Rates of Reaction MCQs – 40 Practice Questions with Answers and Explanations

How fast a reaction goes is one of the most heavily tested topics in chemistry exams. This page gives you 40 original practice questions on rates of reaction – from measuring rates and collision theory at GCSE and IGCSE level, through rate graphs, to rate equations, orders of reaction and the Arrhenius equation for A-Level and AP students. Every question has a worked answer, so click “Show Answer” only after you have tried it yourself.

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Measuring Rates & Collision Theory

Q1. Magnesium reacts with dilute hydrochloric acid and 60 cm3 of hydrogen gas is collected in 30 s. What is the mean rate of reaction over this time?

  • (A) 0.50 cm3/s
  • (B) 2.0 cm3/s
  • (C) 30 cm3/s
  • (D) 1800 cm3/s
Show Answer

Answer: B – Mean rate = quantity of product formed / time taken = 60 / 30 = 2.0 cm3/s. Option A comes from dividing the wrong way round (30 / 60).

Q2. A student follows a reaction by recording the loss in mass of the flask, in grams, against time in seconds. What is the correct unit for the rate?

  • (A) g/s
  • (B) g s
  • (C) s/g
  • (D) g/cm3
Show Answer

Answer: A – Rate is always “amount per unit time”, so mass loss in grams divided by time in seconds gives g/s. Option D is a unit of density, not rate.

Q3. Which method is the most suitable way to follow the rate of a reaction that produces a gas?

  • (A) Timing how long the mixture takes to change colour
  • (B) Measuring the temperature rise of the mixture
  • (C) Collecting the gas in a gas syringe and recording the volume at regular time intervals
  • (D) Weighing the solid reactant before the reaction and again at the end
Show Answer

Answer: C – A gas syringe gives a volume reading at any moment, so the rate can be followed continuously. Weighing before and after (D) only gives the total change, not how fast it happened.

Q4. According to collision theory, why does increasing the temperature make a reaction faster?

  • (A) It lowers the activation energy of the reaction
  • (B) It only makes the particles collide more frequently
  • (C) It only makes the particles move with more energy
  • (D) Particles collide more frequently AND a much larger fraction of the collisions have energy greater than or equal to the activation energy
Show Answer

Answer: D – Both effects matter, but the big one is the second: the proportion of collisions with energy ≥ Ea rises sharply with temperature. Temperature does not change the activation energy itself.

Q5. The same mass of calcium carbonate is reacted with excess dilute hydrochloric acid. The powdered solid reacts much faster than large lumps because the powder…

  • (A) has a larger surface area, so more particles are exposed and collisions happen more frequently
  • (B) has a lower activation energy than the lumps
  • (C) contains more moles of calcium carbonate
  • (D) acts as a catalyst for its own reaction
Show Answer

Answer: A – Only surface particles can collide with the acid. Grinding exposes far more particles, so successful collisions are more frequent. The chemistry, mass and activation energy are unchanged.

Q6. What does a catalyst do to a reaction?

  • (A) It is used up and increases the yield of a reversible reaction
  • (B) It increases the kinetic energy of all the particles
  • (C) It provides an alternative reaction pathway with a lower activation energy, and it is chemically unchanged at the end
  • (D) It makes every collision a successful collision
Show Answer

Answer: C – A catalyst lowers Ea by offering a different route, so more collisions succeed. It is not consumed and it does not change the position of equilibrium or the yield.

Q7. The activation energy of a reaction is…

  • (A) the average kinetic energy of the particles
  • (B) the minimum energy that colliding particles must have for a reaction to occur
  • (C) the energy released when products form
  • (D) the difference in energy between the products and the reactants
Show Answer

Answer: B – Collisions with less energy than Ea simply bounce off unchanged. Options C and D describe the overall energy change of the reaction, not the barrier that must be climbed.

Q8. A flask on a balance loses 0.90 g of mass in 150 s as carbon dioxide escapes. What is the mean rate of mass loss?

