062 – The influence of contact surface on reaction rate 2

Limestone is attacked by acid wherever the two meet, and the consequences are visible on a scale from the industrial to the geological. Flue gases are scrubbed with finely ground limestone because a powder reacts in minutes where a lump would take days; carved stonework on old buildings is eaten away by acidified rain at rates that depend on how weathered and porous the surface has become; and whole cave systems are dissolved out of limestone by groundwater carrying nothing stronger than dissolved carbon dioxide. In each case the reaction is the same one, and what differs is how fast it runs.

Three things govern that rate: how much surface the solid presents, how concentrated the acid is, and how readily the acid gives up its protons. The first two act on the number of collisions per second between acid and carbonate, the third on how many of the acid molecules present are actually dissociated at any instant. All three can be separated experimentally by changing one of them at a time and timing the reaction, which is what this laboratory does.

You will run four reactions of calcium carbonate with acid on a stirred hot plate and time each one to completion. Run A — 2.71 g of powdered calcium carbonate in 100 mL of 2 mol/L hydrochloric acid — is the reference, and each of the other three changes exactly one thing about it: run B replaces the strong acid with ethanoic acid at the same concentration, run C replaces the powder with chunks of the same material, and run D halves the concentration of the hydrochloric acid. The acid is in large excess in every run, so all four go to completion and the only quantity that differs is the time taken. Comparing each run with A in turn isolates the effect of acid strength, of contact surface and of concentration.

Educational Goals

Contact surface and reaction rate

• Explain why a reaction between a solid and a solution can only proceed at the interface, and why dividing the solid therefore changes the rate without changing the amount of product.
• Compare the time taken by powder and by chunks of the same mass, and state what the comparison establishes and what it cannot.

Concentration and reaction rate

• Determine, from the two hydrochloric acid runs, how the reaction time responds to halving the concentration, and express the result as an order of reaction with respect to hydronium ion.
• Verify by calculation that the acid is in excess in every run, so that no run is limited by the amount of acid available.

Strong and weak acids

• Calculate the hydronium concentration of a 2 mol/L solution of ethanoic acid from its acid dissociation constant, and contrast it with that of 2 mol/L hydrochloric acid.
• Explain why the weak acid nevertheless reacts only about twice as slowly rather than hundreds of times more slowly, in terms of the reservoir of undissociated acid.
• Distinguish the strength of an acid from its concentration, and give a definition of each that the other cannot satisfy.

Experimental design and measurement

• Identify the controlled variable in each of the three comparisons and say why the other variables have to be held fixed for the comparison to be valid.
• Judge the end point of a reaction from the cessation of effervescence.

Laboratory practice

• Weigh a solid to a stated mass using the tare function, and transfer powder and chunks without losing material.
• Set up a magnetic stirrer, and clear each beaker into the recovery bin — including undissolved solid, using the tongs — before starting the next run.

Protocol

Preparation

1. Pour 100 mL into each of the 250 mL beakers, as follows (the acid is then in large excess over the 0.027 mol of calcium carbonate used):

• Beaker A: hydrochloric acid (HCl) at 2 M (2 mol/L)
• Beaker B: ethanoic acid (CH₃COOH) at 2 M (2 mol/L)
• Beaker C: hydrochloric acid (HCl) at 2 M (2 mol/L)
• Beaker D: hydrochloric acid (HCl) at 1 M (1 mol/L)

Experience 1

1. Using the balance, weigh exactly “TARE” before).
2. Place beaker A (hydrochloric acid at 2 M) on the hotplate.
3. Place the magnetic stirrer in the beaker on the hotplate.
4. Start the magnetic stirrer using the button on the hotplate.
5. Start the stopwatch to measure the duration of the reaction.
6. Deposit the samples of calcium carbonate in beaker A.
7. Observe the duration of the reaction in beaker A. Stop the stopwatch when the reaction is finished.
8. Stop the magnetic stirrer.
9. Remove the magnetic stick from the beaker.
10. Empty the contents of beaker A into the recovery bin. If there is powder left at the bottom of the beaker, tilt it further above the tray. If solid pieces remain, place them in the recovery bin using the tongs.
11. Reset the stopwatch.
12. Repeat steps 1 to 10, using beaker B (ethanoic acid at 2 M)
13. Compare the results obtained between beaker A (hydrochloric acid at 2 M) and beaker B (ethanoic acid at 2 M).

Experience 2

1. Weigh a piece of calcium carbonate of 2.71 g in the weighing pan, the same mass as the powder used in Experience 1, so that only the physical form differs.
2. Place beaker C (hydrochloric acid at 2 M) on the hotplate.
3. Place the magnetic stirrer in the beaker on the hotplate.
4. Start the magnetic stirrer using the button on the hotplate.
5. Start the stopwatch to measure the duration of the reaction.
6. Deposit the sample of calcium carbonate in beaker C.
7. Observe the duration of the reaction in beaker C. Stop the stopwatch when the reaction is finished.
8. Stop the magnetic stirrer.
9. Remove the magnetic stick from the beaker.
10. Empty the contents of beaker C into the recovery bin. If there is powder left at the bottom of the beaker, tilt it further above the tray. If solid pieces remain, place them in the recovery bin using the tongs.
11. Reset the stopwatch.
12. Compare the results obtained between beaker A (2,71 g of CaCO₃(s) in powder) and beaker C (2,9 g CaCO₃(s) in chunks).

