Welcome to Reversible Reactions and Equilibrium!
Hello future chemists! This chapter is one of the most interesting parts of chemistry because it deals with reactions that don't just go one way—they can go backwards too! Imagine a tug-of-war where neither side ever wins; that's the essence of chemical equilibrium. Understanding how to control these reactions is vital for massive industrial processes, like making fertilizers or sulfuric acid.
We will learn how to identify these reactions and, crucially, how scientists use special rules to push them in the direction they want.
1. Reversible Reactions (The Two-Way Street)
What is a Reversible Reaction? (Core Content)
Most reactions you’ve learned about go to completion (they are irreversible). Once the reactants turn into products, that's it.
A reversible reaction is a chemical reaction where the products can react together to reform the original reactants.
- We show a reversible reaction using the symbol double half-arrow: \(\rightleftharpoons\).
- The reaction going to the right is the forward reaction.
- The reaction going to the left is the reverse reaction.
Analogy: Think of a treadmill. You can run forward (forward reaction), but if you stop and let the belt pull you backward, that’s the reverse reaction. In a reversible reaction, both processes are happening at the same time!
1.1 Core Examples of Reversible Reactions
The IGCSE syllabus focuses on two classic examples involving dehydration (removing water) and hydration (adding water), often shown by striking colour changes. These experiments are usually carried out in open systems (not at equilibrium, just showing reversibility).
A. Hydrated Copper(II) Sulfate
Forward Reaction (Heating/Dehydration – Endothermic):
When you heat the blue crystalline solid hydrated copper(II) sulfate, the water molecules chemically combined within the crystal structure are driven off. This is an endothermic process (it takes in heat).
\(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s)\) (blue) \(\rightleftharpoons \text{CuSO}_4(s)\) (white) \(+ 5\text{H}_2\text{O}(g)\)
- Observation: Blue crystals turn into a white powder.
Reverse Reaction (Adding Water/Hydration – Exothermic):
If you add water to the resulting white powder (anhydrous copper(II) sulfate), the blue crystals reform. This is an exothermic process (it gives out heat—the test tube will feel warm).
- Observation: White powder turns back into blue crystals.
Quick Tip: Anhydrous $\text{CuSO}_4$ (white) is used as a laboratory test for the presence of water.
B. Cobalt(II) Chloride
Forward Reaction (Heating/Dehydration – Endothermic):
Heating the pink solid or solution drives off water, forming anhydrous cobalt(II) chloride.
\(\text{CoCl}_2 \cdot 6\text{H}_2\text{O}(s)\) (pink) \(\rightleftharpoons \text{CoCl}_2(s)\) (blue) \(+ 6\text{H}_2\text{O}(g)\)
- Observation: Pink changes to blue.
Reverse Reaction (Adding Water/Hydration – Exothermic):
Adding water causes the reverse reaction.
- Observation: Blue changes back to pink.
🔑 Key Takeaway (Core)
A reversible reaction uses the \(\rightleftharpoons\) symbol. The key practical examples show a reversal of color when water is added or removed (hydration/dehydration).
2. The State of Dynamic Equilibrium (Extended Content)
Definition of Equilibrium
For a reversible reaction to reach equilibrium, two key conditions must be met:
- The reaction must be in a closed system. (This means nothing can enter or leave, especially gases. The pressure and temperature are usually kept constant.)
- It must be dynamic.
When a reversible reaction reaches dynamic equilibrium:
- Rate Condition: The rate of the forward reaction is exactly equal to the rate of the reverse reaction.
- Concentration Condition: The concentrations of the reactants and products are no longer changing (they remain constant).
Wait, if the rates are equal, does that mean the reaction stops?
No! This is why it’s called dynamic (moving). Although the concentrations appear constant, the forward and reverse reactions are still happening simultaneously at the same speed. For every molecule of reactant turning into product, one molecule of product is turning back into reactant.
Analogy: The Elevator Ride
Imagine an escalator going up (forward reaction) and people walking down the same escalator (reverse reaction). Once the number of people going up per minute equals the number of people going down per minute, the total number of people on the escalator stays constant—even though everyone is still moving. This is dynamic equilibrium.
3. Le Châtelier's Principle (Extended Content)
If a system is at equilibrium, we can "stress" it by changing the conditions. Le Châtelier’s Principle explains how the system reacts:
"If a change in condition (a stress) is applied to a system in dynamic equilibrium, the system shifts its position of equilibrium to counteract the change."
The reaction will temporarily favor one direction (forward or reverse) until a new equilibrium is established.
3.1 Effect of Changing Concentration
If you change the concentration of any component, the equilibrium will shift to try and use up the substance you added or replace the substance you removed.
- Add Reactants: The equilibrium shifts to the right (forward reaction is favoured) to use up the extra reactants, making more products.
- Add Products: The equilibrium shifts to the left (reverse reaction is favoured) to use up the extra products, making more reactants.
- Remove Products: The equilibrium shifts to the right to replace the missing product. (This is often done in industry to maximize yield).
3.2 Effect of Changing Temperature
Temperature changes affect the equilibrium position because reactions are either endothermic or exothermic.
Consider a general exothermic forward reaction:
\(A + B \rightleftharpoons C + D\) (\(\Delta H\) = negative, Forward is Exothermic)
If the forward reaction is exothermic, the reverse reaction must be endothermic.
- Increase Temperature (Add Heat): The system shifts in the direction that absorbs heat (the Endothermic direction) to cool itself down.
- In this case, the equilibrium shifts Left, decreasing the yield of C and D.
- Decrease Temperature (Remove Heat): The system shifts in the direction that releases heat (the Exothermic direction) to warm itself up.
