posted by Tri .
An intermediate step in the production of battery acid, sulfuric acid, invoves the information of sulfur trioxide gas from sulfur dioxide and oxygen gases. (this reaction is also the usual first step in the production of acid rain).
The Kp for this reaction at 830degreeC is 0.13. In one experiment, 96.9g of sufur dioxide and 34.6g of oxygen were added to a rxn flask. What must the total pressure in the flask be at eqilibrium if the reaction has an 80% yield of sulfur trioxide. Assume the volume is 3.5L.
As I see the the first problem.
2SO2 + O2 ==> 2SO3.
Convert g SO2 to mols, convert that to mols SO3, take 80%. That is mols SO3 we should have at the end of the reaction for an 80% yield. Then calculate SO2 remaining, O2 used up (at 80%) and O2 remaining. Add mols of each and use PV = nRT to determine total P.
Check my thinking.
Taking away 80% means what you have at 100% - 80% ?
Since the molar ratios are the same for S03 and S02, are the mols going to be the same? So that it would make 0.3+0.3?
Will the total mols be: mols of O2 + 0.3+ 0.3?
Thanks for your help much appreciated!
I don't think so.
Mols SO2 = about 0.3 but mols SO3 will be what formed at 80% yield which is not 0.3. I have that as about 1.2 or so. Then O2 used in the reaction is 1/2 SO3 formed so that is about 0.6 (1.2 * 1.2) and that subtracted from the O2 initially is 1.08 - 0.6 = about 0.48 O2 remaining unreacted. You need to refine all of the numbers and don't round too much in doing so. I think the total mols is more like 0.3 something for SO2, 0.48 or so for O2, and about 1.2 or so for SO3. Check my thinking.