Question

4.80 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 1.30 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

Answer:

A = 3.5 moles

Explanation:

In order to do this, we need first to write the innitial conditions. The reaction is as follow:

A(s) <-------> B(g) + C(g)

From here, we can write an expression for the equilibrium constant which is:

Kc = [B]*[C]

We don't take A in the expression, because is solid, and solid and liquids do not contribute in the equilibrium.

We know the concentration of B, which is the same for C, because the moles of C are coming from A when it's decomposed, so if B is 1.3 M we can assume C is the same.

Kc for this innitial condition is:

Kc = (1.3)*(1.3) = 1.69

Now, in the second part of reaction the volume is doubled, which means that concentrations are halved:

A: 4.8/2 = 2.4 M

B and C: 1.3/2 = 0.65 M

So conditions for the second part will be like this:

   A(s) <----> B(g) + C(g)   Kc = 1.69

I:   2.4           0.65   0.65

C:  -x              +x       +x

E: 2.4-x         0.65+x   0.65+x

Replacing in Kc we have:

1.69 = (0.65+x)²   solving for x here

√1.69 = 0.65+x

1.3 = 0.65+x

x = 1.3 - 0.65

x = 0.65 M

Therefore, the remaining moles of A would be:

A = 2.4 - 0.65

[A] = 1.75 M

moles A = 1.75 * 2 = 3.5 moles