- In this case, the equilibrium shifts Right, increasing the yield of C and D.
3.3 Effect of Changing Pressure (Gases Only)
Pressure only affects equilibrium if there are gases involved and there is a difference in the total number of gas molecules on each side of the equation.
Count the moles of gas molecules (using the big numbers/coefficients in the equation).
- Increase Pressure: The system shifts towards the side with the fewer total number of gas molecules to reduce the pressure.
- Decrease Pressure: The system shifts towards the side with the greater total number of gas molecules to increase the pressure.
Example: \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\)
Left side has \(1 + 3 = 4\) moles of gas. Right side has \(2\) moles of gas. Increasing pressure favors the side with 2 moles (the right side), thus increasing ammonia yield.
What if the moles are equal? If the total moles of gas reactants equals the total moles of gas products, changing the pressure has no effect on the position of equilibrium.
3.4 Effect of Using a Catalyst
A catalyst increases the rate of reaction by lowering the activation energy ($\text{E}_\text{a}$).
Important Rule: A catalyst increases the rate of the forward reaction and the reverse reaction by the same amount.
- A catalyst helps the system reach equilibrium faster.
- A catalyst has no effect on the position or percentage yield of the equilibrium.
🧠 Le Châtelier's Mnemonic
Concentration: Shift to Use Up what you added.
Temperature: Shift to Oppose the change (Exo or Endo direction).
Pressure: Shift to the side with Fewer Moles (to reduce volume).
4. Industrial Equilibrium: The Haber Process (Extended Content)
The Haber process is crucial for manufacturing ammonia ($\text{NH}_3$), which is then used to make vital nitrogen-based fertilizers.
4.1 Reaction Details
Symbol Equation:
\(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\) (\(\Delta H\) = negative, Exothermic)
Sources of Reactants:
- Nitrogen ($\text{N}_2$): Obtained directly from air (which is about 78% nitrogen).
- Hydrogen ($\text{H}_2$): Usually obtained from reacting methane (natural gas) with steam.
Conditions Used:
- Temperature: 450 °C
- Pressure: 20,000 kPa (or 200 atm)
- Catalyst: Iron (Fe)
4.2 Explaining the Compromise Conditions
We apply Le Châtelier's Principle to determine the best conditions, keeping in mind safety and economics (cost of energy/equipment).
A. Temperature
Le Châtelier's Ideal: Since the forward reaction is exothermic, a low temperature would favor the right side and give a high yield of $\text{NH}_3$.
Reality (The Compromise): A very low temperature makes the reaction rate too slow to be profitable. The 450 °C used is a compromise temperature:
- It's low enough to maintain a reasonable equilibrium position (good yield).
- It's high enough to ensure a fast reaction rate.
B. Pressure
Le Châtelier's Ideal: The reaction goes from 4 moles of gas to 2 moles of gas. Therefore, a very high pressure would favor the right side and give a high yield.
Reality (Economics & Safety): Pressurizing the system is very expensive and requires extremely strong (and therefore costly) piping and vessels, and poses serious safety risks if equipment fails. The 20,000 kPa is a compromise pressure:
- It's high enough to shift the equilibrium significantly to the right (good yield).
- It’s a pressure that is still economically viable and safe to maintain.
C. Catalyst (Iron)
The iron catalyst is used to increase the rate of reaction at the moderate 450 °C temperature, allowing the system to reach equilibrium quickly, saving time and energy costs, without affecting the final yield.
5. Industrial Equilibrium: The Contact Process (Extended Content)
The Contact process is used to manufacture sulfur trioxide ($\text{SO}_3$), which is the key step in making sulfuric acid ($\text{H}_2\text{SO}_4$).
5.1 Reaction Details
The reversible step in this process is the conversion of sulfur dioxide to sulfur trioxide:
Symbol Equation:
\(2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)\) (\(\Delta H\) = negative, Exothermic)
Sources of Reactants:
- Sulfur dioxide ($\text{SO}_2$): Obtained by burning sulfur in air, or by roasting sulfide ores (like zinc blende).
- Oxygen ($\text{O}_2$): Obtained directly from air.
Conditions Used:
- Temperature: 450 °C
- Pressure: 200 kPa (or 2 atm)
- Catalyst: Vanadium(V) oxide ($\text{V}_2\text{O}_5$)
5.2 Explaining the Compromise Conditions
A. Temperature
Le Châtelier's Ideal: The forward reaction is exothermic, so low T gives high yield.
The Compromise: Just like the Haber process, 450 °C is used as a compromise. At lower temperatures, the $\text{V}_2\text{O}_5$ catalyst is not effective, and the rate is too slow.
B. Pressure
Le Châtelier's Ideal: The reaction goes from 3 moles of gas to 2 moles of gas. High pressure favors the right side (fewer moles) for high yield.
The Compromise: Unlike the Haber process, the yield is already over 99% even at atmospheric pressure (1 atm) because the equilibrium lies so far to the right naturally! A very high pressure is unnecessary. Therefore, only a moderate pressure of 200 kPa (2 atm) is used, which is cheap and safe.
C. Catalyst ($\text{V}_2\text{O}_5$)
The vanadium(V) oxide catalyst is used to ensure a high reaction rate at 450 °C.
✅ Quick Review: Industrial Processes
- Both processes (Haber and Contact) use a moderate temperature (450 °C) to maximize the rate, even though a lower temperature would give a better yield (Exothermic reaction).
- Pressure is chosen based on a balance between achieving a good yield, and managing the cost and safety of the equipment.
- Catalysts are used solely to increase the rate and save time/money.