Reply 1
12
Q1. Sulphur dioxide and oxygen were mixed in a 2:1 mol ratio and sealed in a flask with a catalyst. The following equilibrium was established at temperature T1 2SO2(g) + O2(g) 2SO3(g) ΔH = –196 kJ mol–1 The partial pressure of sulphur dioxide in the equilibrium mixture was 24 kPa and the total pressure in the flask was 104 kPa. (a) Deduce the partial pressure of oxygen and hence calculate the mole fraction of oxygen in the equilibrium mixture.
Partial pressure of oxygen ...........................................................................
Mole fraction of oxygen ................................................................................
(3) (b) Calculate the partial pressure of sulphur trioxide in the equilibrium mixture. ......................................................................................................................
Partial pressure of oxygen ...........................................................................
Mole fraction of oxygen ................................................................................
(3) (b) Calculate the partial pressure of sulphur trioxide in the equilibrium mixture. ......................................................................................................................
Reply 2
12
Where I'm stuck is that for oxygen, you only divide 24 by 2. however for sulfur trioxide all of a sudden you have to subtract from total pressure. why can't I use the same principle and say that sulfur trioxide has a partial pressure of 24 because there are two moles?
Reply 3
Original post
by mitostudentWhere I'm stuck is that for oxygen, you only divide 24 by 2. however for sulfur trioxide all of a sudden you have to subtract from total pressure. why can't I use the same principle and say that sulfur trioxide has a partial pressure of 24 because there are two moles?
The sum of all partial pressures must equal the total pressure.
Partial pressures are proportional to how much of a given gas is in the system, but I think your confusion arises from the more familiar cases where the reaction goes to completion and you use the initial amounts of reagents to calculate how much product you can expect to make.
In this case, they’ve told you that SO2 has a partial pressure of 24 kPa at equilibrium, which means you have some SO2 left - this tells you nothing about the initial amount of SO2, only that the reaction doesn’t quite go to completion.
Reply 4
The division of 24 kPa by 2 when finding p(O2), however is the correct thing to do. They’ve said the mixture of SO2 and O2 contains twice as much SO2 as there is O2 and this is the reacting ratio of the two gases. As such, the ratio of their partial pressures in this case must be 2:1.
(edited 1 year ago)
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