Comparing Rates of Reaction by Mass Loss
A student follows the rate of reaction between calcium carbonate and dilute hydrochloric acid by standing the reaction flask on a balance and recording the loss in mass every 10 s. Each run uses 50.0 cm³ of acid and 2.00 g of calcium carbonate; the acid is in excess in every case. The conditions and the time taken for the mass to become constant are: A: powder, 1.00 mol/dm³ acid, 25 °C - 95 s B: chips, 1.00 mol/dm³ acid, 25 °C - 240 s C: chips, 2.00 mol/dm³ acid, 25 °C - 130 s D: chips, 1.00 mol/dm³ acid, 40 °C - 70 s The final loss in mass is 0.88 g in all four runs. The student writes: 'Temperature has the biggest effect on rate because run D was fastest, and doubling the concentration has the same effect as raising the temperature by 15 °C.'
1. Explain why the final mass loss is the same in all four runs, and calculate the amount of carbon dioxide responsible.
Read the worked answer
n(CaCO3) = 2.00 / 100 = 0.0200 mol. Acid present: 0.0500 mol in A, B and D; 0.100 mol in C. The equation CaCO3 + 2HCl → CaCl2 + CO2 + H2O needs 0.0400 mol of acid for 0.0200 mol of carbonate, so the acid is in excess in every run and calcium carbonate is always limiting.
n(CO2) = 0.0200 mol, so m = 0.0200 × 44 = 0.88 g.
The balance records carbon dioxide leaving, not the rate at which the solid disappears. Changing surface area, concentration or temperature changes how quickly the 0.88 g is lost, not how much is lost, because the same amount of limiting solid reacts in each case.
2. Evaluate the student's two claims using the data, and explain the effect of temperature and concentration in terms of collision theory.
Read the worked answer
The claims are only partly sound. Valid comparisons change one variable at a time: B versus C changes concentration (chips, 25 °C, same volume); B versus D changes temperature (chips, 1.00 mol/dm³). Run A differs from B in surface area and should not be mixed into either comparison.
Collision theory: doubling the concentration puts more acid particles in the same volume, so collisions with the carbonate surface are more frequent and the rate rises. Raising the temperature gives particles more kinetic energy, so a larger fraction of collisions exceed the activation energy; this affects the fraction of successful collisions, not just their frequency, which is why it normally has the larger effect.
On the student's own numbers, the 15 °C rise cuts the time by about 3.4 times (240 → 70 s), while doubling the concentration cuts it by about 1.8 times (240 → 130 s). The direction of the first claim is supported, but the second is false ('same effect' is not justified), and time to completion is not a rate. A fair test would compare initial rates from the tangent at t = 0, and would need several temperatures to generalise.
3. Give one reason why measuring mass loss may misrepresent the rate, and one improvement to the method. Why might the powder run give a misleadingly fast start?
Read the worked answer
Mass loss is indirect. Carbon dioxide dissolves in the acid, especially at 25 °C, acid spray can be carried out with the escaping gas, and draughts or a warm flask affect the balance reading. Run D, at 40 °C, holds less dissolved CO2, so it may appear to start faster for a reason unconnected with the reaction itself.
Improvement: collect the gas in a gas syringe (or over water) and record the volume at fixed intervals, using a bung and delivery tube so that spray cannot escape; this measures the CO2 produced directly.
The powder is misleading at the start because the very fine particles react so quickly that froth can rise into the neck of the flask, and some powder sticks to the dry upper wall and reacts late. The measured initial rate is therefore very high, while the recorded 'completion' includes a slow tail from the stuck solid, so the run does not represent a clean comparison with the chips.