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.

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.

    posted by DrBob222

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