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24 January, 02:35

How do I do this?

Consider the chemical reaction 2NH3 (g) N2 (g) + 3H2 (g). The equilibrium is to be established in a 1.0-L container at 1000 K, where Kc = 4.0 * 10-2. Initially, 1.22 moles of NH3 (g) are present. What is the equilibrium concentration of H2 (g) ?

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  1. 24 January, 05:09
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    The chemical reaction is as follows:

    2NH3 (g) = N2 (g) + 3H2 (g)

    We do the ice method to answer this.

    NH3 N2 H2

    I 1.22 0 0

    C - 2x x 3x

    E 1.22 - 2x x 3x

    Kc = [h2]^3 [n2] / [nh3]^2

    4.0 * 10^-2 = 27x^3 (x) / (1.22 - 2x) ^2

    x = 0.259 mol

    H2 = 0.259/1 = 0.259 mol/L
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