Reversible reactions and equilibrium

1. Overview

In many chemical reactions, reactants are completely converted into products. However, some reactions are reversible, meaning the products can react together to reform the original reactants. Understanding how to control these reactions is essential for large-scale industrial processes like the production of fertilizers and acids.

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Key Definitions

• Reversible Reaction: A reaction that can go both forwards (left to right) and backwards (right to left).
• Dynamic Equilibrium: A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.
• Closed System: A reaction vessel where no reactants or products can escape, but energy can be transferred.
• Anhydrous: A substance containing no water.
• Hydrated: A substance chemically combined with water (water of crystallisation).

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Core Content

The Reversible Reaction Symbol

Reversible reactions are represented by the symbol: $\rightleftharpoons$

Water and Hydrated Salts

You can change the direction of certain reversible reactions by adding or removing water or heat.

1. Copper(II) sulfate

• Word Equation: hydrated copper(II) sulfate $\rightleftharpoons$ anhydrous copper(II) sulfate + water
• Symbol Equation: $CuSO_4 \cdot 5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(l)$
• Observation:
  • Forward reaction (Endothermic): Heating blue hydrated copper(II) sulfate crystals turns them into a white anhydrous powder.
  • Reverse reaction (Exothermic): Adding water to white anhydrous copper(II) sulfate turns it back to blue and releases heat.

2. Cobalt(II) chloride

• Word Equation: hydrated cobalt(II) chloride $\rightleftharpoons$ anhydrous cobalt(II) chloride + water
• Symbol Equation: $CoCl_2 \cdot 6H_2O(s) \rightleftharpoons CoCl_2(s) + 6H_2O(l)$
• Observation:
  • Forward reaction: Heating pink hydrated cobalt(II) chloride turns it into blue anhydrous cobalt(II) chloride.
  • Reverse reaction: Adding water to blue anhydrous cobalt(II) chloride turns it pink.

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Extended Content (Extended Curriculum Only)

Characteristics of Equilibrium

In a closed system, a reversible reaction reaches equilibrium when:

1. The rate of the forward reaction is equal to the rate of the reverse reaction.
2. The concentrations of reactants and products remain constant (though not necessarily equal).

Predicting the Position of Equilibrium (Le Chatelier’s Principle)

If a change is made to the conditions of a system at equilibrium, the system shifts to counteract that change.

Change	Effect on Equilibrium Position
Increase Temperature	Shifts in the direction of the endothermic reaction.
Decrease Temperature	Shifts in the direction of the exothermic reaction.
Increase Pressure	Shifts to the side with the fewer moles of gas.
Decrease Pressure	Shifts to the side with the more moles of gas.
Increase Concentration	Shifts to consume the added substance.
Add Catalyst	No effect on position; it increases the rate of both reactions equally.

How to apply Le Chatelier’s Principle (step by step) Consider the Haber process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ (forward = exothermic)

What happens if you increase the pressure?

1. Count the moles of gas on each side: left = $1 + 3 = 4$ moles, right = $2$ moles.
2. The system will shift to reduce pressure, so it moves towards the side with fewer gas moles.
3. Equilibrium shifts right (towards NH₃) — more ammonia is produced.

What happens if you increase the temperature?

1. The forward reaction is exothermic (gives out heat).
2. Increasing temperature adds heat. The system shifts to remove that extra heat.
3. It favours the endothermic (reverse) direction — less ammonia, more N₂ and H₂.
4. So why use 450°C? Because even though yield drops, the rate increases dramatically. A lower temperature would give more ammonia but take too long. 450°C is the compromise.

This mole-counting and heat-direction approach works for any equilibrium question.

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The Haber Process (Ammonia Production)

• Equation: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ (Forward reaction is exothermic)
• Sources:
  • Nitrogen: Obtained from the air.
  • Hydrogen: Obtained from methane (natural gas) reacted with steam.
• Conditions:
  • Temperature: 450°C
  • Pressure: 20,000 kPa / 200 atm
  • Catalyst: Iron (Fe)
• Explanation of Conditions:
  • Pressure: High pressure favors the right side (4 moles gas → 2 moles gas), increasing yield. 200 atm is a compromise between high yield and the high cost/safety risk of equipment.
  • Temperature: Low temperature favors the exothermic forward reaction (higher yield), but the rate would be too slow. 450°C is a compromise temperature to get a reasonable yield at a fast enough rate.

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The Contact Process (Sulfuric Acid Production)

• Equation: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ (Forward reaction is exothermic)
• Sources:
  • Sulfur Dioxide: Burning sulfur or roasting sulfide ores (e.g., zinc blende).
  • Oxygen: Obtained from the air.
• Conditions:
  • Temperature: 450°C
  • Pressure: 200 kPa / 2 atm
  • Catalyst: Vanadium(V) oxide ($V_2O_5$)
• Explanation of Conditions:
  • Pressure: High pressure favors the right side (3 moles gas → 2 moles gas), but the yield is already so high (~98%) at 2 atm that higher pressure is not economically necessary.
  • Temperature: 450°C is used to ensure a fast reaction rate without shifting the equilibrium too far to the left.

