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Course: AP®︎/College Chemistry > Unit 7
Lesson 5: Calculating equilibrium concentrationsWorked example: Calculating equilibrium concentrations from initial concentrations
In this video, we'll learn how to use initial concentrations along with the equilibrium constant to calculate the concentrations of reaction species at equilibrium. Created by Jay.
Want to join the conversation?
Um, I feel like he did the problem wrong because I got x=0.39. I did not square the problem like he did and used the quadratic formula to solve.(3 votes)
The answer is still 0.34 if you solve it with the quadratic formula. You actually find two answers with the formula (because it's a quadratic) which means x could equal 0.34 and 2.46. But only 0.34 works since 2.46 would create negative molarities for the reactants at equilibrium. Not sure how you got 0.39 though.
Hope that helps.(5 votes)
Okay...so I might have missed when Jay was discussing this...but...
why are we putting 'minus x's' for both Bromine and Chlorine?
I think I'm missing something to be honest....
Smh.
- THE WATCHER(3 votes)- The x's represent essentially the change in concentration for the reactants and products. Since the reaction in moving in the forward direction, the concentration of the reactants will decrease while the concentration of the product will increase which explains the signs. We don't exactly know by how much the concentration changes though yet so we represent that with the variable.
Hope that helps.(2 votes)
Video transcript
- [Instructor] For the
reaction bromine gas plus chlorine gas goes to BrCl, Kc is equal to 7.0 at 400 Kelvin. If the initial concentration
of bromine is 0.60 molar and the initial concentration of chlorine is also 0.60 molar, our
goal is to calculate the equilibrium concentrations
of Br2, Cl2 and BrCl. To help us find the
equilibrium concentrations, we're gonna use an ICE table, where I stands for the
initial concentration, C stands for the change in concentration and E stands for
equilibrium concentration. For the initial concentrations, we have 0.60 molar for bromine, 0.60 molar for chlorine, and if we assume the reaction hasn't started yet, then we're gonna put a zero
in here for our product, BrCl. Next, we think about Br2
reacting with Cl2 to form BrCl. Some of the bromine is going to react, but we don't know how much, so we're gonna call that amount x, and we're gonna lose some of that bromine when we form our product,
so we're gonna write minus x under bromine in our ICE table. Next, we think about mole ratios. In the balanced equation,
it's a one to one mole ratio of bromine to chlorine. Therefore, if we're losing x for bromine, we're also going to lose x for chlorine. So I can write here minus x
under chlorine in the ICE table. When Br2 and Cl2 react
together, we lose our reactants, and that means we're gonna
gain some of our products. To figure out how much, we
need to look at mole ratios. So the mole ratio of bromine
to BrCl is one to two, therefore if we're losing x for Br2, we must be gaining two x for BrCl. So I can go ahead and write
plus two x under BrCl. Next, let's think about
equilibrium concentrations. If the initial concentration
of bromine is 0.6 and we're losing x, the
equilibrium concentration must be 0.60 minus x. And the same thing for chlorine. It would be 0.60 minus x. For BrCl, we start off with
zero, and we gained two x. Therefore at equilibrium,
the equilibrium concentration would be equal to just two x. The next step is to use
the balanced equation to write an equilibrium
constant expression. So we would write Kc is equal to, and then we look at our balanced equation, and for our product we have BrCl with a two as a coefficient, so Kc would be equal to the
concentration of BrCl squared, and we're gonna divide
that by the concentration of our reactants, which would be Br2, so the concentration of Br2
raised to the first power, because the coefficient of one, times the concentration of Cl2 also raised to the first power. The concentrations in an
equilibrium constant expression are equilibrium concentrations,
therefore we can plug in the equilibrium concentrations
from our ICE table. So the equilibrium concentration
for BrCl was two x, the equilibrium concentration
for Br2 was 0.60 minus x, and the same for chlorine, so
we can plug that in as well. Here we have our
equilibrium concentrations plugged into our equilibrium
constant expression, and also Kc was equal to 7.0
for this reaction at 400 Kelvin so 7.0 is plugged in for Kc. Our goal is to solve for x, and
I've re-written it down here because 0.60 minus x times 0.60 minus x is equal to 0.60 minus x squared. And if you write it this
way, it's a little bit easier to see that we can solve for x by taking the square root of both sides. So let's go ahead and take
the square root of both sides and solve for x. Taking the square root of both sides gives us 2.65 is equal to
two x over 0.60 minus x. To solve for x, we would
then multiply both sides by 0.60 minus x to give us this, and then after a little more algebra, we get 1.59 is equal to 4.65x. So x is equal to 1.59 divided by 4.65, which is equal to 0.34. Now that we know that x is equal to 0.34, we can plug that into our ICE table and solve for our
equilibrium concentrations. So for the equilibrium
concentration of Br2, it's 0.60 minus x, so
that's 0.60 minus 0.34, which is equal to 0.26 molar. So 0.26 molar is the equilibrium
concentration for bromine. For chlorine, it would
be the same calculation, 0.60 minus x would be 0.60 minus 0.34, so the equilibrium
concentration of chlorine is also 0.26 molar. For BrCl, it's two times x
so that's two times 0.34, which is equal to 0.68 molar. So 0.68 molar is the equilibrium
concentration for BrCl.