Experience 3

1. Using the balance, measure exactly 2,71 g of powdered calcium carbonate using the weighing pan.
2. Place beaker D on the hotplate.
3. Place the magnetic stirrer in the beaker on the hotplate.
4. Start the magnetic stirrer using the button on the hotplate.
5. Start the stopwatch to measure the duration of the reaction.
6. Deposit the sample of calcium carbonate in beaker D.
7. Observe the duration of the reaction in beaker D. Stop the stopwatch when the reaction is finished.
8. Stop the magnetic stirrer using the button on the hotplate.
9. Remove the magnetic stirrer from the beaker.
10. Empty the contents of beaker D into the recovery bin. If there is powder left at the bottom of the beaker, tilt it further above the tray. If solid pieces remain, place them in the recovery bin using the tongs.
11. Compare the results: (beaker A) hydrochloric acid at 2 M vs. (beaker D) hydrochloric acid at 1 M.

Note: the reaction is accelerated 10 times faster, to more easily observe the complete reaction

Anticipated Outcomes

The reactions

Calcium carbonate is a base, and both acids dissolve it to give a soluble calcium salt, water and carbon dioxide:

CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
CaCO₃(s) + 2 CH₃COOH(aq) → Ca(CH₃COO)₂(aq) + H₂O(l) + CO₂(g)

Both are the same reaction written twice. Stripping out the spectator ions gives the process that actually occurs at the surface of the solid, and it involves the hydronium ion alone:

CaCO₃(s) + 2 H₃O⁺(aq) → Ca²⁺(aq) + 3 H₂O(l) + CO₂(g)

With M(CaCO₃) = 100.09 g/mol, the 2.71 g weighed out is n = 2.71 / 100.09 = 0.0271 mol and requires 0.0542 mol of hydronium ion. The 100 mL of 2 mol/L acid supplies 0.200 mol and the 1 mol/L acid supplies 0.100 mol, so the acid is in excess by factors of 3.7 and 1.8 respectively. Every run therefore consumes all of its carbonate and releases the same 0.0271 mol of carbon dioxide; only the time differs.

Expected results

Two points about these figures before they are interpreted. The simulation runs the chemistry ten times faster than real life, as the note at the end of the protocol says, so the corresponding bench times would be of the order of eight to eighteen minutes. And run C uses 2.9 g of carbonate rather than 2.71 g, about 7 % more material than the other three; correcting for that puts its equivalent time at about 104 s, so its ratio against the reference is closer to 2.0 than the 2.2 the raw numbers give.

Concentration: run D against run A

This is the cleanest of the three comparisons, because only a number has been changed. Halving the concentration of hydrochloric acid takes the time from 51 s to 104 s, a factor of 2.04. If the rate is proportional to some power n of the hydronium concentration, rate ∝ [H₃O⁺]ⁿ, then the time to consume a fixed amount of carbonate goes as [H₃O⁺]⁻ⁿ, and n follows from the two measurements:

n = log(t_D / t_A) / log([H₃O⁺]_A / [H₃O⁺]_D) = log(104 / 51) / log(2.0 / 1.0) = log(2.04) / log(2) = 1.03

The reaction is first order in hydronium ion, which is what is measured for the dissolution of calcite in acid at these concentrations. Collision theory gives the reason directly: doubling the number of hydronium ions per litre doubles the number arriving at each square centimetre of carbonate every second, and each arrival has the same chance of reacting as before. Nothing about the surface or the mechanism has changed — only the traffic.

Acid strength: run B against run A

Hydrochloric acid is fully dissociated in water, so 2 mol/L of it is 2 mol/L of hydronium ion. Ethanoic acid is not: with K_a = 1.8 × 10⁻⁵, the equilibrium CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺ gives

[H₃O⁺] = √(K_a × C) = √(1.8 × 10⁻⁵ × 2.0) = 6.0 × 10⁻³ mol/L,  pH = 2.22

Here is the interesting part. Run D established that the rate is first order in hydronium ion. Applied naively to run B, that would predict a time longer by the ratio of the concentrations — 51 s × (2.0 / 0.0060) = 17 000 s, nearly five hours. The observed time is 93 s. The first-order rate law is not wrong; the mistake is in assuming that the free hydronium concentration in the bulk solution is what governs the surface reaction.

The beaker holds only 6.0 × 10⁻⁴ mol of free hydronium at any moment, and the carbonate needs 0.0542 mol — ninety times more than is present. Every hydronium ion consumed at the surface is therefore replaced immediately by the dissociation of another ethanoic acid molecule, and there are 0.200 mol of those waiting. Because the undissociated acid diffuses to the surface and dissociates there, what limits the rate is the delivery of total acid rather than of pre-formed hydronium. That limit is the same for both acids, which is why run B takes 1.8 times as long as run A rather than 330 times: the answer lies close to the total-acidity limit of 1.0 and nowhere near the free-hydronium limit of 330. A student who predicts hours and measures 93 s has learned something more valuable than one who predicts correctly for the wrong reason.

