Equilibrium question Help!

EDIT: look at post below to see question.

So what I know is that equilibrium is first established at time 3 minutes into the reaction because I see that at this time the rate of forward and backward reactions are occurring at equal rates as well as concentrations of both reactant and product being kept the same.

Now how would I sketch the conc of SO2 on the graph, I think definitely the curve would level off after time 3 minutes (as this is where equilibrium is established) but how would I figure out the concentration that it would go up to.

I know also that when sketching the line, it should begin from the origin because when starting an equilibrium mixture you start off with 0 moles of product.

I tried using the molar ratio from the Q, where the molar ratio between SO3 and SO2 is 2:2, so would the conc go up to 0.1 mol dm-3, or instead would I look at the molar ratio between SO2 and O2 which is 2:1 so the conc would go up to 0.3 mol dm-3 since 0.15 * 2 = 0.3.

I'm not really sure what the answer could be here? Any help would be really nice!

Thanks!

Use the molar ratio of the products, so you will get 0.3 mol dm^-3
You're correct that the concentration will be reached in 3 mins. At each of time point up to 3 mins you could plot the SO₂ conc as being exactly twice the O₂ conc - this will help you draw the curve for SO₂

I see, why would it be incorrect to use the molar ratio for 2SO3 to 2SO2 i.e. a molar ratio of 1:1?

It's not incorrect to do so, just not as straightforward (in my opinion).

But if we did, does that not mean the concentration of SO2 would be 0.1 mol dm-3 and not 0.3 mol dm-3?

No. You will end up with the same answer by any valid method.
The concentration is the concentration regardless of which method we use to calculate it. Our method of doing the maths can't influence what is happening in the vessel.

Yes, but I’m just a bit confused with how the equation was set up.

Firstly, molar ratio for SO3 to SO2 is 1:1, if the equilibrium concentration for SO3 is 0.1 mol dm-3, shouldn’t the same concentration apply to SO2 I.e. 0.1 * 1 = 0.1 so SO2 has conc 0.1 mol dm-3

Secondly, molar ratio for SO2 to O2 is 2:1, and the concentration of O2 is 0.15 mol dm-3, so concentration for SO2 is 0.15 * 2 = 0.3 hence the conc for SO2 is 0.3 mol dm-3

These are giving two different answers. I’m sorry but I’m not sure if I understand what method to use that gives the concentration of SO2 to always be the same no matter what.

Sorry if I’m complicating things more than it has to be.

You need to consider that as the SO₃ concentration falls, the SO₂ concentration increases because SO₂ is formed from the decomposition of SO₃. To work out the final SO₂ concentration you have to calculate how much the SO₃ concentration has decreased by.
The SO₃ con started at 0.4 and finished at 0.1, a decrease of 0.3. Therefore the SO₂ concentration increases by 0.3 from zero to 0.3.