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The following reaction has an equilibrium constant of 0.020 at a given temperature.

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

If you have 1.00 mol HI(g) in a 0.750-L container initially, how many moles of HI(g) will be present when the system reaches equilibrium?

  • Chemistry -

    Technically you don't have an equation because you omitted the arrow. You MUST include an arrow for us to know where the reactants stop and the products start.
    2HI ==> H2 + I2

    K = 0.020 = (H2)(I2)/(HI)^2
    Set up an ICE CHART and solve.
    H2 = O
    I2 = 0
    HI = 1.00 mole/0.750L = 1.33

    H2 = +x
    I2 = +x
    HI = -2x

    H2 = +x
    I2 = +x
    HI = 1.33-2x

    Substitute into the K expression and solve for x.

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