There are reactions that look like nothing else is taking place no matter how long the reaction is taking place. This is when the reactions are at equilibrium. Let’s help you master this topic for your grade 12 physical sciences.
Content:
1. Introduction
In chemical reactions, it does not always happen that reactants are used up. These are times when there are reactants available but products are no longer forming . While on it, some reactions can occur in both directions. These kinds of reactions are called the reversible reactions. They are presented with the sign \(\rightleftharpoons\). See the reaction below:
$${\normalsize aA+bB \rightleftharpoons cC+dD}$$This reaction occurs in both directions:
\(aA+bB \rightarrow cC+dD\) forward reaction
\(cC+dD \rightarrow aA+bB\) reverse reaction
Did you know nature and biology also has some reversible reactions? Examples:
- Evaporation and condensation.
- Reaction between hemoglobin and oxygen between lungs and tissue.
Some key definitions on this topic:
- Open system:
- A system that continuously interacts with its environment.
- Closed system:
- A system that is isolated from its surroundings.
- Reversible reactions:
- A reaction that can proceed in both the forward and the reverse direction under the same conditions.
- Dynamic equilibrium:
- A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the macroscopic properties remain constant.
- Chemical equilibrium:
- The state of a reversible reaction in a closed system when the rates of the forward and reverse reactions are equal.
- Le Chatelier’s principle:
- When a system at equilibrium is disturbed (by change in concentration, temperature or pressure), the system responds by opposing the change and a new equilibrium is established.
2. Chemical equilibrium
In section 1, we defined chemical equilibrium as the state of a reversible reaction in a closed system when the rates of the forward and reverse reactions are equal . This defines equilibrium using rates of reactions. This is the same as dynamic equilibrium. Now let's define chemical equilibrium using amounts or concentration of substances.
- Chemical equilibrium :
- is the stage where concentrations/ quantities of reactants and products remain constant.
So, for chemical equilibrium to exist, there must be a closed system, i.e., matter should not enter or leave the system. Now, what if matter enters the system? In this case, chemical equilibrium will never be reached. This is an open system.
Generally, reactions in equilibrium are reversible and are represented as:
$${\normalsize aA+bB \rightleftharpoons cC+dD}$$Revisiting the examples of evaporation and condensation - and haemoglobin and oxygen:
In a closed container that is sealed, evaporation and condensation occur as:
$${\normalsize H_{2}O(l) \rightleftharpoons H_{2}O(g)}$$The forward reaction represents evaporation while the reverse represents condensation.
In our body, haemoglobin and oxygen react as follows:
$${Hb(aq)+O_2(aq) \rightleftharpoons HbO_2(aq)}$$The forward reaction takes place in the lungs while the reverse take place in the body tissues.
The reactions above involve substances in the same phase. Equilibrium established by such substances are termed the homogeneous equilibrium . One established by substances in different phases is called the heterogeneous equilibrium.
3. Le Chatelier’s principle
When a system in equilibrium is disturbed, a new equilibrium will be formed. To understand how the system will react to the disturbance, we use Le Chatelier's principle.
- Le Chatelier’s principle:
- When a system at equilibrium is disturbed (by change in concentration, temperature or pressure), the system responds by opposing the change and a new equilibrium is established.
Factors affecting chemical equilibrium:
- concentration
- Pressure
- Temperature
A catalyst can be added to this list but its effect is nothing but to ensure that equilibrium is reached faster.
These three factors affect equilibrium in different ways. Let's how they affect the equilibrium.
i. Concentration
- Concentration:
- the amount of solute present per unit volume of solution.
Another definition of concentration, \(c\), is that it is the number of moles of a substance,\(n\) per cubic decimetre of solution, \(V\). It has the formula:
$${\normalsize c = \frac{\text{n}}{V}}$$We must note that concentration is directly proportional to amount of substance, provided is constant. So a change in concentration is a direct change in amount of substance.
Increasing concentration of the reactant
- When the concentration of the reactant is increased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by decreasing the concentration of the reactant.
- The reaction that uses(decreases) the reactants is the forward reaction.
- The forward reaction is favoured.
- The amount/concentration of the product will increase, while reactants’ will decrease.
Decreasing concentration of the reactant
- When the concentration of the reactant is decreased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by increasing the concentration of the reactant.
- The reaction that increases the reactants’ concentration is the reverse reaction.
- The reverse reaction is favoured.
- The amount/concentration of the product will decrease, while reactants’ will increase
Increasing concentration of the product
- When the concentration of the product is increased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by decreasing the concentration of the product.
- The reaction that uses(decreases) the products is the reverse reaction.
- The reverse reaction is favoured.
- The amount/concentration of the product will decrease, while reactants’ will decrease
Decreasing concentrating of the product
- When the concentration of the product is decreased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by increasing the concentration of the product.
- The reaction that increases the products concentration is the forward reaction.
