Topic X: Reversible Reactions & Dynamic Equilibrium
Hello! Welcome to the fascinating world of chemical equilibrium. Ever wondered why some reactions don't seem to 'finish'? Or how big chemical plants can produce huge amounts of products so efficiently? The answers lie in understanding reversible reactions and a special state called dynamic equilibrium.
In this chapter, we'll explore reactions that can go both forwards and backwards. We'll learn what it means for a reaction to be 'in balance' and discover how chemists can cleverly push a reaction in the direction they want. Don't worry if this sounds complicated; we'll break it down with simple examples and analogies. Let's get started!
Reversible vs. Irreversible Reactions
Most reactions you've learned about so far go in one direction. When you burn a piece of paper, it turns into ash. You can't easily turn the ash back into paper. This is an irreversible reaction.
Example: Burning magnesium in air. $$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$$
However, many chemical reactions are two-way streets. The products can react with each other to re-form the original reactants. These are called reversible reactions. We use a special double arrow ($$\rightleftharpoons$$) to show this.
The reaction going from left to right is called the forward reaction.
The reaction going from right to left is called the reverse reaction (or backward reaction).
Example: Heating hydrated copper(II) sulphate. The blue crystals turn into a white powder when heated, but if you add water to the white powder, it turns blue again! $$CuSO_4 \cdot 5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(g)$$ (Blue hydrated solid) $$ \rightleftharpoons $$ (White anhydrous solid) + (Water vapour)
Key Takeaway
Irreversible reactions go in one direction (→). Reversible reactions can go in both directions ($$\rightleftharpoons$$), with a forward and a reverse reaction happening at the same time.
What is Dynamic Equilibrium?
Imagine a busy shop. People are constantly entering, and other people are constantly leaving. If the number of people entering per minute is the same as the number of people leaving per minute, the total number of people inside the shop stays constant. From the outside, it might look like nothing is changing, but inside, there is constant movement. This is a great analogy for dynamic equilibrium!
In a reversible reaction, when the rate of the forward reaction becomes equal to the rate of the reverse reaction, the system is in a state of dynamic equilibrium.
- Dynamic: This means the reactions haven't stopped! Both forward and reverse reactions are still happening.
- Equilibrium: This means there is no overall change. The amounts (concentrations) of reactants and products remain constant because they are being formed at the same rate they are being used up.
Characteristics of Dynamic Equilibrium
For a system to be in dynamic equilibrium, it must have these features:
- The rates of the forward and reverse reactions are equal.
- The concentrations of all reactants and products are constant.
- It can only happen in a closed system, where no substances can enter or leave.
- Macroscopic properties (like colour, pressure, concentration) do not change over time.
Quick Review: Important Distinction!
Common Mistake: Thinking that the reaction has stopped at equilibrium.
Correction: The reactions are still going! They are just perfectly balanced, like two equally matched tug-of-war teams pulling with the same force.
Key Takeaway
Dynamic equilibrium is a state in a reversible reaction where the forward and reverse reaction rates are equal, resulting in constant concentrations of reactants and products. It's a state of balance, not inactivity.
The Equilibrium Constant (Kc)
At equilibrium, the concentrations of reactants and products are constant, but that doesn't mean they are equal. Sometimes there are lots of products and very few reactants, and sometimes it's the other way around. The equilibrium constant (Kc) is a number that tells us the position of the equilibrium.
How to Write the Kc Expression
For a general reaction:
$$aA(aq) + bB(aq) \rightleftharpoons cC(aq) + dD(aq)$$
The equilibrium constant expression is written as:
$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
Let's break that down:
- The square brackets [ ] mean "concentration of" in mol dm⁻³.
- The concentrations of the products always go on top (numerator).
- The concentrations of the reactants always go on the bottom (denominator).
- The stoichiometric coefficients (the numbers a, b, c, d from the balanced equation) become the powers for each concentration.
VERY IMPORTANT RULE: We DO NOT include pure solids (s) or pure liquids (l) in the Kc expression. This is because their "concentration" (density) is essentially constant and doesn't change during the reaction.
Example: For the reaction $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$, the expression is: $$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$
What Does the Value of Kc Tell Us?
- If Kc is large (e.g., > 1000): The top part of the fraction is much bigger than the bottom. This means at equilibrium, there are many more products than reactants. We say the equilibrium lies to the right.
