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23 June, 14:03

Suppose a 250.0 mL flask is filled with 1.3mol of I2 and 1.0mol of HI. The following reaction becomes possible:

H2 (g) + I2 (g) = 2HI (g)

The equilibrium constant K for this reaction is 0.983 at the temperature of the flask.

Calculate the equilibrium molarity of HI. Round your answer to one decimal place.

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  1. 23 June, 14:49
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    The molarity of HI at the equilibrium is 2.8M

    Explanation:

    Step 1: Data given

    Volume of the flask = 250.0 mL = 0.250L

    Number of moles I2 = 1.3 mol

    Number of moles HI = 1.0 mol

    Kc = 0.983

    Step 2: The balanced equation

    H2 (g) + I2 (g) ⇆ 2HI (g)

    For 1 mole I2 consumed, we need 1 mole H2 to produce 2 moles HI

    Step 3: Calculate initial concentrations

    Initial concentration I2 = 1.3mol / 0.25L

    Initial concentration I2 = 5.2 M

    Initial concentration HI = 1.0 mol / 0.25L

    Initial concentration HI = 4.0 M

    Step 4: Calculate concentrations at equilibrium

    The concentration at equilibrium is:

    [I2] = (5.2+x) M

    [HI] = (4.0 - x) M

    [H2] = xM

    Kc = [HI]²/[H2][I2]

    0.983 = (4-x) ² / (x * (5.2+x))

    0.983 = (4-x) ² / (5.2x + x²)

    5.1116x + 0.983 x² = 16 - 8x + x²

    -0.017x² + 13.1116x - 16 = 0

    x = 1.222 = [H2]

    [HI] = 4.0 - 1.222 = 2.778M ≈ 2.8 M

    [I2] = 5.2 + 1.222 = 6.422 M ≈ 6.4 M

    To control we can calculate:

    [2.778]² / [1.222][6.422] = 0.983 = Kc

    The molarity of HI at the equilibrium is 2.8M
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