- The forward reaction is favoured.
- The amount/concentration of the product will increase, while reactants’ will decrease
ii. Pressure
If pressure of a gaseous system is changed, the equilibrium will shift to minimise that change.
NB: Pressure does not affect liquids and solids.
At constant temperature, using Boyle’s law:
$${\normalsize P \propto \frac{1}{V}}$$From this, we know that: when pressure is increased, volume is decreased and when pressure is decreased, volume is increased.
To accurately predict the direction of the equilibrium, get the total moles (sum of coefficients) of the gas reactant and of the gas product. Then compare these in ratio, i.e., total gas reactant: total gas product.
For pressure to have an impact, both sides must have unequal moles of gas.
For effect of pressure:
- Increasing pressure
- favours the reaction that reduces the number of moles of gas, i.e., the reaction will proceed in the direction of the side with fewer moles of gas.
- Decreasing pressure
- favours the reaction that increases the number of moles of gas, i.e., the reaction will proceed in the direction of the side with a greater number moles of gas.
Consider the Haber process, for example:
$${\normalsize N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{2(g)}}$$Increasing pressure
- When the pressure of the system is increased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by decreasing the pressure.
- This favours the side with fewer number of moles of gas, thereby decreasing system pressure.
- There are fewer moles of gas in the products. So the forward reaction is favoured.
- The equilibrium will shift to the right.
- The yield of \(\normalsize NH_3\) will increase while the amounts of \(\normalsize N_2\) and \(\normalsize H_2\) decrease.
Decreasing pressure
- When the pressure of the system is decreased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by increasing the pressure.
- This favours the side with a greater number of moles of gas,thereby increasing system pressure.
- There is a greater number of moles of gas in the reactants. So the reverse reaction is favoured.
- The equilibrium will shift to the left.
- The yield of \(\normalsize NH_3\) will decrease while the amounts of \(\normalsize N_2\) and \(\normalsize H_2\) increase.
iii. Temperature
The effect of temperature on the equilibrium depends on whether the reaction is endothermic or exothermic.
- Exothermic reaction
- is a chemical reaction that releases heat. It has a negative change in enthalpy.
- Endothermic reaction
- is a chemical reaction that absorbs energy. It has a positive change in enthalpy.
Alternatively, these can be shown with heat as a reactant for endothermic reactions and as a product for an exothermic reaction.
A guideline for the effect of temperature on equilibrium:
- Increasing temperature
- favours the endothermic reaction.
- Decreasing temperature
- favours the exothermic reaction.
The first step to understanding the effect of temperature is to determine whether the forward reaction is endothermic or exothermic. This is derived from enthalpy change, \(\Delta{H}\), of the reaction. Then, the reverse will be the other one (exothermic if forward was endothermic).
Let's try an example of the following reaction: $${\normalsize CO_{2}(g)+H_{2}(g) \rightleftharpoons H_{2}O(g)+CO(g)\text{, }\Delta{H}\gt0}$$
This reaction is endothermic. This means the forward reaction is endothermic and the reverse is exothermic.
Increasing temperature
- When the temperature of the system is increased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by decreasing the temperature.
- Increasing temperature favours the endothermic reaction because the system will try to decrease temperature by absorbing heat (cooling down).
- The forward reaction is the endothermic reaction. So the forward reaction is favoured.
- The equilibrium will shift to the right.
- The amounts of products will increase while the amounts of reactants decrease.
Decreasing temperature
- When the temperature of the system is decreased, according to Le Chatelier’s principle, the system will act to oppose the disturbance by increasing the temperature.
- Decreasing temperature favours the exothermic reaction because the system will try to increase temperature by releasing energy (heating up).
- The reverse reaction is the exothermic reaction. So the reverse reaction is favoured.
- The equilibrium will shift to the left.
- The amounts of products will decrease while the amounts of reactants increase.
4. Equilibrium constant
Reversible reactions occur in both directions. Engineers and chemists would like to know which of the reactions, the forward or reverse will have a higher concentration of products or reactants. To do this, they use equilibrium constant, \(K_c\).
- Equilibrium constant:
- the ratio of the concentration of products to the concentration of the reactants.
For a balanced equation:
$${\normalsize aA+bB \rightleftharpoons cC+dD}$$ $${\normalsize K_c = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}}$$The square brackets ”[]” represent the concentrations in \(mol \cdot dm^{-3}\) at equilibrium and the powers are coefficients in a balanced reaction. This only applies to aqeuous and gaseous substances only.
\(K_c\) is unitless. So what do the values of \(K_c\) mean?
- \(0 \lt K_c \lt 1 \) Reverse reaction is dominant (Equilibrium moves to the left). More reactants than products.
- \(K_c \gt 1\): Forward reaction is dominant (Equilibrium moves to the right). More products forming than reactants.
