So we have looked at how to simplify the algebra in cases where you have an x cubed and you're not able to do the algebra brute force to solve an equilibrium problem. But sometimes there are equilibrium problems that need to be solved with x is small approximations, but you're not able to do so. So let's set this one up. I have my reaction here.
Initially, I have zero of my reactant. I have one atmosphere of NO and 0.5 atmospheres of O2. And I'm just going to go straight forward through this and see what happens.
So fill in my change row, carry it down to my equilibrium row, and look at my equilibrium constant in comparison to what I'm starting with based on my initial conditions. And I cannot assume that x is small. So I have a really small equilibrium constant, which means at equilibrium, I should end up with very little product. And what I have is 100% product. So the way that this reaction is going to move is that it is going to move a lot to the left.
It's going to move the majority of the products back to reactant when it goes to equilibrium. based on the value of my equilibrium constant. So okay, I can't assume x is small. Let's set up my equilibrium constant expression with the values plugged in. And when I do this, the top of this equation is where you can see that the x cubed is going to pop up, right?
I'm going to have an x cubed. The only way that I know how to solve this is by assuming that x is small, but we cannot use it from this set of conditions. So comparing my initial conditions to my equilibrium constant, I am not close to equilibrium at all, which is the requirement for being able to use x is small. So the way that we go about this, now I'm going to show you what it looks like in this one case.
You can use it in in a lot of cases. But this is just for this example. If I have a small equilibrium constant, let's get it into a position where it will have a small amount of change.
So I've restarted my table with my initial row. So I am going to choose to push this reaction completely to the left. This is going to be if 100% of my products reacted to form the reactants, where would my reaction be?
So I know that based on my equilibrium constant and the current initial conditions that 99.9% of it is going to turn into reactant. Let's turn 100% of it into reactant and then let it go to equilibrium. So let's push it past the equilibrium point and then let it go back to equilibrium. So with the mole ratios that I have here, if I completely react my O2, so subtract 0.5 atmospheres of O2 from the product side, that would perfectly react one atmosphere of NO, and it would make one atmosphere of NO2.
This is based on the mole ratios in the reaction. And in this case, the mole ratios are perfect, right? I'm going to use up 100% of both of my products.
That might not always be true. So, right, you might have to watch for a limiting reactant. You will have to look at the mole ratios to figure out what's being used up and what's being made. Now, my new initial conditions are one atmosphere of reactant and no product. So my reaction is going to go to the right to get to equilibrium.
I have my equilibrium row set up and I will still have an x cubed in the top of this expression. However, I'm starting with a high concentration of reactant and my equilibrium constant is really small. So I can now correctly assume that x is small.
This leaves me with 4x cubed is equal to 5.9 times 10 to the minus 13th. I'm still in x cubed, but I can solve this on these calculators. x is 5.284 times 10 to the minus fifth. If you test this, it's a great assumption.
The percentage is times 10 to the minus third percent. So this x value works, and I was initially asked for the pressure of NO at equilibrium, which would be 2 times x, and that means my pressure at equilibrium is 1.057 times 10 to the minus fourth. atmospheres and it does say NO.
I did give the pressure for NO. That is just a handwriting mistake. That should not say NO2.