predict shift in equilibrium position and effect on amount of chlorine gas when volume is doubled at constant temperature.
PCl3 (g) + Cl2(g) -> PCl5 (g)
2 answers
OK. You need to get Le Chatlier's Principle CLEARLY in mind. In plain, but not very fancy, words, it says that a system in equilibrium will move to the left or to the right so as to undo what we do to it. If volume is doubled that means the pressure is halved (meaning it is less that it was). You have 2 mols gas on the left and 1 mol on the right. An INCREASE in P means the reaction will move to the side with fewer mols; the opposite if you have a DECREASE in P. So moving to the left means (PCl5) gets smaller and (Cl2) and (PCl3) become larger.
Here is a much longer way to do it but it has numbers and you may get the idea better.
.......PCl3 + Cl2 ==> PCl5
Let me just make up some numbers. Let's say PCl5 was 20, PCl3 was 10 and Cl2 was 10 when the system was at equilibrium. Then what would Keq be? That is
K = (PCl5)/(PCl3)(Cl2). Substituting the values I made up we get a K value of 0.2. That is from (20/(10)(10) = 0.2.
Now if we double the volume that means the concentrations are halved which makes the K expression = (10)/(5)(5) and that is 0.4. But remember that K is a constant and can't change. That's why the reaction must shift so as to keep the K from changing (in this case keep it at 0.2. So we have a value of 0.4 which is too high. That means the numerator must get smaller and/or the denominator must get larger(or both) in order to make that 0.4 go back to 0.2. So PCl5 must get smaller; Cl2 and PCl3 must get larger and that means a shift to the left.
.......PCl3 + Cl2 ==> PCl5
Let me just make up some numbers. Let's say PCl5 was 20, PCl3 was 10 and Cl2 was 10 when the system was at equilibrium. Then what would Keq be? That is
K = (PCl5)/(PCl3)(Cl2). Substituting the values I made up we get a K value of 0.2. That is from (20/(10)(10) = 0.2.
Now if we double the volume that means the concentrations are halved which makes the K expression = (10)/(5)(5) and that is 0.4. But remember that K is a constant and can't change. That's why the reaction must shift so as to keep the K from changing (in this case keep it at 0.2. So we have a value of 0.4 which is too high. That means the numerator must get smaller and/or the denominator must get larger(or both) in order to make that 0.4 go back to 0.2. So PCl5 must get smaller; Cl2 and PCl3 must get larger and that means a shift to the left.