Thank you to anyone who answers correctly ;w; (and if you tried) What is the Kp for the decomposition of ammonia (if the reactio

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Thank you to anyone who answers correctly ;w; (and if you tried)

What is the Kp for the decomposition of ammonia (if the reaction is at 500°C)?

2 NH3(g) N2(g) + 3 H2(g)​

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Bơ 4 years 2021-08-31T04:46:27+00:00 1 Answers 13 views 0

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  1. Nick
    0
    2021-08-31T04:47:52+00:00
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    Answer:

    Kp = 4.5 x 10⁻⁹

    Explanation:

    from equilibrium of gas phase rxns, Kp = Kc(R·T)^-Δn

    2NH₃(g) ⇄ 3H₂(g) + N₂(g)

     2Vm          3Vm      1Vm

    Δn = ∑Vm(products) – ∑Vm(reactants) = (3Vm + 1Vm) – 2Vm = 2Vm = +2

    Kc = Kb = [H₂]³[N₂]/[NH₃]² = 1.8 x 10⁻⁵M²

    R = 0.08206 L·atm/mol·K

    T = 500°C = (500 + 273)K = 773K

    Kp = Kb(R·T)^-Δn = 1.8 x 10⁻⁵M²[(0.08206L·atm/mol·K)(773K)]⁻⁽⁺²⁾

    = (1.8 x 10⁻⁵mol²/L²)(63.43atm/mol)⁻²

    = 4.5 x 10⁻⁹atm

    Hope this helps. 🙂

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