  • (A) 6.0 g/s
  • (B) 0.60 g/s
  • (C) 167 g/s
  • (D) 0.0060 g/s
Show Answer

Answer: D – Mean rate = 0.90 / 150 = 0.0060 g/s. Option C is the inverted calculation (150 / 0.90), a very common slip.

Q9. Increasing the concentration of a reactant solution increases the rate because…

  • (A) there are more particles in the same volume, so collisions are more frequent
  • (B) the particles move faster at higher concentration
  • (C) the activation energy decreases
  • (D) each particle becomes larger
Show Answer

Answer: A – Concentration changes the number of particles per unit volume, not their speed or size. More particles in the same space means more collisions per second.

Q10. In the “disappearing cross” experiment, sodium thiosulfate solution reacts with hydrochloric acid and the time for a cross to vanish is recorded. Why does this time measure the rate?

  • (A) The acid is used up the moment the cross disappears
  • (B) The cross reacts with the sulfur that forms
  • (C) A precipitate of sulfur slowly clouds the solution; the faster the reaction, the sooner the cross is hidden, so rate is proportional to 1 / time
  • (D) The solution warms up and evaporates until the cross dries out
Show Answer

Answer: C – Sulfur forms as a fine precipitate that makes the mixture opaque. A shorter time to reach the same cloudiness means a faster reaction, which is why 1/t is used as a measure of rate.

Rate Graphs & Experimental Methods

Q11. On a graph of volume of gas produced against time, the rate of reaction at a particular moment is given by…

  • (A) the volume of gas at that moment
  • (B) the gradient of the tangent to the curve at that moment
  • (C) the total volume divided by the total reaction time, at every point
  • (D) the area under the curve
Show Answer

Answer: B – Rate = change in volume / change in time at that instant, which is the gradient of the tangent. Total volume / total time only gives the overall mean rate for the whole reaction.

Q12. A tangent drawn to a volume-time curve at t = 0 passes through (0, 0) and reaches 48 cm3 at 12 s. What is the initial rate?

  • (A) 0.25 cm3/s
  • (B) 2.0 cm3/s
  • (C) 4.0 cm3/s
  • (D) 576 cm3/s
Show Answer

Answer: C – Initial rate = gradient of the tangent at t = 0 = 48 / 12 = 4.0 cm3/s. Option A is the upside-down gradient.

Q13. Two experiments use identical amounts of reactants. Curve X rises steeply and levels off; curve Y rises gently but levels off at exactly the same volume. What can you conclude?

  • (A) Experiment X produced more product
  • (B) Experiment X started with more reactant
  • (C) Experiment Y was the faster reaction
  • (D) Experiment X was faster, but both produced the same total amount of product
Show Answer

Answer: D – Steepness shows rate; the plateau height shows total product. Same plateau = same amounts of reactant; steeper curve = faster reaction (for example warmer, more concentrated or catalysed).

Q14. A volume-time graph eventually becomes flat (horizontal). Why?

  • (A) The reaction has become reversible
  • (B) The limiting reactant has been completely used up, so the reaction has finished
  • (C) The catalyst has stopped working
  • (D) The gas syringe has cooled down
Show Answer

Answer: B – Once one reactant runs out, no more product can form and the volume stops changing. The reactant that runs out first is the limiting reactant.

Q15. Magnesium ribbon reacts with excess dilute acid. In a second run, the acid concentration is doubled, with the magnesium (the limiting reactant) kept exactly the same. How does the second curve differ?

  • (A) It is steeper at the start but levels off at the same volume
  • (B) It has the same gradient but levels off at double the volume
  • (C) It is less steep and levels off at the same volume
  • (D) It is steeper and levels off at double the volume
Show Answer

Answer: A – Doubling the acid concentration doubles the collision frequency, so gas is produced faster. But the same mass of magnesium makes the same total volume of hydrogen, so the plateau is unchanged.

Q16. A student investigates the effect of concentration on the thiosulfate “disappearing cross” reaction. Which plan is correct?

  • (A) Change the temperature each run and measure the mass of sulfur formed
  • (B) Change the volume of acid each run and record the final colour
  • (C) Change the concentration of the thiosulfate, measure the time for the cross to disappear, and keep temperature, volumes and acid concentration the same each run
  • (D) Use a bigger conical flask for each new concentration
Show Answer

Answer: C – Only the variable being tested (concentration) may change; everything else must be controlled. The time for the cross to disappear is the measured outcome.

