So, you were given this equilibrium reaction
A_((aq)) rightleftharpoons B_((aq)) + C_((aq))
Now, an equilibrium reaction is comprised of two reactions that take place at the same time - a forward reaction, which is the one given to you, and a reverse reaction, which looks like this
B_((aq)) + C_((aq)) rightleftharpoons A_((aq))
This aspect is important because the forward reaction will be spontaneous if K_("forward") >1; likewise, the reverse reaction will be spontaneous if K_("reverse") = 1/K_("forward") >1.
A reaction's spontaneity is determined by the sign of its Gibbs free energy, DeltaG^@. If DeltaG^@ is negative for the forward reaction, then the forward reaction will be spontaneous. If DeltaG^@ is positive for the forward reaction, the forward reaction will not be spontaneous, but the reverse reaction will.
So, the equilibrium constant for the forward reaction is
K_("forward") = ([B] * [C])/([A]) = (0.0200M * 0.1 M)/([A]) = 0.100
The expression for DeltaG^@ at equilibrium is
DeltaG^@ = -RT * ln(K)
Notice that in order for DeltaG^@ to be negative, the natural log of K must be positive, which atutomatically means that K must be greater than 1.
Since no information about temperature is given, the equilibrium constant will remain unchanged during the reaction, which means that the reverse reaction will be spontaneous, since
K_("rverse") = 1/K_("forward") = 1/0.100 = 10.0
This means that
K_("reverse") = ([A])/([B] * [C]) = ([A])/(0.0200 * 0.1) = ([A])/(0.002) = 10.0
[A] = 0.002 * 10.0 = "0.02 mol/L", or
[A] = "20 mmol/L"
Now, don't forget about the reaction quotient, Q_c. If Q_c is greater than K_("forward"), the reaction will shift to the left, favoring the formation of more reactant. SInce the reverse reaction is spontaneous, this means that you could get any value for the concentration of A that satisfies this inequality
Q_c = (0.002)/([A]) > 0.1=> [A] < 0.02 M
