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.

Answer 1

3.40 mol

Step-by-step 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