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11 May, 07:18

5.80 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant.

A (s) equilibrium B (g) + C (g)

Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?

Hint: Start by finding the value of K. Make a table that expresses the initial and final amounts of each species. We can use moles in the table so long as we convert back to concentrations before plugging into the K expression. Fill in the missing values in this table, then calculate Kc. Hint: You can use the stoichiometry of the reaction to determine the changes in A and C.

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Answers (1)
  1. 11 May, 09:41
    0
    3.40 mol

    Explanation:

    By the reaction equation, the number of moles of B and C is equal, so their concentration must be equal. To determinate the equilibrium constant, the solids are not put in the expression, because their activity is equal to 1, so:

    Kc = [B]¹x[C]¹

    Kc = 1.2 x 1.2 = 1.44

    By the Le Chatilers principle, a change in the volume will change the pressure in the system and will shift the equilibrium. But Kc only changes with the temperature, so the concentration of B and C remains 1.2 mol/L. To find the number of moles, we multiply the concentration by the new volume (2.00 L)

    nB = nC = 1.2 x 2 = 2.4 mol

    So, because the stoichiometry is 1 mol of A: 1 mol of B: 1 mol of C, it was consumed 2.4 mol of A, then remains:

    nA = 5.80 - 2.40 = 3.40 mol
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