Q17. The same reaction is run with and without a catalyst, using identical reactants. How does the catalysed curve compare on a volume-time graph?

  • (A) It levels off at a lower volume, reached sooner
  • (B) It is identical but takes longer to level off
  • (C) It levels off at a higher volume
  • (D) It is steeper and reaches the same final volume in a shorter time
Show Answer

Answer: D – A catalyst changes speed only. The same reactants give the same total amount of product, so the plateau is the same height – it is just reached sooner.

Q18. Equal masses of powdered and lump calcium carbonate are each reacted with excess acid. Comparing the two volume-time curves:

  • (A) The powder curve levels off at a higher volume
  • (B) Both curves level off at the same volume, but the powder curve gets there sooner
  • (C) The lumps curve reaches its plateau sooner
  • (D) The powder curve levels off at a lower volume
Show Answer

Answer: B – Equal masses of CaCO3 contain equal moles, so equal volumes of CO2 form in total. The powder’s larger surface area only changes how quickly that volume is reached.

Q19. The gradient of a volume-time curve gets smaller as the reaction proceeds. The main reason is that…

  • (A) the concentration of the reactants falls, so collisions become less frequent
  • (B) the temperature of every reaction always falls as it runs
  • (C) the products begin to catalyse the reverse reaction
  • (D) the particles permanently lose their activation energy
Show Answer

Answer: A – Reactants are being used up, so there are fewer particles to collide and the rate drops. This is why most reactions are fastest at the start.

Q20. From a volume-time graph, the volume of gas is 30 cm3 at 20 s and 45 cm3 at 50 s. What is the mean rate between 20 s and 50 s?

  • (A) 1.5 cm3/s
  • (B) 0.90 cm3/s
  • (C) 0.50 cm3/s
  • (D) 2.0 cm3/s
Show Answer

Answer: C – Mean rate = (45 − 30) / (50 − 20) = 15 / 30 = 0.50 cm3/s. Remember to subtract both the volumes and the times.


Rate Equations & Orders of Reaction

Q21. A reaction has the rate equation rate = k[A][B]2. What is the overall order of the reaction?

  • (A) 1
  • (B) 2
  • (C) 3
  • (D) 4
Show Answer

Answer: C – Add the powers: first order in A (power 1) plus second order in B (power 2) gives an overall order of 3.

Q22. Initial-rate experiments at constant [B] give: [A] = 0.10 mol/dm3, rate = 0.020 mol/dm3/s; [A] = 0.20 mol/dm3, rate = 0.040 mol/dm3/s. The order of reaction with respect to A is…

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) 3
Show Answer

Answer: B – [A] doubles (×2) and the rate also doubles (×2), so rate is proportional to [A]: first order in A.

Q23. In another set of experiments at constant [A], doubling [B] from 0.10 to 0.20 mol/dm3 increases the rate from 0.030 to 0.120 mol/dm3/s. The order with respect to B is…

  • (A) 0
  • (B) 1
  • (C) 3
  • (D) 2
Show Answer

Answer: D – [B] doubles but the rate goes up ×4 (0.030 → 0.120). Since 22 = 4, the reaction is second order in B.

Q24. What are the units of the rate constant k for a first-order reaction?

  • (A) s-1
  • (B) mol/dm3/s
  • (C) dm3/mol/s
  • (D) mol/dm3
Show Answer

Answer: A – For first order, rate = k[A], so k = rate / concentration = (mol/dm3/s) / (mol/dm3) = s-1. Option B is the unit of rate itself (and of k for zero order).

Q25. What are the units of k for a reaction that is second order overall?

  • (A) s-1
  • (B) mol/dm3/s
  • (C) dm3/mol/s
  • (D) dm6/mol2/s
Show Answer

Answer: C – k = rate / [A]2 = (mol/dm3/s) / (mol/dm3)2 = dm3/mol/s. Option D is the unit for a third-order rate constant.

