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Math Help - chemistry work

  1. #1
    Senior Member euclid2's Avatar
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    chemistry work

    1. In a closed container, nitrogen and hydrogen react to produce ammonia. the equilibrium constant is 0.050. at a specific time in the process [NH_2(g)] is 2.0 * 10^{-4} mol/L, [H_2(g)] is 4.0 * 10^{-3} mol/L and [NH_3(g)] is 2.2 * 10^{-4} mol/L

    N_2(g) + 3H_2(g) forward reverse 2NH_3(g)

    In which direction must the reaction produce to establish equilibrium?

    I thought;

    0.050= \frac{[NH_3)]^2}{[N_2(g)][H_2(g)]^3}<br />

    0.050 =\frac{[2.2*10^{-4}]^2}{[2.0*10^-4][4.0*10^-3]^3}

    doesen't seem to be getting me anywhere tho

    2. Consider the system

    CO_2(g) + H_2(g) forward reverse CO(g) + H_2O(g)

    Initially, 0.25 mol of water vapor and 0.20 mol of carbon monoxide are placed in a 1.00 L reaction vessel. At equilibrium, spectroscopic evidence shows that 0.10 mol of carbon dioxide is present. calculate K for the reaction.
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  2. #2
    Senior Member
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    This equation will be true at equilibrium. To make this equation true, will you need to increase the numbers in the numerator or denominator? Which way must the reaction proceed to accomplish this?

    2. Consider the system

    + forward reverse +

    Initially, 0.25 mol of water vapor and 0.20 mol of carbon monoxide are placed in a 1.00 L reaction vessel. At equilibrium, spectroscopic evidence shows that 0.10 mol of carbon dioxide is present. calculate K for the reaction.
    Since one CO and one H_2O form one CO_2 and one H_2, 0.10mol of CO and 0.10mol of H_2O form 0.10mol of CO_2 and 0.10mol of H_2
    so
    amount of H_2 = 0.10
    amount of <br />
CO = 0.20-0.10
    H_2O = 0.25-0.10

    K = \frac{[CO][H_2O]}{[CO_2][H_2]}
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