Question

4.60 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 1.10 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 A remain?

Answer 1

Given initial moles of A = 4.60 mol

Volume of the container , V = 1.00 L

So concentration of A in the initial stage is = number of moles / volume in L

                                                                   = 4.60 mol / 1.00 L

                                                                   = 4.60 M

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

initial conc       4.60                0            0

change             -a                 +a           +a

Equb conc   4.60-a                a            a

Given Equb concentration of B is = 1.10 M

    So a = 1.10 M

So Equb concentration of A is = 4.60 - a = 4.60 - 1.10 = 3.50 M

Equilibrium concentration of C = a = 1.10 M

Equilibrium constant , K = [B][C] / [A]

                                      = (1.10 x 1.10 ) / 3.50

                                     = 0.346

When the system is double in volume that means concentrations are halved.     

Since Molarity = number of moles / volume

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

initial conc       3.50             0.55         0.55

change             -a                 +a           +a

Equb conc    3.50-a          0.55+a      0.55+a

Equilibrium constant , K = 0.36 = [B][C] / [A]

                                          0.36 = [(0.55+a)x (0.55+a)] / (3.50 - a )

a² +1.46 a - 0.957 = 0

Solving for a we get a = 0.49 M

So Equilibrium concentration of A is [A] = 3.50 - a = 3.50 - 0.49 = 3.01 M

So the number of moles of A remain is 3.01 mol