5.80 mol of solid A was placed in a sealed 1.00-L container and allowed to decom

Question

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.

Explanation / Answer

By le chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.

In this problem, the volume doubles, that means you would have to double the molarity of B/ C (since B=C.) Since A is a solid it's activity = 1 and is then ignored.

So doubling B would be 2.4M, 2.4 M moles of A was consumed so the work is 5.80 - 2.40 = 3.40 moles of A remaining.

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