Introduction to Kinetics and Equilibria

Q: Does a catalyst change the Maxwell-Boltzmann distribution curve?

No. A catalyst only lowers the activation energy , shifting the threshold left on the existing distribution curve. It does not alter the shape or area of the Maxwell-Boltzmann curve itself. Temperature changes alter the curve shape.

Edexcel International A-Level Chemistry· 9.1–9.11 (Unit 2 WCH12)· 25 min read

1. Collision Theory and Reaction Rate Calculation★★☆☆☆⏱ 5 min

📘 Definition

Collision Theory

A model explaining reaction rate: reactions occur when reactant particles collide with sufficient energy (≥ activation energy, $E_{a}$ ) and correct orientation to break bonds and form products.

Four key factors affect the frequency of successful collisions and thus reaction rate: concentration/pressure (higher values = more particles per unit volume = more collisions), temperature (higher = particles move faster, more high-energy collisions), surface area (larger = more exposed reactant particles available for collision), and presence of a catalyst.

📐 Worked Example

A student investigates the rate of reaction between magnesium ribbon and dilute hydrochloric acid. They time how long it takes for 10 cm³ of hydrogen gas to be produced, recording a time of 25 s. Calculate the relative rate of reaction, and state the effect on rate if the acid is replaced with double the concentration.

1
1. Use the rate = 1/time formula for relative rate calculation:
2
$r = \frac{1}{t} = \frac{1}{25} = 0.04 \, s^{-1}$
3
1. Doubling the concentration of hydrochloric acid increases the number of H+ ions per unit volume. This increases the frequency of successful collisions between Mg and H+ particles.
4
1. The reaction rate will approximately double.

✓ Quick check

Why does increasing the surface area of a solid reactant increase reaction rate?
- It increases the activation energy of the reaction
- It increases the number of exposed reactant particles available for collision
- It increases the average kinetic energy of reactant particles
- It changes the orientation of collisions
Reveal answer
1 —
Incorrect: Surface area does not affect activation energy or particle kinetic energy. It only exposes more particles to collisions.

2. Maxwell-Boltzmann Distribution and Temperature Effects★★★☆☆⏱ 5 min

📘 Definition

Maxwell-Boltzmann Distribution

A graph showing the distribution of kinetic energies of gas (or solution) particles at a fixed temperature. The area under the curve equals the total number of particles, which remains constant for a closed system.

Key features of the curve: the peak represents the most probable kinetic energy, the mean energy is to the right of the peak, and only the fraction of particles with energy ≥ $E_{a}$ can react successfully. When temperature increases: the curve shifts right, the peak becomes lower and broader, the area under the curve stays the same, and the fraction of particles with energy ≥ $E_{a}$ increases significantly, leading to a faster reaction rate.

📐 Worked Example

Sketch and label a comparison of Maxwell-Boltzmann curves for a reaction at 298 K and 310 K, marking the activation energy $E_{a}$ and explaining the effect of the temperature increase on reaction rate.

1
1. Draw the 298 K curve: higher, narrower peak at lower kinetic energy.
2
1. Draw the 310 K curve: lower, broader peak shifted to higher kinetic energy, both curves start at the origin and asymptote to the x-axis at high energy.
3
1. Mark $E_{a}$ as a vertical line to the right of both peaks.
4
1. The area under the curve to the right of $E_{a}$ is much larger for 310 K than 298 K: a greater fraction of particles have sufficient energy for successful collisions, so rate increases.

Exam tip:

Always state that the area under the M-B curve is constant (equal to total number of particles) in your exam answers, this is a common mark point.

3. Catalysts and Reaction Profiles★★★☆☆⏱ 5 min

📘 Definition

Catalyst

A substance that increases the rate of a chemical reaction without being used up in the reaction. It works by providing an alternative reaction route with a lower activation energy $E_{a}$ .

