Collision theory and the four factors
Collision theory: particles must collide with energy at least the activation energy. Rate depends on how often particles collide and the fraction of collisions reaching that energy. Four factors raise it: higher concentration (more particles per volume); higher temperature (faster particles, more frequent and more energetic collisions); larger surface area (more exposed particles); a catalyst (a lower-activation-energy pathway).
Rate graphs — the gradient is the rate
Rate graphs plot mass of reactant (or volume of gas) against time. The gradient equals the rate: steepest at the start (concentration highest), falling to zero when a reactant runs out and the line goes flat. A faster-only change — higher concentration, temperature, surface area or a catalyst — starts the new curve steeper and levels it off sooner, but at the SAME final level (same product). Only changing the limiting reactant amount moves that level.
Catalysts and alternative pathways
A catalyst increases the rate and is chemically unchanged at the end — both parts are needed for the definition. It gives an alternative pathway of lower activation energy, so more collisions succeed. (Extended) A homogeneous catalyst is in the same phase as the reactants (e.g. an acid in solution); a heterogeneous catalyst is in a different phase (e.g. solid iron in the Haber process, or manganese(IV) oxide with hydrogen peroxide).
Drawn from real examiner reports.
Rate and time taken are reciprocals
Rate and time taken move in OPPOSITE directions — reciprocals, not synonyms. Increasing concentration, temperature or surface area raises the rate but DECREASES the time to finish; if the rate doubles, the time halves. Read the command word: if it asks about time taken, answer the time is shorter / decreases, not rate increases.
W22 P31 Q6(b); W22 P32 Q6(e); S23 P32 Q6(a) — candidates answered in terms of rate when asked about the time taken (or vice versa).
Say frequency of collisions, not more
Extended explanations give two marks: increased FREQUENCY of collisions, and a greater PROPORTION reaching the activation energy. Writing there are more collisions is too vague — the marker wants more frequent collisions. The activation-energy mark is almost never stated. For temperature give both; concentration and surface area need frequency only.
W22 P41 Q3(g); W22 P42 Q3(v); S23 P41 Q4(c) — more collisions credited weaker than increased frequency; the activation-energy proportion almost never stated.
Catalyst must be unchanged at the end
The catalyst definition needs BOTH parts: it increases the rate AND is chemically unchanged at the end. Answering only speeds up the reaction scores one mark; adding that it lowers the activation energy also earns credit. Do not say it lowers the temperature, or write it into the overall equation — it is neither reactant nor product.
S23 P41 Q4(a) — candidates stated the catalyst increases rate but omitted that it is chemically unchanged at the end, losing the definition mark.
Rate depends on concentration, not amount
Initial rate depends on CONCENTRATION (particles per unit volume), not on the mass, volume or moles of reactant. A reaction is fastest at the start because concentration is highest then, slowing as reactants are used up. Doubling an acid's volume at the SAME concentration leaves the initial rate unchanged — it only makes more product.
S23 P41 Q4(b)(ii) — rate given as depending on mass, volume or amount of reactant rather than on concentration.
Rate graphs: a curve, not a line
Sketch a rate graph as a CURVE that starts steep and gradually flattens — not a straight line and not an S-shape. A second curve for a faster reaction should deviate from the origin, not follow the original too long. A faster-only change reaches the SAME final level (same product): do not draw the plateau higher or lower unless the amount of limiting reactant changes.
S22 P31 Q8(b) — straight lines instead of curves, curves that follow the original too long, or the new curve drawn to a wrong final level.
A catalyst lowers Ea, not temperature
A catalyst does not heat the mixture or change its temperature. It gives an alternative pathway with a LOWER activation energy, so more of the existing collisions now succeed. Saying a catalyst raises or lowers the temperature, or gives particles more energy, is wrong: the particles move as fast as before; only the energy barrier is lower.
Explain: the three-link chain
Extended Explain how a factor increases rate needs three links: its effect on the particles; collisions become more FREQUENT; a greater PROPORTION reach the activation energy. Temperature needs all three; concentration and surface area need only the frequency link.
Answer rate or time — check which
Underline whether the question asks about RATE or TIME TAKEN. If time, write the time decreases / is shorter — never rate increases. If rate, write rate increases / decreases. They are opposite quantities, and the mark scheme credits only the one asked for.
Drawing a faster-reaction curve
For a faster change (higher concentration, temperature, surface area or catalyst): start the curve steeper and level off sooner, at the SAME final height. Label the axes — mass (g) or gas volume (cm3) against time. Move that height only if the limiting reactant amount changes.
Rate of reaction: How quickly reactants are converted into products — measured as the change in quantity of reactant or product per unit time.
Activation energy (): The minimum energy required for a collision between reactant particles to result in a reaction.
Catalyst (mark-scheme definition): A substance that increases the rate of a chemical reaction and is chemically unchanged at the end of the reaction.
Both elements are required for full marks — "speeds up the reaction" alone scores only one mark.
Collision theory: Reactions occur when reactant particles collide with energy the activation energy. The rate depends on:
A student claims that doubling the volume of hydrochloric acid used (at the same concentration) will double the initial rate of its reaction with magnesium. Evaluate this claim.
Hydrogen peroxide decomposes in the presence of manganese(IV) oxide catalyst.
(a) Write the balanced symbol equation for this reaction, including state symbols.
(b) State two observations you would make during this reaction.
