The Kp of the reaction is 1.5*10^-3
The mole fraction of ammonia at equilibrium is 0.4149. Then the paartial pressure of NH3 is
P NH3 = XNH3 * Ptotal
= 0.4149 * 110atm
= 45.6 atm
The partial pressure of NH3 at equilibrium (45.6 atm) to 2x, and solve for x:
2x = 45.6 atm
x = 22.8 atm
Total pressure is 110 atm.
Solve for Ptotal using the value of x
Ptotal = 110 atm + 2x
= 110 atm + 2(22.8)
= 155.6atm
Then the partial pressures of N2 and H2 are
PN2 = 0.25Ptotal' - x
= 0.25 * 155.6atm - 22.8
= 16.1 atm
PH2 = 0.75Ptotal' - 3x
= 0.75 * 155.6atm - 3(22.8atm)
= 48.3atm
The equation for Kp is
Kp = P^2 NH3 / P N2 P^3H2
Kp = (45.6)^2 / (16.1)(48.3)^3
Kp = 1.15 *10^-3
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