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27 July, 01:37

At a certain temperature, 0.4011 mol of N 2 and 1.501 mol of H 2 are placed in a 3.00 L container. N 2 (g) + 3 H 2 (g) - ⇀ ↽ - 2 NH 3 (g) At equilibrium, 0.1801 mol of N 2 is present. Calculate the equilibrium constant, K c. K c =

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  1. 27 July, 01:55
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    Kc = 16.6

    Explanation:

    Step 1: Data given

    Number of moles N2 = 0.4011 moles

    Number of moles H2 = 1.501 moles

    Volume = 3.00 L

    At equilibrium, 0.1801 mol of N 2 is present

    Step 2: The balanced equation

    N2 (g) + 3H2 (g) ⇆ 2NH3 (g)

    Step 3: Calculate amount of N2 reacted

    moles N2: 0.4011 mol - 0.1801 mol = 0.221 mol

    For 1 mol N2 therewill react 3 moles of H2 to produce 2moles of NH3

    Moles H2 reacted = 3 * 0.221 mol = 0.663 mol es

    Moles NH3 produced = 2 * 0.221 mol = 0.442 mol es

    Step 4: Amount of moles at equilibrium

    Moles N2 = 0.1801 mol

    [N2] = 0.1801 mol / 3.0 L = 0.06003 M

    Moles H2 = 1.501 moles - 0.663 mol = 0.838 moles

    [H2] = 0.838 moles / 3.0 L = 0.2793 M

    Moles NH3 = 0.442 mol

    [NH3] = 0.442 mol / 3.0 L = 0.1473 M

    Step 5: Calculate Kc

    Finally, we can calculate Kc:

    Kc = [NH3]² / ([N2] [H2]³)

    Kc = (0.1473²) / (0.06003 * 0.2793³)

    Kc = 16.6
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