Reaction profiles (energy level diagrams) show the change in energy of reactants as they form products. For an uncatalysed reaction, the peak energy equals the activation energy for the original route. For a catalysed reaction, the peak is lower, and may show an intermediate energy level if the catalyst forms a temporary intermediate with reactants. Catalysts do not affect the enthalpy change of the reaction, or the position of equilibrium: they only increase the rate at which equilibrium is reached.

📐 Worked Example

Draw a reaction profile for an exothermic reaction, showing both catalysed and uncatalysed routes, and label the activation energy for both.

1
1. Draw a horizontal line for reactant energy, a lower horizontal line for product energy (since reaction is exothermic, ΔH is negative).
2
1. Draw a high curved peak from reactants to products for the uncatalysed route: label the vertical difference between reactant energy and the peak as $E_{a}$ (uncatalysed).
3
1. Draw a lower curved peak (or two smaller peaks showing an intermediate) for the catalysed route: label the vertical difference between reactant energy and the lower peak as $E_{a}$ (catalysed).
4
1. Label the enthalpy change ΔH as the vertical difference between reactant and product energy.

4. Dynamic Equilibrium and Le Chatelier's Principle★★★☆☆⏱ 6 min

📘 Definition

Dynamic Equilibrium

A state in a closed reversible reaction system where the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products remain constant (they are not equal, just unchanging).

Le Chatelier's Principle states that if a change is applied to a system at dynamic equilibrium, the position of equilibrium will shift to oppose the change. You only need to apply this qualitatively for homogeneous systems (all reactants and products in the same state) for this topic: 1. Concentration: increase reactant concentration → shift right to use up extra reactants; increase product concentration → shift left to use up extra products. 2. Pressure (for gaseous systems only): increase pressure → shift to the side with fewer moles of gas to reduce pressure; decrease pressure → shift to the side with more moles of gas. 3. Temperature: increase temperature → shift in the endothermic direction to absorb extra heat; decrease temperature → shift in the exothermic direction to release heat.

📐 Worked Example

Consider the homogeneous gaseous equilibrium: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) Δ H = - 92 k J m o l^{- 1}$ . Predict and justify the effect of increasing pressure on the position of equilibrium.

1
1. Count the moles of gas on each side: left side = 1 + 3 = 4 moles of gas, right side = 2 moles of gas.
2
1. Increasing pressure shifts equilibrium to the side with fewer moles of gas to oppose the pressure increase.
3
1. Equilibrium shifts to the right, increasing the yield of ammonia.

5. Industrial Compromise Conditions★★☆☆☆⏱ 4 min

Industrial chemical processes balance three key factors: maximum yield, fast reaction rate, and low operational cost. The optimal conditions are often a compromise between these factors, as conditions that increase rate may reduce yield, or vice versa. For example, the Haber process (ammonia production) uses moderate temperature, high pressure, and an iron catalyst to balance yield, rate and cost.

📐 Worked Example

For the Haber process equilibrium (exothermic forward reaction), explain why a moderate temperature of around 450°C is used instead of a much lower temperature, even though lower temperature gives a higher equilibrium yield of ammonia.

1
1. A lower temperature would shift equilibrium right to give higher ammonia yield, but reaction rate would be very slow, as fewer particles have energy ≥ $E_{a}$ .
2
1. A very high temperature would give a fast rate, but shift equilibrium left, reducing ammonia yield significantly.
3
1. The moderate 450°C temperature is a compromise: it gives a fast enough reaction rate while still producing an acceptable yield of ammonia, especially when combined with an iron catalyst to further increase rate.

6. Common Pitfalls

Wrong move:

Stating that a catalyst shifts the position of equilibrium or increases product yield.

Why:

Catalysts only increase the rate of forward and reverse reactions equally, so they do not affect equilibrium position or yield, only the time taken to reach equilibrium.

Correct move:

State that catalysts reduce activation energy, increasing reaction rate, but have no effect on equilibrium yield or position.

