The Extent of a Reaction•The tendency for a reaction to occur is indicated by the magnitude of the equilibrium constant. •A value of K larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.

•A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.

jA + kB ⇌ lC + mDkj

ml

K]B[]A[

]D[]C[

Reaction Quotient•To determine in which direction a system will shift to reach equilibrium, we use the reaction quotient, Q.•Q is obtained by using the law of mass reaction using initial concentrations instead of equilibrium concentrations.•For example,

N2(g) + 3H2(g) ⇌ 2NH3(g)

Expression for the reaction quotient is

where the subscript zeros indicate initial concentrations.

30202

203

]H[]N[

]NH[Q

To determine in which direction a system will shift to reach equilibrium, compare the values of Q and K.1.Q is equal to K. The system is at equilibrium; no shift will occur.

2.Q is greater than K. To reach equilibrium, a net change of products to reactants must occur. The system shifts to the left, consuming products and forming reactants.

3.Q is less than K. The system must shift to the right, consuming reactants and forming products, to attain equilibrium.

•A typical equilibrium problem involves finding the equilibrium concentrations (or pressures) of reactants and products, given the value of the equilibrium constant and the initial concentrations (or pressures).•Since such problems sometimes become complicated mathematically, we will develop useful strategies for solving them.

•Equilibrium calculations for gaseous reactions can be performed using either Kp or K, but for reactions in solution we must use K.•Whether we deal with concentrations or partial pressures, however the same basic principles apply.•Overall, we can divide equilibrium calculations into two main categories.1.Calculating equilibrium constants from known equilibrium concentrations or partial pressures (we did earlier).2.Calculating one or more equilibrium concentrations or partial pressures using the known value of K or Kp.

1. The only values we can substitute into the equilibrium expression are equilibrium concentrations – the values that appear in the last row of the table.

2. When we enter initial concentrations into the table, they should be in units of moles per liter (mol/L). The initial concentrations are those present in the reaction mixture when it’s prepared; we imagine that no reaction occurs until everything is mixed.

3. The changes in concentration always occur in the same ratio as the coefficients in the balanced equation.

4. In constructing the “change” row, be sure the reactant concentrations all change in the same direction, and that the product concentrations all change in the opposite direction. If the concentrations of the reactants decrease, all the entries for the reactants in the “change” row should have a minus sign, and all the entries for the products should be positive.

•For reactions that have a small K values, the mathematical difficulties can be greatly reduced by making assumptions.•When K is small, the system will not proceed far to the right to reach equilibrium. •For example, gaseous NOCl decomposes to form the gases NO and Cl2. At 35 oC the equilibrium constant is 1.6 x 10-5. In an experiment in which 1.0 mol NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations?

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

The initial concentrations are

Since there are no products initially, the system will move to the right to reach equilibrium. We will define x as the change in concentration of Cl2 needed to reach equilibrium.

22

25

]NOCl[

]Cl[]NO[10x6.1 K

0]Cl[0]NO[M50.0L0.2

mol0.1]NOCl[ 0200

2NOCl(g) ⇌ 2NO(g) + Cl2(g)Initial 0.50 0 0Change -2x +2x +xEquilibrium 0.50 – 2x 2x x

Since K is so small (1.6 x 10-5), the system will not proceed far to the right. That is, x, represents a small number.Therefore, when x is small,

0.50 – 2x ≈ 0.50

2

2

22

25

)250.0(

)()2(

]NOCl[

]Cl[]NO[10x6.1

x

xxK

We can simplify the equilibrium expression to:

Solving for x3 gives:

and x = 1.0 x 10-2.

How valid is this assumption? If x = 1.0 x 10-2, then[NOCl] = 0.50 – 2x = 0.50 – 2(1.0 x 10-2) = 0.48

The change in initial concentration and equilibrium concentration is 0.02.

2

3

2

2

2

25

)50.0(

4

)50.0(

)()2(

)250.0(

)()2(10x6.1

xxx

x

xx

625

3 10x0.14

)50.0)(10x6.1(

x

To check the validity of our assumption, divide the change in initial concentration and equilibrium concentration by the initial concentration and multiply by 100.

This is called the 5% rule. If the value is less than 5%, it’s fine. If it’s greater than 5%, you cannot use the assumption.

Calculate the equilibrium concentrations:[NOCl] = 0.50 – 2x ≈ 0.50 M[NO] = 2x = 2(1.0 x 10-2) = 2.0 x 10-2 M[Cl2] = x = 1.0 x 10-2 M

Check values by plugging into equilibrium expression.

%4100x50.0

)10x0.1(2 2