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Key Equations

Process	Balanced Symbol Equation
Hydrated Copper Sulfate	$CuSO_4 \cdot 5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(l)$
Hydrated Cobalt Chloride	$CoCl_2 \cdot 6H_2O(s) \rightleftharpoons CoCl_2(s) + 6H_2O(l)$
Haber Process	$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Contact Process	$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

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Common Mistakes to Avoid

• ❌ Wrong: Thinking equilibrium means the concentrations of reactants and products are equal.
• ✓ Right: Equilibrium means the concentrations are constant and the rates are equal.
• ❌ Wrong: Saying a catalyst increases the yield of a reaction.
• ✓ Right: A catalyst only increases the rate at which equilibrium is reached; it does not change the position of equilibrium.
• ❌ Wrong: Forgetting state symbols in equations for the Haber or Contact process.
• ✓ Right: Always include $(g)$ for all reactants and products in these industrial gas-phase reactions.

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Exam Tips

• Command Word: Explain: When asked to explain the effect of pressure, always mention the number of moles of gas on each side of the equation.
• Command Word: Describe: When describing the test for water using cobalt(II) chloride, mention the color change from blue to pink.
• Compromise Conditions: If asked why 450°C is used in the Haber process, emphasize it is a compromise between rate (faster at high temp) and yield (higher at low temp for exothermic reactions).
• Real-World Contexts: Expect questions on the Haber process regarding fertilizer production and the Contact process regarding sulfuric acid.
• State Symbols: Always check if the question asks for state symbols. In equilibrium questions, the state $(g)$ is vital for pressure calculations.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [6 marks]

Question:

Cobalt(II) chloride exists as both anhydrous (CoCl2) and hydrated (CoCl2•6H2O) forms. The hydrated form is pink, while the anhydrous form is blue. The following equilibrium is established in a closed system:

$CoCl_2 \cdot 6H_2O(s) \rightleftharpoons CoCl_2(s) + 6H_2O(g)$

(a) State what is meant by a reversible reaction. [1]

(b) Describe what you would observe if the equilibrium mixture is heated. Explain your answer. [3]

(c) State the effect of adding a drying agent to the equilibrium mixture. Explain your answer. [2]

Worked Solution:

(a)

1. A reversible reaction is one that can proceed in both the forward and reverse directions. The reaction can go either from reactants to products or from products to reactants.

How to earn full marks:

• Mention both forward and reverse reactions for the mark.

(b)

1. Heating the equilibrium mixture will shift the equilibrium to the right, favouring the endothermic direction.
2. The forward reaction is endothermic, so increasing the temperature will favour the formation of the anhydrous cobalt(II) chloride.
3. The mixture will turn from pink to blue.

How to earn full marks:

• State the colour change from pink to blue.
• Explain the shift in equilibrium direction due to heating (Le Chatelier's principle).
• State that the forward reaction is endothermic.

(c)

1. Adding a drying agent will remove water vapour from the system.
2. According to Le Chatelier's principle, the equilibrium will shift to the right to counteract the removal of water, forming more anhydrous cobalt(II) chloride.

How to earn full marks:

• State that the equilibrium will shift to the right.
• Explain the shift in equilibrium direction due to the removal of water.

Common Pitfall: Students often forget to mention the direction of the equilibrium shift when explaining the effect of changing conditions. Always state whether the equilibrium shifts to the left or right, and relate this to the change in concentration of reactants and products.

Exam-Style Question 2 — Short Answer [5 marks]

Question:

Sulfur dioxide ($SO_2$) reacts with oxygen ($O_2$) to form sulfur trioxide ($SO_3$) in the Contact process. The reaction is reversible:

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ $\Delta H = -197 , kJ/mol$

(a) State two sources of sulfur dioxide used in the Contact process. [2]

(b) State and explain the effect of increasing the pressure on the equilibrium yield of sulfur trioxide. [3]

Worked Solution:

(a)

1. Sulfur dioxide can be produced by burning sulfur.
2. Sulfur dioxide can also be produced by roasting sulfide ores.

How to earn full marks:

• State both burning sulfur and roasting sulfide ores.

(b)

1. Increasing the pressure will shift the equilibrium to the side with fewer moles of gas.
2. There are 3 moles of gas on the left side (2 moles of $SO_2$ and 1 mole of $O_2$) and 2 moles of gas on the right side (2 moles of $SO_3$).
3. Therefore, increasing the pressure will shift the equilibrium to the right, increasing the yield of sulfur trioxide.

How to earn full marks:

• State that the equilibrium shifts to the side with fewer moles of gas.
• Correctly identify the number of moles of gas on each side of the equation.
• State that increasing pressure increases the yield of $SO_3$.

