inert gas at equilibrium because the total pressure is fixed and we're adding a new gas, the partial pressure of each of these should reduce. And the sum of all the partial pressures that is the partial pressure of A, B, C and the inert gas should be equal to the total partial pressure, which is 3 atm. So to make our calculations easier, we assume that all of the partial pressures are equal and their value is 3 by 4 atm. And again, you can check by adding all four of these that if all of these are 3 by 4 atm, the total pressure will still be 3 atm. So here, the only change from the previous example is that the reaction is a dissociation type reaction. And for these conditions, if we calculate the reaction quotient, it is going to be the partial pressure of C times the partial pressure of B divided by the partial pressure of A. And the new partial pressures, all of them are 3 by 4 atm. So this is going to be 3 by 4 times 3 by 4 divided by 3 by 4. And again, we can cancel out these two. The reaction quotient is 3 by 4 after the introduction of an inert gas in the case of this reaction. And just like before, if we plug in the original values of partial pressures, we find the equilibrium rate constant to be 1. So in this case, the reaction quotient is less than the equilibrium rate constant, which means the equilibrium will shift in the forward direction. So here, the equilibrium will shift in the direction of the products, which means we'll have more of B and C forming. And again, just like before, we have more number of moles on the product side. We have two moles, one mole of B and one mole of C. And on the reactant side, it is only one mole of A. So in this case also, the equilibrium shifts in the direction of increasing number of moles. But in both of these cases that we saw, we did not have equal number of moles on the reactants and product side. So let's take one last example to see what happens when we have equal number of moles on both sides. So we're repeating the same exercise once again, but this time the reaction is such that we have equal number of moles on the reactant and the product side. And again, we are going to assume that the initial partial pressures are all equal to 1 atm. So this time, the total pressure is the sum of all of these partial pressures, which is 4 atm instead of the 3 atm that we were getting before. And this is how we define the equilibrium rate constant for this reaction. So now when we introduce an inert gas, we will have five gases in the mixture. There is four of these and the inert gas. And just like before, we know that because of the introduction of inert gas, the partial pressure of each of these will reduce. We assume that the partial pressure of all of these gases, including the inert gas, is equal to 4 by 5 atm. And then if we check the total of partial pressures from here, that is the sum of partial pressures of all of these gases, we find that it'll be equal to the 4 atm, which is the total pressure. So all of these calculations are just like before. The reaction quotient will be partial pressure of C, which is 4 by 5 times partial pressure of B, which is also 4 by 5, divided by the product of the partial pressures of A and B, which are also 4 by 5. So now if we cancel out everything, we find the reaction quotient to be 1. And just like before, if we look at the equilibrium rate constant, since the initial pressure of all of these was 1 atm, the value of K will also be 1. And so here we see that the reaction quotient is equal to the value of the equilibrium rate constant. So we know that in this case, because these values are equal, we know that the equilibrium will not shift. So if we compare all the calculations that we did before this, whenever we have unequal number of moles between the reactant side and the product side, we saw that the equilibrium shifts to produce more moles of the gas. And when we had equal number of moles on the reactants and the product side, equilibrium position did not shift. So this was for introduction of an inert gas at constant pressure condition. Now let's see what happens if we introduce an inert gas at constant volume and spoiler alert, the number of moles are not going to matter in that case. So we'll need only one calculation in that case. Let's see what that is. So once again, we go back to the same reaction that we took before. But this time we want to look at the constant volume condition and we know that for this reaction, the rate constant is defined like this. And if we look at this partial pressure in detail, if we look at how it is defined, let's say we want to find the partial pressure of A. So that will be defined as the mole fraction of A times the total pressure. And we know that the mole fraction of A is defined as the number of moles of A divided by the total number of moles. And just like before, let's assume that the partial pressures of A, B, and C are all equal to 1 atm at equilibrium. So we know that the total pressure will be the sum of these partial pressures, which is 3 atm. Now we want to know what happens when we add an inert gas at equilibrium. So let's say we add one mole of an inert gas I and let's say the partial pressure of this inert gas is 1 atm. And again, we're assuming these numbers to make our calculations easier. So in this case, we want to know how the values of partial pressures of A, B, and C change. And here this is in constant volume condition. So if you think of this reaction happening in a box initially at equilibrium, this fixed volume container will have a mixture of gas A, gas B, and gas C. And what we're doing is, let's say there's an opening here. We're just opening this container and adding in the inert gas and closing the container, which is our constant volume condition. So now with the inert gas in the mixture, the total pressure will be the sum of all partial pressures of A, B, C, and the inert gas. So it'll be 4 atm. And now if we calculate the value of partial pressure of A after addition of the inert gas, we know that it'll be equal to the mole fraction of A times the total pressure. So the mole fraction of A will be the number of moles of A, which is one mole, divided by the total number of moles. And because we've also added one mole of inert gas, the total number of moles in this case will be 4 times the total pressure, which is 4 atm. So if we solve this, we get the new partial pressure of A to be 1 atm, which is exactly the same as before. And similarly, if we also calculate the partial pressure of B on addition of inert gas, we get that also to be 1 atm. And similarly for C as well. So the point is, even after introducing the inert gas, the individual partial pressures of A, B, and C do not change. And because the partial pressures don't change, the equilibrium rate constant will also not change. So we can say that at constant volume condition, the addition of a inert gas has no effect on the equilibrium. And although we took a simple case here, you can try different values of the amount of inert gas or the partial pressure of the inert gas that is added or even different types of reactions. And if you go through this calculation, you'll hopefully arrive at the same result that at constant volume, the addition of an inert gas has no effect on the equilibrium.