A sample of air consisting of N2 and O2 was heated to 2500 K until the equilibrium N2 (g) + O2 (g) ⇌ 2NO (g) was established with an equilibrium constant Kc=2.1×10−3. At equilibrium, the mole % of NO was 1.8. Estimate the initial composition of air in mole fractions of N2 and O2.
A sample of air consisting of N2 and O2 was heated to 2500 K until the equilibrium N2 (g) + O2 (g) ⇌ 2NO (g) was established with an equilibrium constant Kc=2.1×10−3. At equilibrium, the mole % of NO was 1.8. Estimate the initial composition of air in mole fractions of N2 and O2.
The correct option is A
Right off the bat, the option d looks very temptingly correct doesn't it?Let us see if that is the case indeed.
Look at how the problem is worded. The initial mixture consists only of N2 and O2 gases. So initial mol % of NO (g) = 0
At equilibrium, mol % of NO (g) = 1.8
Can we say, the equilibrium mol % of the mixture of ( N2 (g) + O2 (g) ) = 100 - 1.8 = 98.2 ?Certainly. Do not confuse "mol percentage” with "mol fraction”. In the current context, mol fraction is slightly irrelevant.
Here mol fraction of NO at equilibrium = 1.8100
Likewise mol fraction of the mixture of N2 + O2 = 98.2100
N2 (g) + O2 (g) ⇌ 2NO (g)
We know the mol % of the mixture (N2 + O2 ) = 98.2
Without any loss in generality, we can start with the assumption that there are a total of 100 moles of the gas comprising of N2, O2 and NO. Of these, let us suppose the number of moles of N2 = x
Automatically, the number of moles of O2 = 98.2 - x
Kc=[NO]2[N2][O2]=2.1×10−3
Do note that △n = 0; So the number of moles of the system will always remain constant. This can be verified mathematically.
We can take the volume of the container to be V litre. Hence 1.82x(98.2−x)=2.1×10−3
Solving the quadratic equation, we have x = 78.56 mol
The question doesn't end here. Amount of O2 = 98.2 - 78.56 = 19.64 mol
But read the question carefully again. We want the initial amounts of N2 and O2 in mole fractions! It is just a cleverly worded version of "initial amounts of N2 and O2” as there is no NO.
Amount of N2 = 78.56 mol + 0.9 mol = 79.46 mol
Amount of O2 = 19.64 mol + 0.9 mol = 20.54 mol.
Right off the bat, the option d looks very temptingly correct doesn't it?Let us see if that is the case indeed.
Look at how the problem is worded. The initial mixture consists only of N2 and O2 gases. So initial mol % of NO (g) = 0
At equilibrium, mol % of NO (g) = 1.8
Can we say, the equilibrium mol % of the mixture of ( N2 (g) + O2 (g) ) = 100 - 1.8 = 98.2 ?Certainly. Do not confuse "mol percentage” with "mol fraction”. In the current context, mol fraction is slightly irrelevant.
Here mol fraction of NO at equilibrium = 1.8100
Likewise mol fraction of the mixture of N2 + O2 = 98.2100
N2 (g) + O2 (g) ⇌ 2NO (g)
We know the mol % of the mixture (N2 + O2 ) = 98.2
Without any loss in generality, we can start with the assumption that there are a total of 100 moles of the gas comprising of N2, O2 and NO. Of these, let us suppose the number of moles of N2 = x
Automatically, the number of moles of O2 = 98.2 - x
Kc=[NO]2[N2][O2]=2.1×10−3
Do note that △n = 0; So the number of moles of the system will always remain constant. This can be verified mathematically.
We can take the volume of the container to be V litre. Hence 1.82x(98.2−x)=2.1×10−3
Solving the quadratic equation, we have x = 78.56 mol
The question doesn't end here. Amount of O2 = 98.2 - 78.56 = 19.64 mol
But read the question carefully again. We want the initial amounts of N2 and O2 in mole fractions! It is just a cleverly worded version of "initial amounts of N2 and O2” as there is no NO.
Amount of N2 = 78.56 mol + 0.9 mol = 79.46 mol
Amount of O2 = 19.64 mol + 0.9 mol = 20.54 mol.