- If Kc is small (e.g., < 0.001): The bottom part of the fraction is much bigger. This means there are many more reactants than products. We say the equilibrium lies to the left.
- If Kc is around 1: There are significant amounts of both reactants and products at equilibrium.
Remember, the value of Kc for a reaction only changes if the temperature changes.
Calculations Involving Kc
Example: At a certain temperature, the equilibrium concentrations for the reaction $$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$$ were found to be [H₂] = 0.10 mol dm⁻³, [I₂] = 0.20 mol dm⁻³, and [HI] = 1.04 mol dm⁻³. Calculate Kc.
Step 1: Write the Kc expression. $$K_c = \frac{[HI]^2}{[H_2][I_2]}$$
Step 2: Substitute the equilibrium concentrations. $$K_c = \frac{(1.04)^2}{(0.10)(0.20)}$$
Step 3: Calculate the value. $$K_c = \frac{1.0816}{0.02} = 54.08$$
(Note: Kc usually has no units.)
Key Takeaway
Kc is a ratio of [products] to [reactants] at equilibrium. A large Kc means the reaction favours products, while a small Kc means it favours reactants. Its value is constant unless the temperature changes.
Changing the Position of Equilibrium (Le Châtelier's Principle)
Chemical engineers often want to maximize the amount of product. They can do this by "disturbing" the equilibrium. The system's response is described by a very important idea: Le Châtelier's Principle.
Le Châtelier's Principle states: If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to counteract the change.
Think of it like this: The system doesn't like change. Whatever you do to it, it will try to do the opposite to get back to a balanced state.
Let's look at the two factors you need to know from the syllabus: concentration and temperature.
1. The Effect of Changing Concentration
This does NOT change the value of Kc. It only shifts the position of equilibrium.
Consider the reaction: $$Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$$ (Pale yellow) + (Colourless) $$ \rightleftharpoons $$ (Blood-red)
- Change: Add more reactant (e.g., more Fe³⁺).
System's response: Tries to *use up* the extra Fe³⁺.
Result: The equilibrium shifts to the right. The forward reaction is favoured, more blood-red product is formed. - Change: Remove a product (e.g., remove [Fe(SCN)]²⁺).
System's response: Tries to *replace* the removed product.
Result: The equilibrium shifts to the right. The forward reaction is favoured. - Change: Add more product (e.g., more [Fe(SCN)]²⁺).
System's response: Tries to *use up* the extra product.
Result: The equilibrium shifts to the left. The reverse reaction is favoured, making more reactants.
2. The Effect of Changing Temperature
This is the ONLY factor that changes the value of Kc.
To understand this, we need to know if the reaction is exothermic (releases heat, ΔH is negative) or endothermic (absorbs heat, ΔH is positive). Let's treat 'heat' as a chemical!
For an EXOTHERMIC reaction (heat is a product): $$A + B \rightleftharpoons C + \text{heat}$$
- Change: Increase the temperature (add heat).
System's response: Tries to *use up* the extra heat.
Result: Equilibrium shifts to the left (the endothermic direction). The amount of product decreases, so Kc decreases. - Change: Decrease the temperature (remove heat).
System's response: Tries to *produce more* heat.
Result: Equilibrium shifts to the right (the exothermic direction). The amount of product increases, so Kc increases.
For an ENDOTHERMIC reaction (heat is a reactant): $$A + B + \text{heat} \rightleftharpoons C$$
- Change: Increase the temperature (add heat).
System's response: Tries to *use up* the extra heat.
Result: Equilibrium shifts to the right. The amount of product increases, so Kc increases. - Change: Decrease the temperature (remove heat).
System's response: Tries to *produce more* heat.
Result: Equilibrium shifts to the left. The amount of product decreases, so Kc decreases.
Did you know? The role of a catalyst.
What about adding a catalyst? This is a common exam trick question!
A catalyst speeds up both the forward and reverse reactions by the same amount. This means the system reaches equilibrium FASTER, but it DOES NOT change the position of the equilibrium or the value of Kc. It's like building a new, faster road between two cities – it doesn't change the size of the cities, it just lets you travel between them more quickly.
Key Takeaway
Le Châtelier's Principle helps us predict how an equilibrium system will respond to change. The system always tries to oppose the change you make. Remember, only a temperature change can alter the value of Kc.