Reversible Reactions and Dynamic Equilibrium for the ESAT

Reversible Reactions and the Nature of Equilibrium

Many chemical reactions are reversible, meaning the products formed can react together to re-form the original reactants. This is indicated in a chemical equation by a double arrow symbol: $\rightleftharpoons$. A classic example is the production of ammonia from nitrogen and hydrogen gases:

$3\mathrm{H}_{2}(\mathrm{g}) + \mathrm{N}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_{3}(\mathrm{g})$

In a closed system, where no matter can enter or leave, these reactions eventually reach a state of dynamic equilibrium. At this point, the forward and reverse reactions are happening simultaneously and at the exactly same rate. While the reaction appears to have stopped on a macroscopic level, molecules are still reacting constantly.

An everyday analogy for dynamic equilibrium is a person walking up a descending escalator. If the person walks up at the same speed the escalator moves down, their position relative to the floor remains constant. They are in motion, but there is no net change in their location.

The Position of Equilibrium

The position of equilibrium describes the relative amounts of reactants and products present when equilibrium is reached. If the equilibrium lies to the left, the mixture contains more reactants than products. If it lies to the right, products predominate. When conditions are changed, the system moves to oppose that change, shifting the equilibrium position.

Factors Affecting Equilibrium

1. Changing Concentration

If the concentration of a substance in the equilibrium mixture is changed, the system will shift to either remove the added substance or replace what was lost:

1. Add more reactant: The equilibrium moves to the right to use up the extra reactant and form more product.
2. Remove reactant: The equilibrium moves to the left to replace the lost reactant.
3. Add more product: The equilibrium moves to the left to remove the extra product.
4. Remove product: The equilibrium moves to the right to replace the lost product.

Example: Chromate and Dichromate Ions Yellow chromate ions $\mathrm{CrO}_{4}^{2-}(\mathrm{aq})$ react with hydrogen ions $\mathrm{H}^{+}(\mathrm{aq})$ to form orange dichromate ions $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})$:

$2 \mathrm{CrO}_{4}^{2-}(\mathrm{aq}) + 2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

If acid (adding $\mathrm{H}^{+}$) is introduced, the equilibrium moves to the right to remove the added $\mathrm{H}^{+}$, making the solution more orange. If an alkali is added, it reacts with and removes $\mathrm{H}^{+}$ ions. The equilibrium then shifts to the left to replace them, turning the solution yellow.

2. Changing Temperature

The effect of temperature depends on whether the forward reaction is exothermic (releases heat, negative $\Delta H$) or endothermic (absorbs heat, positive $\Delta H$). The system will always act to restore the original temperature:

1. Increase temperature: The equilibrium shifts in the endothermic direction to absorb the extra heat.
2. Decrease temperature: The equilibrium shifts in the exothermic direction to release more heat.

Worked Example: Sulfur Trioxide Synthesis In the reaction $2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$, the forward reaction is exothermic. If the temperature is increased, the system shifts in the endothermic (reverse) direction to the left. This results in a decrease in the amount of $\mathrm{SO}_{3}$ produced.

3. Changing Overall Pressure

Pressure changes only affect equilibria involving gases. Pressure is determined by the number of gas molecules in a given volume. Increasing the number of gas molecules increases the pressure.

1. Increase pressure: The equilibrium moves to the side with the fewer total number of gas molecules to lower the pressure.
2. Decrease pressure: The equilibrium moves to the side with the more total number of gas molecules to increase the pressure.

Worked Example: Methane and Steam Consider the reaction $\mathrm{CH}_{4}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 3 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{CO}(\mathrm{g})$. On the left, there are 2 gas molecules. On the right, there are 4 gas molecules. If the pressure is increased, the equilibrium shifts to the left (the side with fewer molecules), which decreases the yield of hydrogen gas.