My first language isn't English and I'm not studying in English so I apologize for any mistakes since I used Google translate. If you don't understand the problem, I'll try and explain. This isn't homework btw, it is an example of an exam that I have next week.
3.448 g of oxygen and 896 g of sulfur(IV) oxide were placed in a reaction vessel with a volume of 14.0 dm3 at a certain temperature, whereby sulfur(VI) oxide is formed. Equilibrium was reached when 10% of oxygen was consumed.
3. a) Use the chemical reaction equation to show the described change.
3. b) Calculate the value of the equilibrium concentration constant at the temperature of the experiment for the given reaction.
Can someone please explain how to solve this!? I don't have answers and all problems my teacher gave me on the topic of chemical equilibrium aren't similar to this one. I'm not sure how to go about it? Do I calculate moles of oxygen and sulfur(IV) oxide? Or do I calculate 10%of oxygen and then the moles of both oxygen and sulfur(IV) oxide?
Do you want Kc or Kp? I will assume Kc and you can convert that to Kp if necessary.
896 g SO2
3.448 g O2
mols SO2 = g/molar mass = 896/64 = 14
M SO2 = mols/dm^3 = 14/14 = 1 M
moles O2 = 3.448/32 = 0.1077
M O2 = 0.1077/14 = 0.00770
10% O2 is consumed = 0.00770*0.10 = 0.000770 M
(Note that [(3.448 - 3.448*0.1)/32]/14 = 0.00693 M which is the same equilibrium value for O2 below on the E line so it makes no different if you use 10% value of grams O2 and convert to M or convert to moles first and take 10% of that.
..................2SO2 + O2 ==> 2SO3
I....................1.......0.0077..........0
C.....-2*0.00077....-0.00077.......2*0.00077
E.............0.998......0.00693.....0.00154
Kc = (SO3)^2/(SO2)^2(O2)
Plug in the E line and calculate Kc.
If you want Kp = Kc(RT)^delta n
By the way your post is excellent English, at least excellent USA English. Good luck in your studies.