If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Worked 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.

Want to join the conversation?

  • duskpin ultimate style avatar for user Li
    how did he figure out that P total is 2.35?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user aparnamp966
    At you 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)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Winter
    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)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      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.
      (1 vote)
  • blobby green style avatar for user Bushra Mansour
    what if we're not given the total pressure?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user narteyreube
    when given just the total pressure to be 0.766 to find the kc.What should you do
    (1 vote)
    Default Khan Academy avatar avatar for user

Video 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.