Question

5.60 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 1.40 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

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

Change in volume affects the equilibria that involve gases. Because change in volume changes the pressure exerted by the gases. (Pressure is taken as equivalent to concentration of gases in equilibrium)

1 mole of A gives one mole of each of A and B. That means when 1.4 moles of B are formd (At equilibrium) 1.4 moles of A will be consumed.

So tthe amount of A = 5.6moles - 1.4 moles = 4.2 moles

When the volume is doubled, the pressure will be reduced to half. Because pressure is inversly proportional to Volume. Therefore to bring pressure back to initial value more B and C must be formed. The Amount of B (or C ) that must be formed = 2 X 1.4 = 2.8 moles.

Hence amount of A at new equilibrium conditions = 5.6 - 2.8 = 2.8 moles.