Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided

Chemical Reactions – Reversible Reactions & Equilibrium ⚗️

In a reversible reaction the forward and reverse processes run at the same time.

When the rates become equal, the system is at equilibrium and the concentrations of all species stop changing.

The position of this equilibrium can be shifted by changing temperature, pressure, concentration or by adding a catalyst.

Let’s explore each factor with simple analogies and examples that fit a 15‑year‑old’s everyday life.

1️⃣ Temperature (ΔT)

Consider a see‑saw balanced at the middle.

If you add heat (like turning up the stove), the system behaves like a hot cup of tea – it wants to release energy.

Use Le Chatelier’s Principle to predict the shift:

• Exothermic forward reaction: ΔH < 0 – adding heat pushes the reverse reaction.
• Endothermic forward reaction: ΔH > 0 – adding heat pushes the forward reaction.

Example: The Haber process:

\$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \quad \Delta H = -92 \text{ kJ mol}^{-1}\$

Heat → reverse → less ammonia.

Cool → forward → more ammonia.

Think of it like a cold bath that makes the ammonia “stay” in the reaction vessel.

2️⃣ Pressure (ΔP)

Imagine a crowded classroom.

If the room gets smaller (pressure ↑), people (molecules) squeeze together.

The system reacts by reducing the number of gas molecules to relieve the crowding.

• More gas molecules on the reactant side → increase pressure → shift to products.
• More gas molecules on the product side → increase pressure → shift to reactants.

Example: Haber process again:

Reactants = 4 g mol⁻¹, Products = 2 g mol⁻¹.

Increasing pressure favours the product side (ammonia).

Think of it like a traffic jam: fewer cars (molecules) on the road (product side) eases the jam (pressure).

3️⃣ Concentration (ΔC)

Picture a crowded dance floor.

Adding more dancers (increasing concentration) forces the system to adjust by creating more space (shifting equilibrium).

• Increase reactant concentration → shift to products.
• Increase product concentration → shift to reactants.
• Removing a product (e.g., by distillation) pulls the reaction forward.

Example: In the reaction \$\ce{2NO2(g) <=> N2O4(g)}\$ adding more \$\ce{NO2}\$ pushes the equilibrium toward \$\ce{N2O4}\$, just like adding more people to a room pushes the crowd toward the exit.

4️⃣ Catalyst (ΔCat)

A catalyst is like a helpful referee that speeds up the game but doesn’t change the final score.

It lowers the activation energy for both forward and reverse reactions equally.

• Equilibrium position (ratio of concentrations) remains unchanged.
• Equilibrium is reached faster.

Example: Using a platinum catalyst in the Haber process accelerates ammonia production but does not alter the amount of ammonia at equilibrium.

Think of it like a turbocharger that lets the car reach its top speed faster without changing the maximum speed itself.

📊 Quick Reference Table

🧪 Mini‑Quiz

1. What happens to the equilibrium of an exothermic reaction if the temperature is increased?
2. Why does increasing pressure favour the side with fewer gas molecules?
3. Will adding a catalyst change the ratio of products to reactants at equilibrium?
4. In the reaction \$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\$, which side has more gas moles?

Answers:

1️⃣ Shift to reverse (heat is a product).

2️⃣ Mole crowding – fewer molecules reduce pressure.

3️⃣ No, only the rate changes.

4️⃣ Reactants side (3 moles) > Products side (2 moles).

Happy studying! Remember, equilibrium is like a tug‑of‑war that balances itself out. Adjust the forces (temperature, pressure, concentration, catalyst) and watch how the balance changes. 🚀