17.2  The equilibrium law 
17.2.1  Solve homogenous equilibrium problems using the expression for Kc. Calculate Kc given all equilibrium concentrations. Given Kc and other appropriate concentrations, find an equilibrium concentration. Kp and Ksp are not required nor is use of the quadratic expression
The equilibrium law
The value of Kc can be expressed for a general reaction:
A + 2B  3C + D 
K_{c} = ([C]^{3}[D])/([A][B]^{2})
The concentrations of the products (each to the power of their coefficient) over the concentrations of the reactants (each to the power of their coefficient).
All concentrations are taken when the system has reached equilibrium, and so given all concentrations, K_{c} can be calculated, or given K_{c} and all but one of the concentrations, the final concentration can be calculated.
The units for K_{c} can also be calculated by replacing each concentration with mols dm^{3} (remembering to take exponents into account) and cancelling out.
Solving problems
This usually involves working with teh stoichiometry of a reaction to determine either the concentrations at equilibrium, or using the concentrations at equilibrium to work out the equilibrium constant.
If we are given the equation:
N_{2}O_{4}(g) 2NO_{2}(g)
And we are told that 2 moles of N_{2}O_{4}(g) are introduced into a 1dm^{3} flask at a given temperature and left to establish equilibrium. At equilibrium it is found that only 1.2 moles of N_{2}O_{4}(g) gas remain.
From this we can determine the equilibrim constant at the given temperature by looking at the coefficients of the balanced equation and working out the molar quantities of each gas at equilibrium.
Initially  moles of N_{2}O_{4}(g) = 2, therefore concentration = 2 mol dm^{3}
Initially  moles of NO_{2}(g) = 0
@ equilibrium  moles of N_{2}O_{4}(g) = 1.2, therefore concentration = 1.2 mol dm^{3}
@ equilibrium  moles of NO_{2}(g) = 1.6 (this comes from doubling the 0.8 moles of N_{2}O_{4}(g) reacted, according to the stoichiometry of the balanced equation), therefore concentration = 1.6 mol dm^{3}
The we substitute values into the equilibrium law equation to obtain Kc.
Kc = [NO_{2}(g)]^{2}/[N_{2}O_{4}(g)] = (1.6)^{2}/1.2 = 2.13
This procedure is sometimes called ICE after the three steps.
 1. Initial moles
 2. Concentrations
 3. Equilibrium moles and concentrations
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