For example, let’s consider the decomposition reaction of POCl₃ to POCl and Cl₂ gases.

            POCl₃(g) ⇌ POCl(g) + Cl₂(g)     K_c = 0.650

If the following amounts of reactants and products were mixed, what will the equilibrium concentration of all components be?

[POCl₃] = 0.650 M, [POCl] = 0.450 M, and [Cl₂] = 0.250 M

Solution

These are the initial concentrations of the components and, most likely, they are different than what we’ll have at equilibrium. They would be equal to equilibrium concentrations if Q = K and this is what we need to find out.

So, the first step is to determine the reaction quotient and thus the direction of the reaction.

\[Q\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[Q\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.450 }} \times \;{\rm{0}}{\rm{.250}}}}{{0.650}}\; = 0.173\]

Because Q < K, the reaction will proceed to the right. Remember, the tendency is to reach equilibrium, and therefore, Q “wants to” increase and become equal to the K. Q will increase if more products are formed since their concentrations appear in the numerator.

This means, the concentrations of POCl and Cl₂ are going to increase, while the concentration of POCl₃ will decrease.

Once we know the direction of the reaction, we are going to set up what is called an ICE table. ICE is an abbreviation for Initial, Change, Equilibrium concentrations of the reactants and products.

This is how we label the rows and columns in the table:

POCl3(g) ⇌ POCl(g) + Cl₂(g)

We know the initial concentrations ([POCl₃] = 0.650 M, [POCl] = 0.450 M, and [Cl₂] = 0.250 M) as they are given in the problem. The “change” is the amounts of the components that react before the equilibrium is established.

Since we don’t know this, we assign x mol/L of POCl₃ as the reacted amount. Next, we determine how much of each product will be formed if x mol/L of POCl₃ decomposes (reacts). This is done based on the stoichiometric ratio of the components, and because it is 1 : 1 mole ratio between all the components, there is going to be x mol/L of POCl(g) and x mol/L Cl₂(g) formed.

We can put x on top/bottom of each component in the chemical equation:

x               x              x
POCl₃(g) ⇌ POCl(g) + Cl₂(g)

Now, instead of simply using x, we also add a sign to it. In this case, for the reactant(s), it is going to “-x” because its concertation is decreasing, and for the products, it is “+x” since their concertation is increasing:

-x               +x            +x
POCl₃(g) ⇌ POCl(g) + Cl₂(g)

At his point, we know the initial concentrations (I in the table), and the change in concentrations (c in the table) which is how much the concentration of the reactant has decreased, and the ones of the products have increased.

So, let’s write these in the table:

POCl3(g) ⇌ POCl(g) + Cl₂(g)

The last part to fill up the table is determining the equilibrium concentrations. And, since we have already set the sign for x, all you need to do here is add the values in the “initial” and “change” cells:

POCl3(g) ⇌ POCl(g) + Cl₂(g)

Now that we have the expressions for equilibrium concentrations, we can plug them in the equation for the equilibrium constant and find the concentrations by solving for the x:

\[K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[K\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.450 + x}}} \right)\left( {{\rm{0}}{\rm{.250 + x}}} \right)}}{{0.650 – x}}\; = 0.650\]

We have a quadratic equation which is what you are going to work with for most problems on equilibrium concentrations. First, we simplify it to be able to use the formula for quadratic equations:

0.1125 + 0.450x + 0.250x + x² = 0.4225 – 0.650x

0.1125 + 0.700x + x² = 0.4225 – 0.650x

x² + 1.35x – 0.310 = 0

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = 1, b = 1.35, c = -0.310

Therefore,

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – 1.35\, \pm \,\sqrt {1.8225\; – \,4\; \cdot \;1\; \cdot \;( – 0.310)} }}{{2 \cdot \;1}}\]

x = 0.200 or x= -1.55

x=-1.55 doesn’t work because that would indicate increasing the concentration of POCl_3, and therefore,

x=0.200

Now, this is how much POCl₃ has reacted, and to find the equilibrium concatenations, we go based on the expressions in the ICE table:

[POCl₃] at equilibrium is 0.650 – 0.200 = 0.450 mol/L

[POCl] at equilibrium is 0.450 + 0.200 = 0.650 mol/L

[Cl₂] at equilibrium is 0.250 + 0.200 = 0.450 mol/L

It is always a good idea to plug this numbers in the expression for K_c and check if they are correct:

\[K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\; = \;\frac{{(0.650)(0.450)}}{{(0.450)}}\; = \;0.650\;\;\]