Q26. Which concentration-time graph is expected for a reaction that is zero order in the reactant being plotted?

  • (A) A curve that gets steadily shallower and never quite reaches zero
  • (B) A straight line with a constant negative gradient
  • (C) A horizontal straight line
  • (D) A curve that gets steeper with time
Show Answer

Answer: B – Zero order means the rate is constant and independent of concentration, so concentration falls by the same amount every second: a straight line with gradient = −k. A horizontal line (C) would mean no reaction at all.

Q27. The half-life of a reactant is measured as 40 s starting from 0.80 mol/dm3, and again as 40 s starting from 0.40 mol/dm3. The reaction is…

  • (A) zero order
  • (B) second order
  • (C) third order
  • (D) first order
Show Answer

Answer: D – A half-life that stays constant whatever the starting concentration is the fingerprint of a first-order reaction. For zero order the half-life shrinks as concentration falls; for second order it grows.

Q28. A first-order reaction has k = 0.0347 s-1. What is its half-life? (t1/2 = 0.693 / k)

  • (A) 20 s
  • (B) 0.050 s
  • (C) 35 s
  • (D) 200 s
Show Answer

Answer: A – t1/2 = 0.693 / 0.0347 = 20 s (to 2 significant figures). Option B comes from inverting the calculation (k / 0.693 roughly) – always check your answer is sensible: a bigger k means a shorter half-life.

Q29. A proposed mechanism is: Step 1 (slow): A + B → intermediate. Step 2 (fast): intermediate + C → products. What is the rate equation?

  • (A) rate = k[A][B][C]
  • (B) rate = k[C]
  • (C) rate = k[A][B]
  • (D) rate = k[intermediate][C]
Show Answer

Answer: C – The rate is set by the slow, rate-determining step, so only the species in Step 1 (A and B) appear. C takes part after the slow step, so it does not appear in the rate equation.

Q30. Doubling [A] makes the rate four times faster, but doubling [B] has no effect on the rate. The rate equation is…

  • (A) rate = k[A][B]
  • (B) rate = k[A]2
  • (C) rate = k[B]2
  • (D) rate = k[A]4
Show Answer

Answer: B – ×2 in [A] giving ×4 in rate means second order in A; no change when [B] doubles means zero order in B, so B does not appear in the equation.

The Arrhenius Equation (A-Level / AP stretch)

Q31. In the Arrhenius equation k = A e−Ea/RT, the factor A is…

  • (A) the activation energy
  • (B) the gas constant
  • (C) the value of k at 0 K
  • (D) the pre-exponential (frequency) factor, linked to how often collisions occur with the correct orientation
Show Answer

Answer: D – A measures the frequency of properly oriented collisions and has the same units as k. Ea is the activation energy (A), R is the gas constant (B), and at 0 K k would be zero, not A.

Q32. For a reaction with Ea = 50 kJ/mol, what is the gradient of a graph of ln k against 1/T? (R = 8.31 J/K/mol; gradient = −Ea/R)

  • (A) −6.0 K
  • (B) −6.0 × 103 J
  • (C) +6.0 × 103 K
  • (D) −6.0 × 103 K
Show Answer

Answer: D – Gradient = −50 000 / 8.31 = −6.0 × 103 K. Note the two traps: Ea must be in joules (50 000, not 50), and the gradient of an ln k vs 1/T plot is always negative.

Q33. A reaction with Ea = 50 kJ/mol has k = 2.0 × 10-3 s-1 at 300 K. Estimate k at 310 K. (ln(k2/k1) = (Ea/R)(1/T1 − 1/T2), R = 8.31 J/K/mol)

  • (A) 2.1 × 10-3 s-1
  • (B) 3.8 × 10-3 s-1
  • (C) 2.0 × 10-2 s-1
  • (D) 4.0 × 10-4 s-1
Show Answer

Answer: B – (50 000 / 8.31) × (1/300 − 1/310) = 6017 × 1.075 × 10-4 = 0.647, so k2/k1 = e0.647 = 1.91 and k2 = 1.91 × 2.0 × 10-3 = 3.8 × 10-3 s-1. This is why a 10 K rise roughly doubles many reaction rates. Always use kelvin, never °C.

