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25 June, 10:48

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0 - L vessel at 300 K. The following equilibrium is established:

2NO (g) + 2H2 (g) - -->/<---N2 (g) + 2H2O (g)

At equilibrium [NO]=0.062M

Calculate the equilibrium concentration of N2.

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  1. 25 June, 10:55
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    The concentration of N2 at the equilibrium will be 0.019 M

    Explanation:

    Step 1: Data given

    Number of moles of NO = 0.10 mol

    Number of moles of H2 = 0.050 mol

    Number of moles of H2O = 0.10 mol

    Volume = 1.0 L

    Temperature = 300K

    At equilibrium [NO]=0.062M

    Step 2: The balanced equation

    2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)

    Step 3: Calculate the initial concentration

    Concentration = Moles / volume

    [NO] = 0.10 mol / 1L = 0.10 M

    [H2] = 0.050 mol / 1L = 0.050 M

    [H2O] = 0.10 mol / 1L = 0.10 M

    [N2] = 0 M

    Step 4: Calculate the concentration at the equilibrium

    [NO] at the equilibrium is 0.062 M

    This means there reacted 0.038 mol (0.038M) of NO

    For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

    This means there will also react 0.038 mol of H2

    The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

    There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

    There will be produced 0.038/2 = 0.019 moles of N2

    The concentration of N2 at the equilibrium will be 0.019 M
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