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Course: AP®︎/College Chemistry > Unit 7
Lesson 5: Calculating equilibrium concentrationsWorked example: Calculating an equilibrium constant from initial and equilibrium pressures
Given an initial partial pressure and the total pressure at equilibrium, we can use Dalton's law to determine the equilibrium partial pressures of the gases in a reaction mixture. Once we know the equilibrium partial pressures, we can calculate the equilibrium constant for the reaction. Created by Jay.
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how did he figure out that P total is 2.35?(5 votes)
It was given in the problem at0:23before he began the calculations.(8 votes)
1. Why is it true that, if PCl5 is losing x partial pressure, the products must each be gaining x partial pressure? What is the relationship between partial pressure and molar ratio?
2. If PCl3 and Cl2 each gains x partial pressure but PCl5 loses x partial pressure, doesn't it indicate that mass was not conserved? How is this possible?
3. When should we use the method of solving with the total partial pressure as opposed to the reaction quotient?(1 vote)- We’re assuming that these are ideal gases, which means the moles of gas from each chemical exerts the same amount of pressure. Even though we have three different chemicals, as far as the pressure is concerned, their molecules are identical. Temperature is constant at 500°C, and we’ll assume it is an inflexible container so volume is constant too. The only thing then that can affect pressure is the number of gas moles according to the ideal gas law. As the reaction progresses, the number of moles of gas changes too resulting in the change in pressure.
The pressure will increase because we’re making more gas moles, but that’s pressure, not mass. Mass is conserved because the chemical equation is balanced.
Well Jay is using equilibrium, but just using partial pressures and the reaction’s Kp instead of Kc and molarities. He’s using the partial pressure equilibrium simply because these are gases which are more easily measured using pressure.
If you actually meant the reaction quotient, well we only use that when the reaction is not at equilibrium and to determine which way to the reaction will progress. But we’re not trying to do that in this problem, we’re actually trying to calculate the equilibrium constant.
Hope that helps.(2 votes)
At1:26you said PCl5 will lose some of it and wrote -x. How can we say that product formation (or PCl3 and Cl2 is formed) is favored without knowing equilibrium constant? What if the equilibrium constant is negative?(1 vote)- what if we're not given the total pressure?(1 vote)
- Then it will have two variables which would hinder the calculation.
In short we must either be given the Total pressure or the partial pressure of each gas.(1 vote)
- when given just the total pressure to be 0.766 to find the kc.What should you do(1 vote)
0:23Video transcript
- [Instructor] Let's say
we have a pure sample of phosphorus pentachloride, and we add the PCl5 to a
previously evacuated flask at 500 Kelvin. And let's say the initial
pressure of the PCl5 is 1.66 atmospheres. Some of the PCl5 is going
to turn into PCl3 and Cl2. Once equilibrium is
reached, the total pressure, let's say, is measured
to be 2.35 atmospheres. Our goal is to calculate the
equilibrium partial pressures of these three substances,
so PCl5, PCl3 and Cl2. And from those equilibrium
partial pressures, we can also calculate the Kp value for this reaction at 500 Kelvin. To help us find the
equilibrium partial pressures, we're gonna use an ICE
table where I stands for the initial partial
pressure in atmospheres, C is the change in partial pressure, and E stands for the
equilibrium partial pressure. We already know we're starting with a partial pressure of
1.66 atmospheres for PCl5. And if we assume that the
reaction hasn't started yet, we're starting with zero
for our partial pressures of PCl3 and Cl2. Some of the PCl5 is going to decompose. And since we don't know how much, we're gonna call that amount x. So, we're gonna write minus x here, since we're gonna lose some PCl5. Next, we need to look at mole ratios. So, the mole ratio of PCl5
to PCl3 is one to one. So, for losing x for PCl5, we must be gaining x for PCl3. The same idea with Cl2, the coefficient in the
balanced equation is a one. So, if we're losing x for PCl5, that means we're gaining x for Cl2. Therefore, the equilibrium
partial pressure for PCl5 would be 1.66 minus x. The equilibrium partial pressure for PCl3 would be zero plus x, which is just x. And the equilibrium
partial pressure for Cl2 would be zero plus x, which is also x. To figure out what x is, we're
going to use Dalton's law. And Dalton's law says
that the total pressure of a mixture of gases is equal to the sum of the individual partial pressures of the gases in the mixture. So, we said that the total pressure of all the gases at equilibrium is equal to 2.35 atmospheres. So, we can plug that into Dalton's law. And then, we can take the
equilibrium partial pressures from our ICE table and plug
those into Dalton's law as well. So, we're gonna plug in 1.66 minus x for the equilibrium
partial pressure of PCl5, x for the equilibrium
partial pressure of PCl3, and x for the equilibrium
partial pressure of PCl2. So, let's plug in 1.66 minus x. And then, we have plus x and plus x. And let's solve for x. Notice how we have a minus x and a plus x. So, that cancels out. So, we simply subtract 1.66 from 2.35 and we find that x is equal to .69. So, if x is equal to .69, the equilibrium partial pressure
of Cl2 is .69 atmospheres. And the equilibrium partial pressure PCl3 is also .69 atmospheres. And 1.66 minus .69 gives us the equilibrium partial pressure of PCl5 and that's equal to .97 atmospheres. Now that we have our
equilibrium partial pressures for all three gases, we can calculate the value
for the equilibrium constant for this reaction at 500 Kelvin. First, we need to write an
equilibrium constant expression. So, we would write Kp is equal
to, and for our products, we have PCl3, so this would be
the partial pressure of PCl3 times the partial pressure
of our other product, which is Cl2, so let's put in there the partial pressure of Cl2. And all of that is divided by
the partial pressure of PCl5. So, this would be the
partial pressure of PCl5. Next, we plug in our
equilibrium partial pressures. So, the equilibrium partial
pressure of PCl3 is .69. The equilibrium partial
pressure of Cl2 is also .69. And the equilibrium partial
pressure of PCl5 is .97. Once we plug our numbers in and we solve, we get that Kp is equal
to .49 at 500 Kelvin.