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Question

A 6.00 mol sample 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)↽−−⇀B(g)+C(g)

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



Answer 1

A(s) ==> B(g) + C(g)

6.00........0.........0........initial

-x..........+x.........+x.....change

6-x.........x...........x......equilibrium

At equilibrium, we have the following:

A = 6 - 1.20 = 4.80 mol

B = 1.20

C = 1.20

Solving for Keq:

Keq = [B][C] = (1.20)(1.20) = 1.44

Doubling the volume of the container will reduce the concentration of B and C by 1/2 to 0.60 each

A(s) ==> B(g) + C(g)

4.80........0.6.....0.6.......initial

-x.............+x......+x.......change

4.8-x.....0.6+x...0.6+x...equilibrium

K = [B][C]

1.44 = (0.6-x)(0.6-x)

Solve for x and then subtract that value from 4.8 to get the moles of A remaining.