Wrong move:

Claiming that increasing temperature increases the total number of particles with energy ≥ $E_{a}$ by increasing the total number of particles.

Why:

The total number of particles in a closed system is constant. Higher temperature increases the fraction of particles with energy ≥ $E_{a}$ , not the total number of particles.

Correct move:

Explain that higher temperature shifts the Maxwell-Boltzmann distribution right, increasing the fraction of particles with energy ≥ $E_{a}$ , leading to more successful collisions.

Wrong move:

Stating that at dynamic equilibrium, the concentrations of reactants and products are equal.

Why:

At dynamic equilibrium, concentrations are constant (unchanging), not equal. The ratio of concentrations depends on the reaction and conditions.

Correct move:

Define dynamic equilibrium as equal forward and reverse reaction rates, leading to constant reactant and product concentrations.

Wrong move:

Applying pressure change effects to equilibrium systems containing only solids or liquids.

Why:

Pressure changes only affect the volume (and thus concentration) of gaseous particles; solids and liquids have negligible volume changes with pressure.

Correct move:

Only apply pressure shift rules to homogeneous gaseous equilibrium systems, counting moles of gas only on each side of the equation.

Wrong move:

Calculating reaction rate from time as rate = time, instead of rate = 1/time for relative rate.

Why:

Faster reactions take less time, so rate is inversely proportional to time taken for a fixed amount of product to form.

Correct move:

Calculate relative rate as the inverse of time taken, with units of $s^{- 1}$ if time is measured in seconds.

7. Quick Reference Cheatsheet

Concept	Key Exam Fact
Collision Theory	Successful collisions require ≥ Eₐ and correct orientation
Rate calculation	Relative rate = 1/time; rate = gradient of tangent to conc-time graph
Maxwell-Boltzmann (higher T)	Curve shifts right, lower peak, higher fraction of particles ≥ Eₐ
Catalyst effect	Lowers Eₐ, no change to M-B curve, no effect on equilibrium yield
Dynamic Equilibrium	Forward rate = reverse rate, constant (not equal) reactant/product concentrations
Le Chatelier: Pressure increase	Shift to side with fewer moles of gas (gaseous systems only)
Le Chatelier: Temperature increase	Shift in endothermic direction to absorb heat
Industrial compromise	Balance yield, rate, and operational cost for optimal conditions

8. Frequently Asked

Do I need to calculate rate constants for this Edexcel IAL U2 topic?

No. Rate equations, rate constants, reaction orders and half-life are out of scope for this Unit 2 topic, they are covered in Unit 4 Topic 11 only. You only need to calculate rate using $1/ t im e$ or tangent gradient of a concentration-time graph.

Does a catalyst change the Maxwell-Boltzmann distribution curve?

No. A catalyst only lowers the activation energy $E_{a}$ , shifting the $E_{a}$ threshold left on the existing distribution curve. It does not alter the shape or area of the Maxwell-Boltzmann curve itself. Temperature changes alter the curve shape.

What counts as a valid Le Chatelier's principle explanation for exam marks?

You must state three parts for full marks: 1. The change applied to the equilibrium system, 2. The direction the equilibrium shifts to oppose the change, 3. The observable effect (e.g. higher product yield, darker solution colour).

Going deeper

syllabusEdexcel IAL Chemistry Unit 2 SpecificationOfficial specification for Topic 9 (9.1-9.11)

What's Next

Now that you have mastered the qualitative fundamentals of kinetics and equilibria for Edexcel IAL Chemistry Unit 2, you are ready to move on to more advanced chemistry topics in the unit, including group chemistry and halogenoalkane reactions. This content also forms the foundation for quantitative kinetics and equilibrium calculations you will encounter in Unit 4 of the IAL Chemistry course, where you will learn to use rate equations, rate constants, and equilibrium constants to solve numerical problems. Ensure you practice past paper questions on this qualitative content to familiarize yourself with common exam phrasing and mark scheme requirements.