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Law of Chemical Equilibrium Tutorial

Key Concepts

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Writing Equilibrium Constant Expressions (Expressions for K): Homogeneous Systems

Consider the following reaction at equilibrium:

aA(g) + bB(g) ⇋ cC(g) + dD(g)

The mass-action expression, Q, for this reaction is:

Q = [C(g)]c[D(g)]d
[A(g)]a[B(g)]b

At equilibrium the rate of the forward reaction is the same as the rate of the reverse reaction so that the concentration of each species remains constant, that is:

[C(g)] = a constant
[D(g)] = a constant
[A(g)] = a constant
[B(g)] = a constant

So

Q = [a constant]c[a constant]d
[a constant]a[a constant]b
= a constant
= equilibrium constant
= K

The expression for the equilibrium constant, K, is the same as the mass-action expression, Q, that is:

K = [C(g)]c[D(g)]d
[A(g)]a[B(g)]b

Similarly, if all the species were present in aqueous solution, the chemical reaction could be represented as

aA(aq) + bB(aq) ⇋ cC(aq) + dD(aq)

The mass-action expression, Q, for this reaction is:

Q = [C(aq)]c[D(aq)]d
[A(aq)]a[B(aq)]b

At equilibrium the rate of the forward reaction is the same as the rate of the reverse reaction so that the concentration of each species remains constant, that is:

[C(aq)] = a constant
[D(aq)] = a constant
[A(aq)] = a constant
[B(aq)] = a constant

So

Q = [a constant]c[a constant]d
[a constant]a[a constant]b
= a constant
= equilibrium constant
= K

And once again, the expression for the equilibrium constant, K, is the same as the mass-action expression, Q, that is:

K = [C(aq)]c[D(aq)]d
[A(aq)]a[B(aq)]b

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Writing Equilibrium Constant Expressions (Expressions for K): Heterogeneous Systems

But what happens if one of the reactants or products is a solid?

aA(aq) + bB(aq) ⇋ cC(s) + dD(aq)

The mass-action expression, Q, for this reaction is:

Q = [C(s)]c[D(aq)]d
[A(aq)]a[B(aq)]b

The concentration of the solid, C(s), is always the same! Even if the system is NOT at equilibrium, the concentration of C(s) will still not change.
The concentration of a solid is fixed by its density, that is, the amount of solid substance in a given volume is always the same at constant temperature and pressure, so the concentration of a solid is a constant.
We could then re-write the mass-action expression as:

  Q  
[C(s)]c 		=     [D(aq)]d    
[A(aq)]a[B(aq)]b

And, at equilibrium:

Q/[C(s)]c = a constant/a constant
          = K (the equilibrium constant)

So, the expression for the equilibrium constant is:

K =     [D(aq)]d    
[A(aq)]a[B(aq)]b

Similarly, if one of the reactants or products is a liquid, the concentration of the liquid species does not change.
The concentration of a liquid is also fixed by its density, that is, the amount of liquid substance in a given volume at constant temperature and pressure, is always the same, so the concentration of a liquid is a constant.

So, for the reaction

aA(aq) + bB(l) ⇋ cC(aq) + dD(aq)

The mass-action expression, Q, for this reaction is:

Q = [C(aq)]c[D(aq)]d
[A(aq)]a[B(l)]b

but since [B(l)] is always constant because B is a liquid, we can re-write the expression as:

Q[B(l)]b = [C(aq)]c[D(aq)]d
[A(aq)]a

and at equilibrium,

Q[B(l)]b = a constant × a constant
= constant
= K (the equilibrium constant)

So the expression for the equilibrium constant, K, is

K = [C(aq)]c[D(aq)]d
[A(aq)]a

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Worked Examples: Writing Equilibrium Constant Expressions

Reactions Involving Aqueous Species

  1. All species are present in aqueous solution. For example:

    Ag+(aq) +2NH3(aq) ⇋ Ag(NH3)2+(aq)

    K = [Ag(NH3)2+(aq)]
    [Ag+(aq)][NH3(aq)]2

  2. Solvent, such as water, takes part in reaction. For example:

    NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq)

    K = [NH4+(aq)][OH-(aq)]
    [NH3(aq)]

    The concentration of the liquid water as solvent is said to be constant and is incorporated into the value of K.

  3. A solid is present in the reaction. For example a precipitation reaction:

    Pb2+(aq) + 2I-(aq) ⇋ PbI2(s)

    K = 1
    [Pb2+(aq)][I-(aq)]2

    The concentration of the solid (the precipitate) is said to be constant and is incorporated into the value of K.

Reactions Involving Gases

  1. All species are present as gases. For example:

    CO(g) + 2H2(g) ⇋ CH3OH(g)

    K = [CH3OH(g)]
    [CO(g)][H2(g)]2

  2. A solid is present in the reaction. For example:

    CaCO3(s) ⇋ CaO(s) + CO2(g)

    K = [CO2]

    The concentration of solids are said to be constant and are incorporated into the value of K.

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Problem Solving: Write the Expression for the Equilibrium Constant, K

Question: Solid silver sulfate, Ag2SO4(s), is only slightly soluble in water.

When some solid silver sulfate is added to water, some of it dissolves producing an aqueous solution of silver(I) ions, Ag+(aq), and sulfate ions, SO42-(aq), which is in equilibrium with solid silver sulfate according to the balanced chemical equation given below:

Ag2SO4(s) ⇋ 2Ag+(aq) + SO42-(aq)

Write the expression for the equilibrium constant, K, for this reaction.

Solution:

(using the StoPGoPS approach to problem solving)

STOP STOP! State the Question.
  What is the question asking you to do?

Write the expression for K

PAUSE PAUSE to Prepare a Game Plan
  (1) What information (data) have you been given in the question?

Ag2SO4(s) ⇋ 2Ag+(aq) + SO42-(aq)

(2) What is the relationship between what you know and what you need to find out?

(i) For the general equation: aA + bB ⇋ cC + dD the equilibrium expression is given by K = [C]c[D]d/[A]a[B]b

(ii) Note that the concentration of a solid or liquid is a constant and is incorporated into the value of K

GO GO with the Game Plan
 

(i) Ag2SO4(s) ⇋ 2Ag+(aq) + SO42-(aq)
Terms to be used:
reactant: [Ag2SO4(s)]
products: [Ag+(aq)]2 and [SO42-(aq)]

(ii) Ag2SO4(s) is a solid so its concentration is constant and is incorporated into the value of K

K = [Ag+(aq)]2[SO42-(aq)]

PAUSE PAUSE to Ponder Plausibility
  Have you answered the question?

Yes, we have written an expression for the equilibrium constant, K

Is your answer plausible?

Work backwards: Given the expression for K, write a balanced chemical equation:

product (numerator): [Ag+(aq)]2 term means the stoichiometric coefficient for Ag+(aq) is 2, that is, 2Ag+(aq)

product (numerator): [SO42-(aq)] term means the stoichiometric coefficient for SO42-(aq) is 1, that is, 1SO42-(aq)

reactant (denominator): does not appear in the expression so the reactant(s) is either a solid or a liquid, and its formula is probably Ag2SO4 based on the ratio of products above.

Therefore the chemical equation is either:

Ag2SO4(l) ⇋ 2Ag+(aq) + SO42-(aq)

Ag2SO4(s) ⇋ 2Ag+(aq) + SO42-(aq)

Since the second chemical equation agrees with that given in the question we are reasonably confident that our answer is plausible.

STOP STOP! State the Solution
 

K = [Ag+(aq)]2[SO42-(aq)]