In previous study guides we have assumed that all reactions go to completion. For example, the reaction between the very reactive sodium metal and toxic chlorine gas produce sodium chloride which is table salt. The reaction is written as:

 
The reaction essentially goes to completion. You can put a container of NaCl in a container and come back 10 years or even 20 years later, and the sodium chloride is still there. It does not revert back to pure sodium metal or chlorine gas. This is because the NaCl is much more stable than either of the reactants.

Consider the following reaction for the decomposition of N₂O₄. The double arrow in the equation means the reaction is reversible. In other words, the reaction does not go to completion. The N₂O₄ will start to decompose to 2 NO₂, and at some point during the reaction the NO₂ will start to reform the N₂O₄. Once the rates of the forward and the reverse reaction are equal, the concentrations of the reactant and the products are constant. At that point, the system is at equilibrium.

 
For the reaction above, N₂O₄ is colorless and NO₂ is brown. At equilibrium, the color of the gaseous solution is brown. The forward and reverse reactions do not stop once a system reaches equilibrium, but the rates of the forward and reverse reactions are equal and the concentration of reactants and products stay constant. Equilibrium is a dynamic process.

The following plot shows the decomposition of N₂O₄ and NO₂ as a function of time. The concentrations are easily read from the plot. The initial concentration of N₂O₄ is 0.04 M and NO₂ is 0.00 M. For every one mole of N₂O₄ that decomposes, two moles of NO₂ is formed.

 
The NO₂ is formed twice as fast as N₂O₄ decomposes. There is a 1:2 mole ratio of N₂O₄ to NO₂. The reaction starts with N₂O₄ decomposing to NO₂, and at some point in time, the NO₂ reforms N₂O₄. As you can see from the plot, once the rates of the forward and reverse reaction are equal, the system is at equilibrium and the concentrations do not change.

For the reaction, N₂O₄ (g) ⇄ 2 NO₂ (g), when the forward and reverse reactions are equal, we can set the rate laws for the forward and reverse reaction equal to each other.

Rates of Reaction and the Equilibrium Constant

For the reaction, N₂O₄ (g) ⇄ 2 NO₂ (g), the forward and the reverse rates are equal at equilibrium.

Divide both sides of the equation by k_r and [N₂O₄].

The right side of the equation is the equilibrium constant expression for the forward reaction which is equal to the equilibrium constant, K_c. The ratio of the rate constants for the forward and reverse reactions is equal to the equilibrium constant, K_c.

For the general reaction,

the equilibrium constant expression is written as,

The square brackets indicate the molar concentration, and the subscript c indicates concentration. The value of K_c is obtained when the product concentrations, raised to the power of the stoichiometric coefficients, are divided by the reactant concentrations raised to the power of the stoichiometric coefficients. Only gases, molecules, and ions can be in the equilibrium constant expression. Solids and pure liquids are never used in the equilibrium constant expression because their concentrations stay almost constant. All concentrations and pressures are equilibrium concentrations and pressures.

At equilibrium:

The equilibrium constant expression can also be written in terms of pressures. For example,

The subscript p in the equilibrium constant indicates pressure.

The equilibrium constant expression whether in terms of concentrations or pressures is unitless. This is because measured concentrations and pressures are divided by their thermodynamic standard state quantities. Recall, standard state means a concentration of 1 M and 1 atm of pressure.

Consider the following reaction, and write the equilibrium constant expression in terms of concentration.

SnO₂ (s) + 2 CO (g) ⇄ Sn (s) + 2 CO₂ (g)

Note, SnO₂ and solid Sn do not appear in the equilibrium constant expression because they are both solids.

If the value of K_c is less than 1, then there will be more reactants in the vessel once equilibrium has been reached. If K_c is greater than 1, there will be more products in the vessel at equilibrium. Below are some K_c values to remember.

To determine the equilibrium constant, K_c or K_p, the balanced chemical equation and the equilibrium concentrations are needed. More on this in a future study guide.

Worksheet: Introduction to Equilibrium

Exercises

Exercise 1. The rate of a forward reaction is 2.6 x 10^-4 s^-1 and the rate of the reverse reaction is 1.9 x 10^-3 s^-1. Calculate K_c and draw a reaction diagram.

Check Solution/Answer to Exercise 1

Exercise 2. Write K_p for the following reaction.

NH₄Cl (s) ⇄ NH₃ (g) + HCl (g)

Check Solution/Answer to Exercise 2

Exercise 3. Write K_c for the following reactions.

a) CO₂ (g) ⇄ CO₂ (s)
b) H₂O (l) + SO₃ (g) ⇄ H₂SO₄ (aq)
c) N₂ (g) + H₂O (g) ⇄ NO (g) + H₂ (g)

Check Solution/Answer to Exercise 3

Exercise 4. Chlorine gas is prepared by the following 2-step reaction.

Check Solution/Answer to Exercise 4

Exercise 5. Write the equilibrium constant expression and indicate if products or reactants are favored at equilibrium for each of the following reactions.

Check Solution/Answer to Exercise 5

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