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24 June, 10:11

The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? 4.7 * 10-4 atm 2.2 * 10-7 atm 2.1 * 103 atm 4.4 * 106 atm

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  1. 24 June, 11:29
    0
    Given:

    Kp = 1.5*10³

    partial pressure pN2 = 0.10 atm

    partial pressure pH2 = 0.15 atm

    To determine:

    Partial pressure pNH3 at equilibrium

    Explanation:

    The decomposition reaction is:-

    2NH3 (g) ↔N2 (g) + 3H2 (g)

    Kp = [pH2]³[pN2]/[pNH3]²

    pNH3 = √ [ (pH2) ³ (pN2) / Kp]

    pNH3 = √ (0.15) ³ (0.10) / 1.5*10³ = 4.74*10⁻⁴ atm

    Ans (a)
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