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chrisbphd
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#1
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The reaction

PCl5(g) --> <-- PCl3(g) +Cl2(g)

(--> <-- means a reversible reaction)

has been studied by mixing one mole of PCl3 and one mole of Cl2 at 1 bar total pressure, allowing them to react and noting the pressure attained at equilibrium; the volume of the container is fixed.

The observed equilibrium pressures, in bar, were 0.53 and 0.82 at temperatures of 400 and 500 K, respectively.

Assuming that all the species behave as ideal gases, determine the mole fraction of PCl5 present at equilibrium in each case, and consequently calculate K(p) and ΔG(standard) at the two temperatures.

Assuming that it is approximately constant over the temperature range in question, estimate ΔS(standard) for the reaction. Is the value in accordance with what you would expect ?

(K(p) is the pressure equilibrium constant, G(standard) is the standard molar Gibbs energy, S(standard) is the standard molar entropy).

Rep for the first one to complete the whole lot correctly...it's difficult, and long.
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madmazda86
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*brain explodes* Zarking hell, there was me expecting to find an A2 Chemistry query and it's a degree-level thing! :rolleyes: I'm sorry, I can't help you because the only thing I understand in that is the Kp because we're learning about that in Chem at the moment. I hope someone is able to help you
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chrisbphd
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You will meet all of these things in your A2 chemistry course, but you won't be shown the way in which they all link together to provide a solution to all of this.

Its nice relationships like ΔG(standard) = -RT ln (K(p)) that make this question work, but I keep ending up with weird answers.
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elpaw
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(Original post by chrisbphd)
You will meet all of these things in your A2 chemistry course, but you won't be shown the way in which they all link together to provide a solution to all of this.

Its nice relationships like ΔG(standard) = -RT ln (K(p)) that make this question work, but I keep ending up with weird answers.
we didnt do G or S on the edexcel A2
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rIcHrD
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Question doesn't make sense because you can't produce 0.53 moles from 2 moles in that question. To reduce the number of moles, you'd have to react PCl3 with Cl2. As there's no PCl3 to begin with, that's impossible.
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elpaw
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(Original post by rIcHrD)
Question doesn't make sense because you can't produce 0.53 moles from 2 moles in that question. To reduce the number of moles, you'd have to react PCl3 with Cl2. As there's no PCl3 to begin with, that's impossible.
one mole of PCl3 was mixed with one mole of Cl2 in the beginning....
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rIcHrD
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sorry - eyes are sleepy tonight and seems part of my brain has already started its snooze. The question becomes quite easy then
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Ben.S.
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(Original post by rIcHrD)
sorry - eyes are sleepy tonight and seems part of my brain has already started its snooze. The question becomes quite easy then
Oh, really? Care to post your fully worked solution?

Ben
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rIcHrD
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For PCl5 --- PCl3 + CL2
Initially no. of moles are:
0 1 1
at equilibria
400K, 0.53 bar pressure
x 1-x 1-x

Using PV=nRT, V from initial conditions (2 moles at 400K give 1x10^5Pa pressure), thus V=nRT/P=0.06648m^3.

As V and T are constant, P is proportional to n. Thus total moles at equilibrium=x + 1-x + 1-x = 2-x=PV/RT. From this x=0.94. There's an easier, faster and more intuitive way of doing this but this shows the empirical steps.

From that K_p is (1-x)^2 * 1/x * 0.82x10^5 after simplification. This comes to 203.0. G (standard)=-RTln(K_p)=-17,600 J per mole.

Same methodology can be used at 500K, to give x=0.36, K_p = 93,300. G(standard)=-47,500 J per mole.

Simultaneously equating G=H + TS for both Gs and Ts, cancels H and shows S(standard)=+299 J per K per mole.

This is as expected as the forward reaction increases number of moles (1 to 2) without changing state, thus resulting in increased entropy which would result in a positive S as is apparent.
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chrisbphd
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Thanks and well done. They match the required answers. Rep is on its way when it becomes available.

EDIT:

Sorry...this is me being a muppet and believing Ben. They are not the right answers. The volume of the container is fixed...you can't change it in the middle of the calculation. You can't calculate Kp in the manner that you have. To use it in Delta G standard = -RTlnKp you must divide all of the partial pressures by the standard pressure so that Kp is dimensionless.

No rep!!!!
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chrisbphd
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#11
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I've edited my reply above.
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