Describe and explain the effect on the rate of reaction of: (a) changing the concentration of solutions (b) changing the pressure of gases (c) changing the surface area of solids (d) changing the temperature (e) adding or removing a catalyst, includi

Chemical reactions – Rate of reaction

(a) Changing the concentration of solutions

Collision theory says that the more molecules you have, the more collisions happen per second.

If the concentration of a reactant increases from \$[A]\$ to \$2[A]\$, the number of collisions roughly doubles, so the rate increases by a factor of 2 (for a first‑order reaction).



Analogy: Think of people in a crowded room. The more people (molecules) there are, the more chances they bump into each other.



Exam tip: Remember that the rate law is \$rate = k[A]^m[B]^n\$. Changing concentration changes the rate by the power of that concentration in the law.

(b) Changing the pressure of gases

🔥 Raising the pressure of a gas compresses the molecules into a smaller volume, increasing the number of collisions per unit volume.

For a reaction involving gases, the rate often increases proportionally to the pressure (for a first‑order gas reaction).



Analogy: Imagine a crowded subway car. The more people (higher pressure), the faster they bump into each other.



Exam tip: Use the ideal gas law \$PV = nRT\$ to relate pressure changes to concentration changes for gases.

(c) Changing the surface area of solids

🧩 When a solid reactant is broken into smaller pieces, its surface area increases, giving more sites for collisions.

The rate increases because more molecules can collide with the solid at once.



Analogy: Think of a pizza. A whole pizza (small surface area) is slower to eat than many thin slices (large surface area).



Exam tip: For heterogeneous reactions, rate ∝ surface area. Remember to mention “surface area” in your answer.

(d) Changing the temperature

🌡️ Increasing temperature gives molecules more kinetic energy, so they collide faster and with more energy.

The Arrhenius equation \$k = A e^{-E_a/RT}\$ shows that the rate constant \$k\$ rises exponentially with temperature.



Analogy: Like a hot summer day, people move faster and bump into each other more often.



Exam tip: Write that the rate increases with temperature and explain it using the Arrhenius equation or collision theory.

(e) Adding or removing a catalyst (including enzymes)

🏃‍♂️ A catalyst provides an alternative reaction pathway with a lower activation energy \$E_a\$.

It does not change the equilibrium but speeds up the approach to equilibrium.

Enzymes are biological catalysts that bind reactants (substrates) in a “lock‑and‑key” pocket, further lowering \$E_a\$.



Analogy: Think of a traffic cop (catalyst) who directs cars (molecules) through a shortcut, reducing the time to reach the destination.



Exam tip: Mention that catalysts lower \$Ea\$, increase \$k\$ (but not \$K{eq}\$), and that enzymes are specific and can be inhibited.

Collision theory – Quick recap

  1. Reactant molecules must collide.
  2. Collisions must have enough energy (≥ \$E_a\$).
  3. Collisions must have the correct orientation.

Summary table of factors affecting rate

FactorEffect on RateKey Point
ConcentrationIncreases (proportional to concentration power in rate law)\$rate = k[A]^m[B]^n\$
Pressure (gases)Increases (∝ pressure for first‑order)Use \$PV = nRT\$ to relate pressure to concentration
Surface area (solids)Increases with more surface areaRate ∝ surface area for heterogeneous reactions
TemperatureExponential increase (Arrhenius)\$k = A e^{-E_a/RT}\$
Catalyst / EnzymeIncreases rate by lowering \$E_a\$Does not change equilibrium constant