Chemistry

Consider the reaction A+B->C+3D. A solution was prepared by mixing 50 ml of .001 M of A, 100 ml of .002 M of B, 10 ml of 1 M of C, and 75 ml of .0015 M of D. At equilibrium, the concentration of D was measured and found to be .0006 M. Calculate the equilibrium concentrations of A, B, C, and D. Calculate the equilibrium constant for the reaction.

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asked by Anonymous

  1. total volume = 100 + 50 + 10 + 75 = 235 mL.

    In the final solution at mixing and before equilibrium
    (A) = 0.001M x (50/235) = ?M
    (B) = 0.002M x (100/235) = ?M
    (C) = 1M x (10/235) = ?M
    (D) = 0.0015M x (75/235) = ?M

    So you know at equilibrium (D) = 0.0006M.
    Therefore, the reaction formed 1/3 * 0.0006M for C. Add that to (C) above.
    (A) at equilibrium = initial concn - 1/3 * 0.0006 and (B) at equilibrium = initial concn B - 1/3 * 0.0006.
    Plug those values into Keq expression and solve for Keq.

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    posted by DrBob222

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