# chemistry help

#1
The question:
When a mixture of 0.345 mol of PCl3 and 0.268 mol of Cl2 was heated in a vessel of fixed volume to a constant temperature, the following reaction reached equilibrium.

PCl3(g) + Cl2(g) ⇌ PCl5(g) H = –93 kJ mol–1

At equilibrium, 0.166 mol of PCl5 had been formed and the total pressure was 225 kPa.

State the effect on the mole fraction of PCl3 in the equilibrium mixture if
(i) the volume of the vessel were to be increased at a constant temperature.

I know the answer is increased . But are we not talking about mole fraction, if equilibrium is shifting to the left which would increase the moles of PCl3, then this would increase moles of Cl2 and decrease moles of PCl5. The mole fraction =moles of PCl3/ total moles. So would the mole fraction not then stay the same???
0
3 weeks ago
#2
(Original post by bluestarr35910)
The question:
When a mixture of 0.345 mol of PCl3 and 0.268 mol of Cl2 was heated in a vessel of fixed volume to a constant temperature, the following reaction reached equilibrium.

PCl3(g) + Cl2(g) ⇌ PCl5(g) H = –93 kJ mol–1

At equilibrium, 0.166 mol of PCl5 had been formed and the total pressure was 225 kPa.

State the effect on the mole fraction of PCl3 in the equilibrium mixture if
(i) the volume of the vessel were to be increased at a constant temperature.

I know the answer is increased . But are we not talking about mole fraction, if equilibrium is shifting to the left which would increase the moles of PCl3, then this would increase moles of Cl2 and decrease moles of PCl5. The mole fraction =moles of PCl3/ total moles. So would the mole fraction not then stay the same???
Imagine in there was 1 mol of PCl5(g) only. The MF would be 1 for PCl5.

Then convert that 1 mol of PCl5 into 1 mol of PCl3(g) and 1 mol of Cl2(g). The MF would be 0.5 for each of them.

OR convert 0.5 mol of the PCl5 into 0.5 mol of each of them. The MF is 0.333 for all three!

Did that help?
0
#3
(Original post by Pigster)
Imagine in there was 1 mol of PCl5(g) only. The MF would be 1 for PCl5.

Then convert that 1 mol of PCl5 into 1 mol of PCl3(g) and 1 mol of Cl2(g). The MF would be 0.5 for each of them.

OR convert 0.5 mol of the PCl5 into 0.5 mol of each of them. The MF is 0.333 for all three!

Did that help?
Sorry I am still confused...
0
3 weeks ago
#4
(Original post by bluestarr35910)
I know the answer is increased . But are we not talking about mole fraction, if equilibrium is shifting to the left which would increase the moles of PCl3, then this would increase moles of Cl2 and decrease moles of PCl5. The mole fraction =moles of PCl3/ total moles. So would the mole fraction not then stay the same???
What you've said sounds pretty reasonable and is partly correct. However, you've made one false assumption.

Let's assume that at equilbrium we have 2 moles of everything present. The mole fraction of PCl3 is 2/6 = 0.333

Now we decrease the pressure and the equilibrium moves to the left. The moles of PCl5 will have to decrease. Let's say it decreases from 2 to 1. If the moles of PCl5 has gone down by 1, the mole of PCl3 and + Cl2 must have each gone up by one. The new moles are:

PCl5 1 mole
PCl3 3 moles
Cl2 3 moles

Now, after the equilibrium has shifted we have seven moles of gas; before the shift we only had six. Your false assumption was that the total moles of gas increased so much that the mole fraction of PCl3 stayed the same. The total number of moles did increase, but not by that much.

The mole fraction of PCl3 is 3/7 = 0.43
1
#5
This makes so much sense. Honestly, thank you so so much. I really appreciate your help. Thank you!!!!
0
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