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 1.20 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 A remain?

Please break down the steps so I can see how to solve. Thank you!

Answer 1

Its a case of hetrogeneous equilibria, where the concentrations ( moles per unit volume ) of pure solid is fixed and cannot vary. Thus the concentration of pure solids or even liquids in hetrogeneous equilibria are not included in equilibrium constant expression. Or, is usually taken as unity.

Let us consider the given reaction at equilibria -   

A_(s) <-----------------> B_(g)   + C_(g)

as per stoichiometry 1.0 mole of A (solid ) decomposes to yield 1.0 mole of B (gas ) & 1.0 mole of C ( gas )

Now, since at equilibrium 1.2 moles of B is formed , hence an equal number of moles ( ie. 1.2 moles ) of C would also be produced & remain present at equilibrium. Further the products ( ie. B & C ) would be obtained by decomposition of 1.2 moles of A (solid )

The number of moles of A( solid ) at equilibrium left in the one litre container = ( 4.6 - 1.2 ) = 3.4

Since the volume of container is 1.00 litre , hence concentrations of reactants & products at equilibrium would be-

[ A_(solid) ] = 1.00 [ B_(g)  ] = 1.20 ; [ C₍₉₎ ] = 1.20 in terms of moles / litre

Apply law of Mass Action to get equilibrium contant K for the reaction at equilibrium state-

K = [ B_(g)  ] [ C_(g) ] / [ A _{( s )}  ]

   ₌   1.2 x 1.2

₌ 1.44   

[ Note that concentrations of pure solids is taken as unity ]

     

Now , when the container volume is doubled ( ie 2.0 litre ) , then let us assume that the number of moles of B (gas ) & C (gas ) present in the two litre reaction vessel is ' x ' moles each

hence their concentrations at equilibrium interms of moles /litre -

[ B_(g)  ] = x / 2 ; [ C_(g)  ] = x /2

& K = (x/2) ( X/2 ) = x²  / 4

1.44 = x²   / 4   

x = 2.40 moles

Hence, number of moles of A _(solid)  left at equilibrium = ( 4.60 - 2.40 )

= 2.20 moles

This `is also in conformation to Le- Chatelier's principle.