There are two ways to determine the value \(K_c\): directly using the equilibrium concentration or using the RICEE (reaction, intial mol, change in mol, equilibrium mol and equilibrium concentration) table. The RICEE table is given below:
$${\normalsize aA+bB \rightleftharpoons cC+dD}$$| Reaction | aA | bB | cC | dD |
|---|---|---|---|---|
| Initial mol | ||||
| Change mol | -ax | -bx | +cx | +dx |
| Equilibrium mol | ||||
| Equilibrium concentration |
In using this table, the \(x\) is used merely as a placeholder. It is negative for reactants and positive for productive. The \(a, b, c, d\) are coefficients of the chemical equation. Write them in pencil so you can erase easily. Because the initial mol and equilibrium mol can be given in a question, use alegebra to solve for \(x\). NOTE:$${\normalsize \text{Initial mol}+ \text{ Change mol}=\text{Equilibrium mol}}$$
Example 1: $${\normalsize N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)}$$ A sealed tube contains \(NO_2\) and \(N_2 O_4\). Equilibrium is reached at 100°C. The concentrations at this stage are 0.00140 \(mol \cdot dm^{-3}\) \(N_2 O_4\) and 0.0172 \(mol \cdot dm^{-3}\) for \(NO_2\). Calculate the \(K_c\).
Solution
\(0 \lt K_c \lt 1 \). Meaning there are more reactants than product at equilibrium. The equilibrium lies to the left.
Example 2: Hydrogen iodide partially decomposes to hydrogen and iodine: $${\normalsize 2HI(g) \rightleftharpoons H_{2}(g)+I_{2}(g)}$$At equilibrium, it was found that the concentrations of HI is \(3.53\times10^{-3}\) \(mol \cdot dm^{-3}\), H2 is \(4.79\times10^{-4}\) \(mol \cdot dm^{-3}\) and I2 \(4.79\times10^{-4}\) \(mol \cdot dm^{-3}\). Determine the value of \(K_c\).
Solution
\(0 \lt K_c \lt 1 \). Meaning there are more reactants than product at equilibrium. The equilibrium lies to the left.
Example 3: $${\normalsize CO(g)+2H_{2}(g) \rightleftharpoons CH_{3}OH(g)}$$An equilibrium at 500K found that the concentrations of \( CH_{3}OH \) is 0.0203 \(mol \cdot dm^{-3}\), CO is 0.085 \(mol \cdot dm^{-3}\) and \(mol \cdot dm^{-3}\). Determine the value of \(K_c\).
Solution
\(K_c \gt 1 \). Meaning there are more product than reactants at equilibrium. The equilibrium lies to the right.
Example 4: Carbon dioxide gas reacts with C(s) according to the following balanced equation: $${\normalsize CO_{2}(g)+C(s) \rightleftharpoons 2CO(g)}$$The system reaches equilibrium at 100°C. Initially, 2,38 mol of \(CO_2\) was reacted with excess carbon in a sealed flask of volume 250 cm3. The equilibrium mixture was found to contain 1,90 mol of CO2. Determine the value of \(K_c\) at this temperature.
Solution
| Reaction | CO2 | C | CO |
|---|---|---|---|
| Initial mol | 2.38 | - | 0 |
| Change mol | -x | - | +2x |
| Equilibrium mol | 1.90 | - | 0.96 |
| Equilibrium concentration | 7.6 | - | 3.84 |
The amounts for C(s) are ignored because we don't consider solid substances for \(K_c\).
To determine the value of \(x\), we use algebra.
$$ \normalsize \begin{align*} 2.38 -x &=1.90\\ -x &= 1.90-2.38\\ -x &= -0.48\\ x &= 0.48 \end{align*} $$To determine the equilibrium concentration:
V=250 cm3 = 0.25 dm3
The completed table:
| Reaction | CO2 | C | CO |
|---|---|---|---|
| Initial mol | 2.38 | - | 0 |
| Change mol | -0.48 | - | +0.96 |
| Equilibrium mol | 1.90 | - | 0.96 |
| Equilibrium concentration | 7.6 | - | 3.84 |
Now, the value of \(K_c\):
$$ \normalsize \begin{align*} K_c&=\frac{[CO]^2}{[CO_2]}\\ &= \frac{3.84^2}{7.6^2}\\ &= 1.94 \end{align*} $$\(K_c \gt 1 \). Meaning there are more product than reactants at equilibrium. The equilibrium lies to the right.
Factors affecting equilibrium constant
While concentration, pressure and temperature affect the equilibrium position, it is only temperature that affects or changes the value of equilibrium constant, \(K_c\).
So, to change the equilibrium constant, we must note the kind of reaction we have. For a larger \(K_c\),\(K_c \gt 1\), temperature must be changed to favour the forward reaction. A smaller \(K_c\) will be favoured by the temperature change that favours the reverse reaction.