[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]

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.

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

P_PCl3 = 0.596*1bar = 0.596bar
P_Cl2 =  0.40*1bar = 0.40bar
P_PCl5 = 0.004*1bar = 0.004bar

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

Hence, K_p = P_PCl5 / (P_Cl2 * P_PCl3) = 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?

[Donaldson Tan]

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

It doesn't matter which way you put it, because the reaction is reversible. Just make sure you use the correct Kp value that corresponds to the expression used.

[Donaldson Tan]

Let V_R be volume of the mixture under constant pressure of 1.0bar

Assuming perfect gas,
K_p = P_PCl5/(P_PCl3*P_Cl2) = (V_R/RT) * n_PCl5/(n_PCl3*n_Cl2)

V_R = RTn_total/P_total where n_total = n_PCl5 + n_PCl3 + n_Cl2

K_p = (V_R/RT) * n_PCl5/(n_PCl3*n_Cl2) = (n_total/P_total) * n_PCl5/(n_PCl3*n_Cl2)


Since the the amount of PCl₃ added not only shares the same pressure of the original mixture, but also temperature and volume, then (assuming perfect gas) the number of moles of PCl₃ added must be N too.

Now we can express the molar quantity of each component in terms of N when PCl₃ was just added.

molar quantity of PCl₃ = 0.596N + N = 1.596N
molar quantity of Cl₂ = 0.40N
molar quantity of PCl₅ = 0.004N
total moles of mixture = 2.0N


let X be number of moles of PCl₃ converted to PCl₅

Hence, n_total = n_{total, initial} + X - X - X = 2.0N - X

Let X = x.N
n_{total, initial} - X = 2.0N - X = (2.0 - x)N

Hence, we must solve K_p = (n_total/P_total) * n_PCl5/(n_PCl3*n_Cl2) where:
n_total = (2.0 - x)N
n_PCl5 = (0.004 + x)N
n_PCl3 = (1.596 - x)N
n_Cl2 = (0.40 - x)N
P_total = 1 bar

This yields x to be 0.0013369, thus the amount of PCl₃ converted is is 0.0013369N

The molar quantities at equilibrium are:
n_total = (2.0 - x)N = 1.9987N
n_PCl5 = (0.004 + x)N = 0.0053N
n_PCl3 = (1.596 - x)N = 1.5947N
n_Cl2 = (0.40 - x)N = 0.3987N

Hence, the partial pressures at equilibrium are:
P_PCl5 = n_PCl5*P_total/n_total = 0.00267 bar
P_PCl3 = n_PCl3*P_total/n_total = 0.79786 bar
P_Cl2 = n_Cl2*P_total/n_total = 0.19946 bar

[edwinksl]

Hmm I have slightly different answers, probably coz of premature rounding-off I guess...

Anyway, how did you solve for x? My way of doing is super long...

[edwinksl]

Yay finally same answer!