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In a vessel two equilibria are established simultaneously at the same temperature: (i) N2(g) + 3H2(g) <=> 2NH3(g) and (ii) N2(g) + 2H2(g) <=> N2H4(g). Initially the vessel contains N2 and H2 in the molar ratio 9: 13. At equilibrium the total pressure is 7*P0, the partial pressure of NH3 is P0 and that of H2 is 2*P0. What are the equilibrium constants Kp(i) and Kp(ii)?

1. Kp(i) = 1/(20*P0²) and Kp(ii) = 3/(20*P0²)
2. Kp(i) = 1/(10*P0²) and Kp(ii) = 1/(20*P0²)
3. Kp(i) = 3/(20*P0²) and Kp(ii) = 1/(20*P0²)
4. Kp(i) = 1/(20*P0²) and Kp(ii) = 1/(10*P0²)

Correct answer: Kp(i) = 1/(20*P0²) and Kp(ii) = 3/(20*P0²)

Solution

Solving conservation gives P(N2) = 2.5*P0, P(H2) = 2*P0, P(NH3) = P0, P(N2H4) = 1.5*P0, leading to Kp(i) = P0²/(2.5*P0 *(2P0)³) = 1/(20*P0²) and Kp(ii) = 1.5*P0/(2.5*P0 *(2P0)²) = 3/(20*P0²).

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