Common Pitfall: When discussing the effect of pressure, make sure you clearly state the number of moles of gas on each side of the equation. This helps to justify the direction of the equilibrium shift.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

The Haber process is used to manufacture ammonia ($NH_3$) from nitrogen ($N_2$) and hydrogen ($H_2$). The reaction is reversible and exothermic:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ $\Delta H = -92 , kJ/mol$

The typical conditions used in the Haber process are 450°C, 200 atm pressure, and an iron catalyst.

(a) State the sources of nitrogen and hydrogen for the Haber process. [2]

(b) Explain why a temperature of 450°C is used, considering both the rate of reaction and the position of equilibrium. [4]

(c) Explain why a pressure of 200 atm is used, considering both safety and the position of equilibrium. [2]

Worked Solution:

(a)

1. Nitrogen is obtained from the air.
2. Hydrogen is obtained from methane.

How to earn full marks:

• State both air for nitrogen and methane for hydrogen.

(b)

1. A higher temperature increases the rate of reaction because molecules have more kinetic energy, leading to more frequent and successful collisions.
2. However, the forward reaction is exothermic. Increasing the temperature will shift the equilibrium to the left, favouring the reactants and decreasing the yield of ammonia.
3. 450°C is a compromise temperature. It's high enough to achieve a reasonably fast rate of reaction but not so high that the equilibrium shifts too far to the left, resulting in a very low yield.
4. A lower temperature would give a higher yield of ammonia, but the reaction would be too slow to be economically viable.

How to earn full marks:

• State that high temperature increases the rate of reaction due to increased kinetic energy and collisions.
• State that the forward reaction is exothermic, so high temperature decreases the yield.
• Explain that 450°C is a compromise between rate and yield.
• Mention the economic viability of the process.

(c)

1. Increasing the pressure shifts the equilibrium to the right because there are fewer moles of gas on the product side (2 moles) compared to the reactant side (4 moles).
2. However, very high pressures are expensive to achieve and maintain, requiring strong and costly equipment. They also pose a significant safety risk due to the potential for explosions.

How to earn full marks:

• State that high pressure shifts the equilibrium to the right (towards ammonia).
• State that very high pressures are expensive and dangerous.

Common Pitfall: When explaining the compromise conditions in the Haber process, remember to discuss both the rate of reaction and the equilibrium position. Don't just focus on one factor; explain why the chosen conditions are a balance between these two competing effects.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student investigates the reaction between nitrogen dioxide ($NO_2$), a brown gas, and dinitrogen tetroxide ($N_2O_4$), a colorless gas. The reaction is reversible:

$2NO_2(g) \rightleftharpoons N_2O_4(g)$ $\Delta H = -57 , kJ/mol$

The student places a sealed tube containing an equilibrium mixture of the two gases in a water bath. They observe that when the water bath is heated, the color of the gas mixture becomes darker brown.

(a) State what is meant by the term equilibrium. [2]

(b) Explain why the color of the gas mixture becomes darker brown when the water bath is heated. [3]

(c) The student then compresses the gas mixture in the sealed tube, keeping the temperature constant. Predict and explain the effect of this compression on:

(i) the concentration of $NO_2$ [2]

(ii) the value of the equilibrium constant, $K_c$ [2]

Worked Solution:

(a)

1. Equilibrium is a state where the rate of the forward reaction is equal to the rate of the reverse reaction.
2. At equilibrium, the concentrations of reactants and products are constant.

How to earn full marks:

• State that the forward and reverse reaction rates are equal.
• State that the concentrations of reactants and products are constant.

(b)

1. The forward reaction is exothermic. Heating the water bath increases the temperature of the gas mixture.
2. According to Le Chatelier's principle, increasing the temperature will shift the equilibrium to the left, favouring the endothermic reaction.
3. The endothermic reaction is the reverse reaction, which produces more nitrogen dioxide ($NO_2$). Since nitrogen dioxide is a brown gas, the mixture becomes darker brown.

How to earn full marks:

• State that the forward reaction is exothermic.
• State that increasing temperature shifts the equilibrium to the left (endothermic direction).
• State that this increases the concentration of $NO_2$, causing the darker brown color.

(c) (i)

1. Increasing the pressure will shift the equilibrium to the right because there are fewer moles of gas on the product side (1 mole of $N_2O_4$) than on the reactant side (2 moles of $NO_2$).
2. However, the concentration of $NO_2$ will increase because the volume of the container has decreased, and concentration = moles/volume. The shift in equilibrium cannot fully compensate for the decrease in volume.

How to earn full marks:

• State that the equilibrium shifts to the right.
• State that the concentration of $NO_2$ increases because the volume decreases, outweighing the equilibrium shift.

(ii)

1. The equilibrium constant, $K_c$, is only affected by temperature.
2. Since the temperature is kept constant, the value of $K_c$ will remain unchanged.

How to earn full marks:

• State that $K_c$ is only affected by temperature.
• State that $K_c$ remains unchanged.

Common Pitfall: Students often confuse the effect of pressure on the equilibrium position with its effect on concentration. While increasing pressure shifts the equilibrium to the side with fewer moles of gas, the concentration of all gases will initially increase due to the reduced volume. Remember that $K_c$ is only affected by temperature changes.