Worked example: Calculating equilibrium concentrations from initial concentrations | Khan Academy

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 going to 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 going to put a zero 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 going to call that amount x. When we form our product, we are going to lose some of that bromine, so we're going to 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 minus x under chlorine in the ICE table.

When Br₂ and Cl₂ react together, we lose our reactants, and that means they're going to gain some of our products. To figure out how much, we need to look at mole ratios. The mole ratio of bromine to BrCl is one to two. Therefore, if we're losing x for Br₂, we must be gaining 2x for BrCl, so I can go ahead and write plus 2x under BrCl.

Next, let's think about equilibrium concentrations. If the initial concentration of bromine is 0.60 and we're losing x, the equilibrium concentration must be 0.60 minus x. The same thing for chlorine; it would be 0.60 minus x. For BrCl, we started off with 0 and we gained 2x; therefore, at equilibrium, the equilibrium concentration would be equal to just 2x.

The next step is to use the balanced equation to write an equilibrium constant expression. So we would write Kc is equal to (concentration of BrCl)² divided by (concentration of Br₂)¹ times (concentration of Cl₂)¹. The concentrations in an equilibrium constant expression are equilibrium concentrations.

Therefore, we can plug in the equilibrium concentrations from our ICE table. The equilibrium concentration for BrCl was 2x, 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 rewritten it down here because (0.60 minus x)² is equal to (0.60 minus x)². If we write it this way, it's a little 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 (2x) / (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.

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 2 times x, so that's 2 times 0.34, which is equal to 0.68 Molar. So 0.68 Molar is the equilibrium concentration for BrCl.