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For: N2(g) + 3H2(g) ⬄ 2NH3(g), Kp = 4.72 x 10–4 at 475 oC. At equilibrium...

For: N2(g) + 3H2(g) ⬄ 2NH3(g), Kp = 4.72 x 10–4 at 475 oC. At equilibrium at 475 oC, PH2 = 0.237 atm and PN2 = 0.582 atm. What is the partial pressure of ammonia at equilibrium?

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Answer #1

Partial pressure of ammonia = 1.91*10-3 atm

N₂(g) + 3H₂(g) ² 2NH₂ (9) PN = 0.582 atm Pa = 0.237 am Kp = 4.728104 Given At equilibrium к: LPo»,1* [PM] [PH] = 4.72x10 = [

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