One mole of N2 and three moles of H2 are placed in a flask at 375 C. Calculate the total pressure of the system at equilibrium if the mole fraction of NH3 is 0.21. The Kp for the reaction is 4.31*10^4.

Question

I assume this to be the reaction
N2 + 3H2 --> 2NH3

Help ?

Forgot to mention: Final answer is 50 atm

Answer 1

...............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.