Q34. A graph of ln k against 1/T has a gradient of −7200 K. What is the activation energy? (R = 8.31 J/K/mol)

  • (A) 59.8 kJ/mol
  • (B) 8.66 kJ/mol
  • (C) 0.87 kJ/mol
  • (D) 866 kJ/mol
Show Answer

Answer: A – Ea = −gradient × R = 7200 × 8.31 = 59 832 J/mol = 59.8 kJ/mol. Option B comes from dividing 7200 by 831 instead of multiplying.

Q35. A rise of only 10 K can roughly double the rate of a reaction. The main reason is that…

  • (A) the frequency factor A roughly doubles
  • (B) every molecule moves about twice as fast
  • (C) the fraction of molecules with energy greater than or equal to Ea increases greatly, because rate depends exponentially on temperature
  • (D) the activation energy is roughly halved
Show Answer

Answer: C – Average molecular speed rises only slightly for 10 K (it depends on the square root of T in kelvin). The exponential term e−Ea/RT is what changes dramatically: many more molecules clear the Ea barrier.

Q36. On a Maxwell-Boltzmann distribution, what happens to the total area under the curve when the temperature is increased?

  • (A) It doubles
  • (B) It stays the same
  • (C) It halves
  • (D) It becomes zero at high energies
Show Answer

Answer: B – The area represents the total number of molecules, which does not change. The curve becomes flatter and shifts to higher energies, but the area – and the molecule count – is constant.

Q37. In Arrhenius terms, a catalyst increases the rate of a reaction mainly because it…

  • (A) decreases the frequency factor A
  • (B) increases the activation energy
  • (C) increases the value of the gas constant R
  • (D) lowers the activation energy, so the exponential term becomes less negative and k increases
Show Answer

Answer: D – The catalysed pathway has a smaller Ea. A smaller Ea means e−Ea/RT is larger, so k (and the rate) increases – even at the same temperature.

Q38. Calculate k at 500 K for a reaction with A = 4.0 × 1013 s-1 and Ea = 100 kJ/mol. (R = 8.31 J/K/mol)

  • (A) 1.4 × 103 s-1
  • (B) 3.6 × 10-11 s-1
  • (C) 1.4 × 10-3 s-1
  • (D) 4.0 × 1013 s-1
Show Answer

Answer: A – Ea/RT = 100 000 / (8.31 × 500) = 24.07, so e−24.07 = 3.6 × 10-11 and k = 4.0 × 1013 × 3.6 × 10-11 = 1.4 × 103 s-1. Option B is the exponential term alone – forgetting to multiply by A is the classic error.

Q39. Two points on an ln k against 1/T graph are (1/T = 3.0 × 10-3 K-1, ln k = −4.0) and (1/T = 3.5 × 10-3 K-1, ln k = −7.0). What is Ea? (R = 8.31 J/K/mol)

  • (A) 5.0 kJ/mol
  • (B) 25 kJ/mol
  • (C) 49.9 kJ/mol
  • (D) 6000 kJ/mol
Show Answer

Answer: C – Gradient = (−7.0 − (−4.0)) / (3.5 × 10-3 − 3.0 × 10-3) = −3.0 / 5.0 × 10-4 = −6000 K. Then Ea = 6000 × 8.31 = 49 860 J/mol = 49.9 kJ/mol. Option D forgets to convert joules to kilojoules – and to multiply by R.

Q40. Which statement about a Maxwell-Boltzmann distribution curve at a higher temperature is correct?

  • (A) The peak shifts to lower energy and becomes taller
  • (B) The total area under the curve increases
  • (C) The most probable energy of the molecules decreases
  • (D) The peak shifts to higher energy and becomes lower, the total area stays the same, and a larger fraction of molecules has energy above Ea
Show Answer

Answer: D – At higher T the distribution spreads out: the peak moves right (higher most probable energy) and drops so the area is conserved. Crucially, the tail beyond Ea grows, which is why the rate rises so sharply.

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