Many reactions are reversible. These reactions will reach a state of equilibrium when the rates of the forward reaction and reverse reaction are equal. The position of equilibrium can be controlled by changing the conditions.

The Equilibrium constant 

  • A rection is in equilibrium in a closed system when the rate of forward reaction equals the rate of backward reaction.
  • The state of dynamic equilibrium is recognized by constant macroscopic properties in a closed system as the concentration of the reactants and products are constant.
  • For a reaction $a\mathrm A + b\mathrm B \rightleftharpoons c\mathrm C + d\mathrm D$
    The concentrations of the reactants and products at equilibrium are related by the equilibrium law:
    $\rm K_{\mathcal c} = \dfrac{[C]^{\mathcal c}[D]^{\mathcal d}}{[A]^{\mathcal a}[B]^{\mathcal b}}$  All concentration are at equilibrium.
    $\mathrm K_c$, the equilibrium constant, is a constant for a given reaction at a specified temperature. It has no units.
    When $\mathrm K_c >> 1$ the equilibrium lies to the right.
    When $\mathrm K_c << 1$ the equilibrium lies to the left.
  • $\rm Q$, the reaction quotient, is a measure of the relative amounts of reactants and products in a reaction mixture at a particular time. It is calculated by substituting non-equilibrium values for reactant and product concentration into the equilibrium expression.
    $\rm Q = \dfrac{[C]^{\mathcal c}[D]^{\mathcal d}}{[A]^{\mathcal a}[B]^{\mathcal b}}$
    For all concentrations.
    If $\rm Q = K_{\mathcal c}$, reaction is at equilibrium.
    If $\rm Q < K_{\mathcal c}$, reaction will move to the right in favour of products.
    If $\rm Q > K_{\mathcal c}$, reaction will move to the left in favour of reactants.

Worked Example

Hydrogen and iodine undergo a reversible reaction.

$\rm H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

  1. Deduce the equilibrium constant $\rm K_1$ for this reaction. 

Solution 

$\rm K_1 = \dfrac{[HI(g)]^2}{ [H_2(g)] [I_2(g)]}$

  1. The same equilibrium can be expressed by the following reaction:

$\rm 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \qquad K_2$

Deduce the equilibrium constant $\rm K_2$ for this reaction.

Solution 

$\rm K_2 = \dfrac{[H_2(g)] [I_2(g)]}{[HI(g)]^2}$

  1. The same equilibrium can be expressed by the following reaction:

$\rm HI(g) \rightleftharpoons ½ H_2(g) + ½ I_2(g)\qquad K_3$

Deduce a relationship between $\rm K_1$ and $\rm K_3$.

Solution

$\rm K_3 = \dfrac{[H_2(g)]^½ [I_2(g)]^½}{[HI(g)]}$
$\rm K_3 = \left(\dfrac{[H_2(g)] [I_2(g)]}{[HI(g)]^2}\right)^½$
$\rm K_3 = \left(\dfrac{1}{K_1}\right)^½$

  1. The value of $\rm K_1$ at $\rm 298~K$ is $794$. Deduce the value of $\rm K_3$ at the same temperature.

Solution

$\rm K_3 = \left(\dfrac{1}{794}\right)^½ = 0.035$

Changing the position of equilibrium 

Worked Example 

Consider the equilibrium: 

$\rm A + B \rightleftharpoons C + D$ 

  1. Describe what happens to the position of equilibrium if the concentration of the reactant $\rm A$ is increased.

Solution

$\rm Q = \dfrac{[C]^{\mathcal c}[D]^{\mathcal d}}{[A]^{\mathcal a}[B]^{\mathcal b}}$ 

If $\rm [A]$ increase $\rm Q$ decreased and becomes less than $\mathrm K_c$.

$\rm Q < K_{\mathcal c}$, reaction will move to the right in favour of products until $\rm Q$ returns to $\mathrm K_c$.

  1. Describe what happens to the position of equilibrium if the concentration of the product $\rm B$ is decreased.

Solution

$\rm Q = \dfrac{[C]^{\mathcal c}[D]^{\mathcal d}}{[A]^{\mathcal a}[B]^{\mathcal b}}$

If $\rm [C]$ increases $\rm Q$ increases and becomes greater than $\mathrm K_c$.

$\rm Q > K_{\mathcal c}$, reaction will move to the left in favour of reactants until $\rm Q$ returns to $\mathrm K_c$.

These results can be generalised as Le Chatelier’s Principle.

Le Chatelier’s Principle  

  • When a change is applied to an equilibrium mixture, the composition of the equilibrium will change to minimize the effect of the change. The new equilibrium mixture will have different concentrations of reactants and products, but the value of $\mathrm K_c$ will be unchanged at the same temperature.

Worked Example 

Consider the reaction below.

$\scriptstyle\rm N_2(g) ~+~ 3H_2(g) ~\rightleftharpoons~ 2NH_3(g)\quad  \Delta H ~=~ –92~KJ~mol^{–1}$

What change in temperature would result in an increase in the amount of $\rm NH_3(g)$?

Solution

The reaction is exothermic.
Consider an increase in temperature by adding heat. 
In line with Le Chatelier’s principle (LCP) the position of equilibrium will adjust absorb the additional heat. It moves in the endothermic direction. 
i.e. the equilibrium will move to the left.    
For the reaction to move in the exothermic direction heat needs to be removed from the system.
The temperature needs to be decreased.

  • The equilibrium constant only depends on the temperature. It increases with temperature for endothermic reactions and decreases with temperature for exothermic reactions.
  • Catalysts increase the rate of the forward and background reaction equally but do not change the yield or the composition of the equilibrium mixture.
  • Optimum conditions for an industrial process are based on equilibrium, kinetic, and economic considerations.