2002 HSC Chem question 28.b

[juliet_garland]

Could someone please help me with this question...

"One of the reactions used to form sulphuric acid is the reaction of oxygen with sulfur dioxide under equilibrium conditions to form sulfur trioxide. Before the reaction, the concentration of the sulfur dioxide was 0.06mol/L and the concentration of oxygen was 0.05mol/L. After equilibrium was reached, the concentration of sulfur trioxide was 0.04mol/L. Calculate the equilibrium constant, K, for the reaction. Show relevant working."

[RuiAce]

Always commence by writing out the relevant chemical equation.

O_2(g) + 2 SO_2(g) ⇌ 2 SO_3(g)

This is a scenario where the volume of the reaction vessel is not explicitly stated. Thus, the formula n=CV is made slightly redundant. Normally, we always work with moles when doing such questions, however we will compromise here. As it so stands, the volume of the vessel will not matter. So we will analyse the question using concentrations.

If you feel pedantic about assuming that V does not matter, try doing the question again with V=1 and V=10.

When doing questions that relate to equilibrium constants, always analyse the scenarios step by step, and in three cases:
Initial concentrations:
[SO₃] = 0 mol L^-1
[SO₂] = 0.06 mol L^-1
[O₂] = 0.05 mol L^-1

Final concentrations:
[SO₃] = 0.04 mol L^-1
[SO₂] = ?
[O₂] = ?

What was involved the reaction:
[SO₃] = ?
[SO₂] = ?
[O₂] = ?

It should be clear that the first thing we can determine is how many moles of sulfur trioxide produced. The concentration of SO₃ produced will just be the final less the initial.

[SO₃] reacted = 0.04 - 0 = 0.04 mol L^-1

Now, recall our mole ratio. 1 mol of SO₃ is produced from 1 mol of SO₂ and 0.5 mol of O₂. Therefore
[SO₂] reacted = 0.04 mol L^-1
[O₂] reacted = 0.02 mol L^-1

So we can thus determine the final concentrations of everything else now
[SO₂] left = 0.06 - 0.04 = 0.02 mol L^-1
[O₂] left = 0.05 - 0.02 = 0.03 mol L^-1
_________________________________________________________
Recollect everything. We ONLY care about the final concentrations.
[SO₂] left = 0.06 - 0.04 = 0.02 mol L^-1
[O₂] left = 0.05 - 0.02 = 0.03 mol L^-1
[SO₃] reacted = 0.04 - 0 = 0.04 mol L^-1

Now we use our formula for K. Recall that we have the products in the numerator and the reactants in the denominator. Also, the moles become the powers.

For such a question, your rounding would only be to 1 s. f.

[juliet_garland]

Oh ok awesome thanks heaps! I was just unsure at first on how to get the final concentrations, but it makes sense now, thanks again