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Chemical Equilibrium Problem with unknown Molar Quantities

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GUnit:
Hi everyone. First post :)

Could someone please help me on the following ASAP -

Physical chem question, 1st year university standard.

http://www.reading.ac.uk/Repol/ExamPapers04/CH1P2%202003-4%20A%20001.pdf

Section B, Question 3 if anyone can help.

Any hints or pointers would be greatly appreciated.

GUnit:
Anyone ???

I've got an exam tomorrow so if someone can please reply today then that would be greatly appreciated.

Donaldson Tan:
I copied the question and pasted it here for quick reference.

In future, please copy and paste the question for our convenient reference

At a temperature of 520 K and a pressure of 1.00 bar, a gas-phase equilibrium mixture of PCl3, Cl2 and PCl5 is found to contain molar percentages of 59.60% PCl3, 40.00% Cl2 and 0.40% PCl5. A volume of pure PCl3 equal to that of the original mixture (and at the same temperature) is then added to the mixture, with the total pressure maintained at 1.00 bar. Calculate the partial pressures of the three components when equilibrium has been re-established

Donaldson Tan:

--- Quote from: geodome on May 26, 2006, 03:48:31 PM --- At a temperature of 520 K and a pressure of 1.00 bar, a gas-phase equilibrium mixture of PCl3, Cl2 and PCl5 is found to contain molar percentages of 59.60% PCl3, 40.00% Cl2 and 0.40% PCl5.
--- End quote ---

First, calculate the equilibrium constant Kp. You are given the total pressure of the system and the molar percentage of each component. Hence the partial pressures at equilibrium are:

PPCl3 = 0.596*1bar = 0.596bar
PCl2 =  0.40*1bar = 0.40bar
PPCl5 = 0.004*1bar = 0.004bar

The chemical equation describing the system is:
PCl3 + Cl2 <-> PCl5

Hence, Kp = PPCl5 / (PCl2 * PPCl3) = 1.67785E-2 bar-1

Now, we have established the value of Kp, so we can proceed to find the new equilibrium composition after some PCl3 is added to the system.

GUnit:
Cheers Geodome :)

One thing though, i think you have the products and reactants the wrong way round as the reaction is PCl5 -----------> PCl3 + Cl2.

It's part (b) i need help with. I understand how to form the quadratic, as the initial partial pressures will now be P(PCl5) = 0.004 and P(Cl2) = 0.40 and when the PCl3 is added the reaction will favour the reactants so PCl5 will be +x whereas the change in partial pressure of PCl3 and Cl2 will be -x. However what will be the initial partial pressure of PCl3?

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