Reversible reactions 

In a nutshell 

Reversible reactions are reactions that can happen both ways. In reactions you have your reactants converting into products. In reversible reactions, your reactants can convert into products and at the same time, your products can turn back into your reactants. 

Dynamic equilibrium

Dynamic equilibrium is when the concentrations of the reactants and products stay constant in an equilibrium reaction. Dynamic equilibrium only occurs in a closed system. A closed system is a system that only exchanges energy with its surroundings. 

Initially, the reactants are used up to form the product. This slows down the forward reaction. This causes the backwards reaction to speed up as there's more product formed. After some time, the forward and backward reactions will occur at the same rate, reaching dynamic equilibrium.

Once dynamic equilibrium is reached, the products and reactants will be formed at the same rate. It will seem like nothing is happening. 

Example

Write the reversible reaction for nitrogen and oxygen.

$$N₂ (g) + O₂ (g) ⇌ 2NO (g)$$

N₂ (g) + O₂ (g) ⇌ 2NO (g)⇌This arrow, ⇌, indicates that the reaction is reversible.

Note: The arrow, ⇌, indicates that the reaction is reversible. This means that the forward reaction and backward reaction are happening at the same time.

Equilibrium constant, $$K_c$$​​

The equilibrium constant, $$K_c$$​, shows which reaction the equilibrium is closer to. It can be calculated with known concentrations of the reactants and products in a closed system. 

Catalysts don't affect the equilibrium constant. This is because the concentrations of the reactants and products aren't affected by the catalyst. Catalysts only speed up the rate at which the dynamic equilibrium is reached.

Homogeneous equilibria and $$K_c$$​​

Homogeneous means everything is in the physical same state. 

The equilibrium constant, $$K_c$$​, can be calculated when a dynamic equilibrium is reached. 

Procedure 

Example

For the example equation, the $$K_c$$ expression is:

$$K_c = \dfrac{[NO]^2} { [N₂]^1 [O₂]^1}$$
​

This is the same as : 

​​$$K_c = \dfrac{[NO]^2}{ [N₂] [O₂]}$$

Substitute values into the equation:

 $$[NO] = 0.1, [N2] = 0.05, [O2] = 0.05$$​​

​$$K_c = \dfrac{[0.1]^2}{[0.05]^1 \times [0.05]^1}$$

Calculate final answer:

​$$K_c = \dfrac{0.01 } { 0.0025}$$

$$K_c = 4$$

The $$K_c$$ expression is $$\underline 4$$.

Heterogeneous equilibria and $$K_c$$​​

Heterogeneous means everything is not in the same physical state.  

When you write the expression for $$K_c$$​, you only include reactants and products that are in the gaseous or aqueous states. 

You can ignore any solid or liquid species because its concentration will remain constant during the reaction.

Example

$$H₂O (g) + C (s) ⇌ H₂ (g) + CO (g)$$​

Carbon is a solid while all the other chemicals are gases. 

$$K_c = \dfrac{[H₂] [CO]}{ [H₂O]}$$

Ignore the carbon when writing the $$K_c$$ expression.