Posted by Chemmy on Wednesday, March 2, 2011 at 8:44pm.
...............N2 + 3H2 ==> 2NH3
begin (mols)..1.0...3.0......0
change.........-x.....-3x.....2x
final........1.0-x..3-3x.....2x
total n = 1.0-x+3-3x+2x = 4-2x
And we know mol fraction NH3 = 0.21; therefore,
(2x)/(4-2x) = 0.21 and solve for x = moles.
Use x to determine moles NH3, moles H2 and moles N2 at equilibrium.
Then determine mole fraction N2, NH3(which you have) and H2.
PNH3 = 0.21*Ptotal
PH2 = XH2*Ptotal
PN2 = XN2*Ptotal
Then substitute partial pressures into Kp expression and solve for P. By the way, I think Kp is too large; in fact, I suspect it should be 10^-4 and not 10^4; i.e., I think you made a typo.
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