Chemistry Forum

The reaction of sulfur dioxide gas with oxygen gas to form sulfur trioxide gas is an important step in the production of sulfuric acid.

a) Write the expression for the equilibrium constant.

b)If a 10.0 L vessel contains 15 mol of SO₃, 2.0 mol of O₂, and 3.0 mol of SO₂ at equilibrium at 1373 K, what is the numerical value for the equilibrium constant?

c) How many moles of sulfur dioxide must be forced into reaction vessel to increase the concentration of SO₃ to 1.6 mol/L?

I am working on this question.

Any chemistry help will be accepted.

(a)\(K= \frac{[SO_3]^2}{[SO_2]^2[O_2]}\)

(b)\(\frac{(\frac{15 mol}{10 L})}{(\frac{3 mol}{10 L})^2(\frac{2 mol}{10 L})}=125 \)

(c) I am working on this question.

a) square at SO₃ missing
b) correct

c) (1,6 mol/l)^2 / (0,3+x)^2* (0,2-0,5x) =125
Solve for x

My answer to (c) is different than your answer.

My attempt to answer (c) =\(K=\frac{[SO_3]^2}{[SO_2]^2[O_2]}= \frac{ (1.6)^2}{(0.3 + 0.10x)^2(0.2-0.05)} \Rightarrow x=0.695041722823 mol SO_2/L\)
∴ (0.69 mol SO₂/L)(10 L)= 6.9 mol of SO₂

Is this answer correct? We have to add SO₂.

Who say that I deduct SO₂ .
(0,3 +x) is an Addition
Also in your equation the square is missing in (0,3 + 0,1×).

I made appropriate changes in my posts as per your suggestions.

Now in answer to (c), There is an increase of 0.1 mol on SO₃. As per chemical reaction 2SO₂ + O₂ →2 SO₃, 50% of 0.10 mol increase in SO₃ will be deducted from 0.20 mol of O₂ and x is the proportion of an increase of 0.10 mol in SO₃.

I am not sure you can use 0.1 for the SO₂. If no Addition then 0.1 SO₃ requires 0.1 SO₂ of the existing amount. But if SO₂ should be added then the 0.1 is taken of the Reservoir and the additional amount x = SO₂ exists + SO₂ New. But it could also nothing is taken and the loss is refilled so the amount doesnt change. Or its only an Addition. Between these is the trues.

I think it is similar to the XeF₄ example.
We increase SO₃ at 0,1 mol , what means the SO₂ would be deducted by 0,1 mol, but we should add SO₂ . (0,3-0,1 + x)^2

My result 0,169 mol

Correct answer, thanks.