 ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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# Equilibrium Constants from Initial Concentrations - Concept

Jonathan Fong ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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Here are some tips and trick for calculating equilibrium concentrations from initial concentrations. So we will take a look at 3 different cases. And the first case we’re going to calculate, equilibrium concentrations when initial concentrations are given and one equilibrium concentration is known. In all these cases we going to use what’s called and ICE table. And the ‘I’ means Initial, and the ‘C’ stands for Change, and ‘E’, which should make sense stands for Squilibrium. So we’re going to use the ICE table method.

And in the ICE table method, all units are in molar, for molarity. So it will be Molar, capital M so don’t forget that. If it's moles you got to convert the litres.

Say for example we’re given that we have 1Molar N2, and then we have 2Molar H2, and then we have 0Molar NH3. And then in our ICE table, we have the equilibrium concentration of and NH3 is known. It's known to be equal to 1M. So we want to find out what the N2 concentration is going to be at equilibrium, and we also what to find out what the H2 concentration is at equilibrium. So what we’re going to do is, in the ICE table methods, we will have our I line because that’s given, and we have our equilibrium concentration. In our C line, what we’re going to do is, we are going to balance our equation, because we always need to do that. So if you balance the equation you will end up with a 3 and the 2 here.

And so what we’re going to do is, since we have initially no NH3, no Ammonia, what we’re going to do is, we know the change is going to be positive in the product side. Does it really matter? No, it doesn’t. So if we actually put minus, then we would actually get a negative change and then we would figure it out.

But from here the change is going to be +x. Now there is going to be a 2 in front of the x, because it matches the coefficient. So it will be +2x, so I have that there.

For the C line for N2, since it’s on the opposite side on the reactant, it’s going to get used up. So I’m going to have –x. And then for the C line for H2, since I have a coefficient at 3, it will be -3x. So remember, the coefficient falls down with the x here. So now what we’re going to use is, we going to use our information here. Because we know that to get the change 2x, we had none to start with. So zero plus 2x should equal 1M. So if zero plus 2x equals 1, and we'll just drop the molar just for the sake of making it easier for right now. Then, we know that 2x should equal the number 1.

So if we divide, x should equal 0.5. So then, we can actually plug in the x’s in our spots and figure it out. So N2, 1 minus 0.5 will get me 0.5M per equilibrium. And then 2 minus 3 times 0.5, it will be 2 minus 1.5 so we would be left with 0.5M also. So that’s how we figure out the equilibrium concentrations when we’re given initial concentrations, and equilibrium concentrations of one species. And so remember, balance your equation and then on your C line, remember your coefficients drop down.

Number 2 is very similar. So calculating when initial concentrations when one change is known. So we have this equation here, and we will use our ICE table nice and quick. We’re given that we have 1Molar of N2, we have 2Molar of Br2 and we have no Molars of HBr, for example. And so say for example, we know this is the change of the H2 is -0.25M. So we want to figure out the equilibrium concentrations of H2, the equilibrium concentration of Br2, and the equilibrium concentration of HBr.

So like we did before remember balance, so we’re going to balance the equations. So we'll put a 2 in front of the HBr. Now in our C line we use our x method. So we add 2x to the HBr, because the coefficient of 2 is there. We would have -x for the Br2 because there is one coefficient and then also, important is the H2, because we know that we would also have –x here. So that means that based on what we have here, x should equal 0.25M because of what we have, based on the change that we know. And so, if we do the math, 1 minus 0.25 is 0.75M for H2. For Br2 it will be 2 minus 0.25, so that would 1.75M. And then for HBr it will be 0 plus 2 times 0.25 so that’s 0.50M. And so that’s how we figure out the equilibrium concentrations when initial concentrations and one change is known.

Now the last part is the trickiest part, and here are some tips and tricks. Calculate the equilibrium concentrations when initial concentrations and the K value is known. So remember, K is the equilibrium constant value that we have. So we’re given that the K of this equation is equal to 794 at 25 degrees Celsius. And we’re given our initial concentrations, so we will make so that we have no Molar of this. And then we’re given that we have 2Molar H2 and 2Molar I2. So those are the things that we're given.

So now we want find the equilibrium concentrations of H2, of I2 and of HI. So we do the same thing; balance, balance the equation. So we'll put a 2 in front of the HI. And then, so we have our I line and that's it. So under our C line, since we have no HI, that means we’re going to use up some of the issue. So -x, since there was only 1 for the coefficient. –x for the I2 and then +2x for the HI.

Now keep in mind on the C line, the –x’s do not need to necessarily add up to the plus x’s on the opposite side. Because if you look at a our first example, from N2 plus 3H2 yields to an H3, if you notice the x’s don’t add up on both sides. I have minus 4x on the left side, and now on the product side, we have +2x. So those x’s do not necessarily equal. So just to answer your question if you had it.

So under the E line, we would calculate that. We would have 2 minus x, to 2 minus x and we have 2x here. So then what we will do is, we would use our K expression. And if you don’t how to write to a K expression, take a look at some of our other videos, and it help you to write a K expression. So the products go on the top on the K expression. So it will be the concentration HI and then it would be squared since the coefficient was 2, over the concentration of H2, times the concentration of I2.

Now we plug our E line into those. So we would have 2x, the quantity squared, because don’t forget to keep that square there, over 2 minus x, over 2 minus x. And so that would be the same thing as saying, 2x squared over 2 minus x squared. Now all that is equal to 794.

Now there are 3 cases that could happen here, and I have them listed as A, B and C. I could have a perfect square, where I can take the square root of both sides, and that’s where my x’s are, I can spacely square root that to simplify that. Then I have letter B where if I can’t square root, then I have some x’s here, and I can actually cancel the x’s out like the 5 plus x and the 2 minus x there. I can cancel those out because those would be small compared to the 5 and compared to the 2. I can’t eliminate that x. The reason why is because the small number is still a small number by itself. But when it’s attached to a 5, or attached to a 2, then I can eliminate those because those would be negligible compare to whatever number I have. After I calculate x then I can compare those to my concentration to figure out if they’re indeed less than 5%. And that’s called the 5% rule that you may have heard about in class.

And then the last one is using a quadratic. If I can’t use the 5% rule, then I would use a quadratic equation, and that would just take time, and then you would just us your quadratic equation like in math class, and then, you can solve for your x that way. So in this case our example, since we have K equals and then 2x the quantity squared, over 2 minus x. So that quantity squared is equals 794. If you take a look here, we do actually have a perfect square, where we could square root both sides, both that and that. And then we can get 2x over 2 minus x equals the square root of 794, and then we can solve for x. I won’t do that here to save yourself some time.

So those are the 3 cases that you would see for calculating equilibrium concentrations from initial concentrations. The first here are pretty easy. All you’re doing is some simple algebra. The last one at is a little trickier. And so hopefully this tips and tricks, help lay out a plan for you to attack and figure out how to calculate those equilibrium concentrations. Have a good one.