Question

Introduced into a 1.70 L flask is 0.120 mol of PCl5; the flask is held at a temperature of 227 C until equilibrium is established.
PCl5(g) <-> PCl3(g) + Cl2(g)
What is the total pressure of the gases in the flask at this point?

Answer 1

Do we know the equilibrium constant for this reaction?

Answer 2

no, that is what was given

Answer 3

Does it say anywhere the enthalpy and the entropy of this reaction or of PCl5/PCl3/Cl2?

We need some way to find K if it hasn't been given, in order to figure out how much of the product is formed. I'm thinking of finding it by calculating the Gibbs free energy, which we can do if we know ∆H and ∆S.

Answer 4

Can't see what you typed^, I see a bunch of question marks in diamonds

Answer 5

They are delta's - just refresh the page you should see it ok.

Answer 6

Would that help? I'm just not sure how to approach this problem

Answer 7

This is exactly what I was hoping for!

Can you just type out the 3 numbers for Cl2 (g) as well? They aren't in the picture you provided.

Answer 8

Cl2: delta H : 0 , delta G: 0 , delta S: 223 J/K

Answer 9

Perfect. Here's what to do:

You're adding PCl5, and because the reaction we're talking about is an equilibrium, we know it needs to decompose to some extent to PCl3 and Cl2. However, to be able to answer this question at all, we need some way of knowing how far the reaction proceeds to the right, in order to know how many moles of gas are present (because this will determine the pressure).

We can figure out how far the reaction proceeds if we know the equilibrium constant for the reaction. Since we're not given it in the question, we can figure it out using the Gibbs free energy (∆G), since it's related to K by ∆G° = -RTln(K). ∆G° is referring to the standard Gibbs free energy, which is what's provided in your table.

To figure out the ∆G° of our reaction, you could use the equation ∆G° = ∆H° - T∆S° (which is why I was asking about enthalpies and entropies earlier). However, since your table gives ∆G° of formation for each individual compound in the reaction, we can use a simpler equation: just subtract ∆G°(f) of the reactants from ∆G°(f) of the reactants to find the "net" ∆G° of the entire reaction:

\[\Delta G^{o}=\Delta G_{products}^{o}-\Delta G_{reactants}^{o}\]\[\Delta G^o=-267.8-(-305.0)=37.2\]
So for this reaction, ∆G°=37.2 kJ/mol=37200 J/mol.

I converted to J/mol because we need to use these units in the equation for calculating K. I'm not done yet, but just wanted to make sure you understand this so far. Any questions?

Answer 10

Not at the moment

Answer 11

Ok. Next we find K using ∆G° = -RTln(K). R is the gas constant (here it needs to be 8.314 J/mol K) and T is the temperature in Kelvin. So:

\[37200=-8.314(500)\ln K\]\[\ln K=-8.95\]\[K=1.3 \times 10^{-4}\]
So now we have our equilibrium constant, so we can find equilibrium amounts of each compound using an ICE table. Can you tell me what you get for the moles of PCl5, PCl3, and Cl2 at equilibrium?

Answer 12

PCl5: 0.116 mol , PCl3/Cl2: 0.00402mol

Answer 13

@matt101 this is how i did it...

Answer 14

Great work - you did it right, you just dropped your negative in the second quadratic formula step!

So the we should have PCl5: 0.116 mol , PCl3/Cl2: 0.00388 mol

Now the final step - calculate the pressure. We know the total number of moles now (just add them up), so just solve for P using PV=nRT. Let me know what you get!

Answer 15

293.14?

Answer 16

Can I see your calculation?

Answer 17

P x 1.7 = 0.11988 x 8.314 x 500
P x 1.7 = 498.34
P = 293.14

Answer 18

Perfect! Just make sure you indicate that the units for pressure are kPa in this case (if you need atm, just use 0.08206 instead of 8.314).

Took a bit of time but I'm glad we worked it through. Do you understand the process?

Answer 19

Yea, thanks for the help. Would you mind taking a look at this problem too?

Answer 20

The answer I got is wrong so I'm sure not where I made a mistake.

Answer 21

Oh and I just noticed your moles should be 0.12376 (you need to add 0.00388 twice)! You're expecting your moles after equilibrium to be higher than what you started with before, so it needs to be higher than 0.12!

Answer 22

Right, good catch

Answer 23

No problem. For the second question, your ∆G needs to be in J/mol, so it should be 69000 instead of 69 in the equation. The 69 you calculated was in kJ/mol.

Answer 24

If you're ever unsure, just look at the units of R and make sure everything else is consistent. If you use 8.314, remember it's units are J/mol K, so your ∆G needs to be in J/mol as well.

Answer 25

But I multiplied it by 10^3?

Answer 26

10^-3*

Answer 27

I get the same answer even if I do e^(-69000/(8.314 x 298))

Answer 28

Oh yes I just noticed you did that. Well then yes, your answer is correct - but not for K(p).

The ΔG equation used here gives you the value of K(c), not K(p). K(c) is defined by molar concentrations, while K(p) is defined by partial pressures.

The conversion to go from K(c) to K(p) is:

\[K_p=K_c(RT)^{Δn}\]
Where R = 0.08206 L atm / mol K and Δn is the change in moles (i.e. moles of products - moles of reactants).

Answer 29

I see, so how would I calculate the change in moles? Since I wasn't given an initial for the problem

Answer 30

You know what - I just thought about the first question for a bit and realized we might have to tweak the answer just a bit. Will explain after we finish your second question.

For the second question, you just need to look at the coefficients of the reaction itself. In this case, it would be 2-1.5=0.5.

Answer 31

For the first question, we actually do need to consider concentrations as opposed to just moles, because this changes our values slightly. If we divide all the moles by 1.7 L to get the concentration for each species, you'll see that only 1 pair of 1.7's reduce out, and we're still left with one in the numerator of the equation for K(c).

Really, the equation should be:

\[K_c={{{x \over 1.7} \times {x \over 1.7}} \over {0.12-x \over 1.7}}\]
A bit messier looking, I know, but now it's correct! To make things easier, I've kept x representing moles, instead of making it represent concentrations, because moles is what we're interested about.

If you solve for x, you will get 0.005 mol. That means you will have 0.115 mol of PCl5 and 0.005 mol of PCl3/Cl2, giving 0.125 mol total. Plug that into PV=nRT and you'll get P=305.7 kPa. I know it's not a big difference from the answer we had above, but this is the most correct answer! I hope that makes sense, but if you have any questions please ask!

Sorry my explanations have been a bit choppy, just haven't done these sorts of problems in a while...

Answer 32

Yea, it makes sense.

Answer 33

Were you able to get the right answer for the second question?

Answer 34

Would the correct answer be 3.97E-12?
Kp = 8.03E-13(0.08206x298)^0.5

Answer 35

Yup looks right to me!

Answer 36

Thanks very much!