For an 80% yield you will need 1.6 mols SO3.
.........2SO2 + O2 ==> 2SO3
I........2 mols..2.......0
C........-2x.....-x......2x
E.......2-2x.....2-x.....2x
Then an 80% yield gives SO3 = 1.6 mol; therefore, 1.6 = 2x and x = 0.8
Solve for 2-2x = 0.4 mol SO2
2-x = 1.2 mols O2
2x = 1.6 mol SO3
Total mols = 0.4+1.2+1.6 = 3.2
XSO2 = 0.4/3.2 = 0.125
XO2 = 1.2/3.2 = 0.375
XSO3 = 1.6/3.2 = 0.5
pSO2 = XSO2*Ptotal
pO2 = XO2*Ptotal
pSO3 = XSO3*Ptotal
Substitute these into Kp expression and solve for P. Post your work if you get stuck. I ran through it very quickly (and could have made a blunder) but something in the area of 330 atm I think for Ptotal.
the formation of SO3 from so2 and o2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. the equilibrium constant Kp for the reaction is 0.13 at 830 degree celcius. in one experiment, 2.00 mol so2 and 2.00 mol o2 were initially present in a flask. what must the total pressure at equilibrium be to have an 80.0 percent yield of so3?
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