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13 March, 09:42

The equilibrium constant Kc for the equation2H2 (g) + CO (g) ⇌ CH3OH (g) is 35 at a certain temperature. If there are 3.21 * 10-2 moles of H2 and 4.87 * 10-3 moles of CH3OH at equilibrium in a 3.63-L flask, what is the concentration of CO?

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  1. 13 March, 12:04
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    Answer: the concentration of [CO] = 0.0532M

    Explanation:

    From The equation of reaction

    2H2 (g) + CO (g) ⇌ CH3OH (g)

    Applying Kc = [CH3OH]/[H2]^2[[CO]

    [CH3OH] = 0.00487

    [CO] = x-0.00487

    [H2] = (0.032-0.00487) ^2=0.0271

    Substitute into formula

    Kc=[CH3OH]/[H2]^2[[CO]

    35 = 0.00487 / (0.0271) ^2 (x-0.00487)

    Simplify

    x-0.00487=0.189

    x = 0.00487+0.189=0.193moles

    [CO] = n/C = 0.193/3.63 = 0.0532M
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