Contact surface: run C against run A

Calcium carbonate has a density of 2.71 g/cm³, so the 2.71 g weighed out occupies almost exactly 1.00 cm³ however it is divided. For a divided solid the area of a fixed volume is A = 6V/d, where d is the particle diameter, so a powder of 100 µm grains presents about 6 × 1.00 / 0.0100 = 600 cm², while the same cubic centimetre as two or three chunks presents under 10 cm². The geometric expectation is therefore that the chunks react tens of times more slowly, and on a bench with marble chips that is roughly what happens.

The simulation gives a factor of about 2.0 once the extra 7 % of mass in run C is allowed for. The direction is right and it is the direction the laboratory exists to establish — more surface, faster reaction — but the magnitude is much smaller than the geometry of the two samples implies, and the ratio should not be quoted as a measurement of the surface-area effect. It is worth comparing with laboratory 061, where the same comparison is made with magnesium powder against magnesium ribbon and gives 1.7: there the small ratio is genuine, because a ribbon is only 0.15 mm thick and the reaction has to eat through very little metal to finish, whereas a chunk of limestone is thousands of micrometres thick and cannot behave the same way.

The underlying idea is the surface-to-volume ratio. The area of a particle varies as the square of its size and its volume as the cube, so area per unit volume varies as 1/d: grinding a solid to a tenth of its diameter multiplies its exposed area by ten. This is why an industrial scrubber grinds its limestone, and why the reactivity of a solid is a property of its preparation as much as of its formula.

The gas, and how the reaction could be followed quantitatively

Each run liberates one mole of carbon dioxide per mole of carbonate, so 0.0271 mol — a volume of V = nRT/P = (0.0271 × 8.314 × 298) / 101 325 = 6.6 × 10⁻⁴ m³, about 0.66 L of gas, and a loss of 0.0271 × 44.01 = 1.19 g of mass from the beaker. The effervescence is what the timing actually measures: the reaction is judged complete when bubbling stops. Since a balance is on the bench, the same experiment could be followed quantitatively by standing the beaker on it and recording the mass every ten seconds, which turns four single timings into four rate curves and lets initial rates be compared rather than finishing times.

The identity of the gas can be confirmed by bubbling it through limewater, a saturated solution of calcium hydroxide, which turns milky as insoluble calcium carbonate precipitates: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l). This is an optional extension — the protocol does not include the test and no limewater is provided — but it is the standard confirmation and worth mentioning, not least because the precipitate it forms is the substance the experiment started with.

Summary of Assignment by Grade Range

Grade 9–10

Focus: observing that surface area, concentration and the choice of acid all change how fast a reaction runs, and learning the vocabulary that describes them.

Activities: write the word equation for the reaction and name the gas produced; predict, for each of the three comparisons, which beaker will finish first; carry out the four runs and record each duration in a table; order the four from fastest to slowest and identify for each comparison the single thing that was changed; describe in words what happens to the piece of limestone as the reaction proceeds, and explain why the bubbling eventually stops.

Grade 11

Focus: quantitative treatment — moles, excess reagent, mean rate, and the distinction between the strength and the concentration of an acid.

Activities: balance both equations and write the net ionic equation; calculate the moles of carbonate and of acid in each beaker and show that the acid is in excess in all four runs; calculate the volume of carbon dioxide released and the mass the beaker loses; express each duration as a mean rate in mol/s and tabulate the three ratios; calculate the hydronium concentration and pH of 2 mol/L ethanoic acid from K_a and compare it with 2 mol/L hydrochloric acid; explain, using collision theory, why halving the concentration of hydrochloric acid roughly doubles the time.

Grade 12 / College Level

Focus: extracting an order of reaction from the data, resolving the weak-acid result, and criticising the design.

Activities: determine the order with respect to hydronium ion from runs A and D and state the assumptions that step requires; use that order to predict the duration of run B from its free hydronium concentration, obtain a figure of several hours, and then resolve the contradiction by comparing the free hydronium available in the beaker with the amount the carbonate consumes; formulate the total-acidity argument and identify what measurement would distinguish it from the free-hydronium model; derive A = 6V/d, estimate the area ratio between powder and chunks; propose a redesign that measures initial rates by mass loss instead of finishing times, and specify the fifth run needed to bring temperature into the study.

Laboratory essentials

Instruments

• Beakers, 250 mL — four, one for each run
• Graduated cylinder (100 mL)
• Electronic balance
• Weighing pan
• Hot plate with magnetic stirrer
• Magnetic stir bar
• Spatula
• Tongs
• Stopwatch
• Recovery bin

Products

• HCl 2 mol/L (solution) — 100 mL for beaker A and 100 mL for beaker C
• HCl 1 mol/L (solution) — 100 mL for beaker D
• CH₃COOH 2 mol/L (solution) — 100 mL for beaker B
• Calcium carbonate (powder) — 2.71 g for each of beakers A, B and D
• Calcium carbonate (chunks) — 2.9 g for beaker C