How Do I Calculate Equilibrium Partial Pressures in a Dissociation Reaction?

[wingsb2]

It's been a while since I've done any equilibrium problems - especially ones at constant temperature so some help would be appreciated.

I have a dissocation reaction A_2 <----> 2A. I am told that at temperature T and pressure P1 = 1atm, there are twice as many moles of A_2 molecules as there are of A atoms. Then the mixture is expanded at constant temperature to 4 times the initial volume. I need to find the equilibrium partial pressures of A_2 and A and the total pressure P2.

Thanks for your help!

 

[Navin A S]

The partial pressure of A_2 at the initial state is P1/3. The partial pressure of A at the initial state is 2P1/3. The total pressure of the system at the initial state is P1.When the system is expanded to 4 times the initial volume, the partial pressures of A_2 and A do not change because they are at constant temperature. However, the total pressure of the system changes to P2 = P1/4. Therefore, the partial pressures of A_2 and A at the new equilibrium state are P2/3 and 2P2/3 respectively.

 

[ravenprp]

Sure, I'd be happy to help with your equilibrium problem. First, let's review the basics of equilibrium. In a reaction like the one you have, A_2 <----> 2A, there is a forward reaction (A_2 breaking apart into A atoms) and a reverse reaction (A atoms coming together to form A_2). At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of A_2 and A remain constant.

Now, let's apply this to your problem. At the initial conditions, we have 2 moles of A_2 and 1 mole of A in a 1 atm pressure. Since there are twice as many moles of A_2 as A, the equilibrium constant (K) for this reaction can be written as:

K = [A]^2/[A_2] = (1/2)^2 = 1/4

Next, we expand the mixture to 4 times its initial volume. This means the concentration of both A_2 and A will decrease by a factor of 4, but the equilibrium constant remains the same. Now, we can set up an ICE (Initial, Change, Equilibrium) table to find the new equilibrium concentrations:

A_2 <----> 2A
I: 2 mol 1 mol
C: -1/4 mol -1/2 mol
E: 1 3/4 mol 1/2 mol

Note that we have to take into account the stoichiometry of the reaction when determining the change in concentration. Now, we can plug these values into the equilibrium constant expression and solve for the equilibrium partial pressures:

K = [A]^2/[A_2] = (1/2)^2 = 1/4
1/4 = (1/2)^2/[A_2]
[A_2] = 1 atm

Since the equilibrium partial pressure of A_2 is 1 atm, the partial pressure of A must be 1/2 atm (due to the 2:1 stoichiometry). Therefore, the total pressure (P2) in the expanded mixture is 1 + 1/2 = 3/2 atm.

I hope this helps! If you have any further questions, please don't hesitate to ask. Equilibrium problems can be tricky, but with practice they become easier to solve. Good luck!