Marble Chips and Acid: A Confounded Rate Experiment
A student investigates the reaction between calcium carbonate and hydrochloric acid by measuring the volume of carbon dioxide collected every 10 s. In Run 1 they use 2.00 g of marble chips and 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid at 20 °C. In Run 2 they use 2.00 g of powdered calcium carbonate and 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid at 40 °C. Run 2 is much faster and produces the same final volume of gas. The student concludes that 'smaller particles always make a reaction faster, so surface area is the only important factor.' A classmate says the comparison is unfair and that the final gas volume should not be the same if the acid concentration has changed.
1. Identify the uncontrolled variables and explain why the student's conclusion is invalid.
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Run 2 changes three variables at once: particle size (chips to powder), acid concentration (1.00 to 2.00 mol/dm³) and temperature (20 °C to 40 °C). Any of these could increase the rate, so the experiment does not show that surface area alone is responsible. To test surface area, the student should keep the mass of calcium carbonate, the volume and concentration of acid, and the temperature the same, and change only the particle size. As written, the comparison is confounded and the conclusion is unsafe.
2. Use collision theory to explain how smaller particles and a higher temperature each increase the rate.
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Smaller particles expose a larger surface area for the same mass. More acid particles can collide with the solid surface per unit time, so the frequency of collisions increases and the rate increases. A higher temperature gives the reacting particles more kinetic energy. A greater fraction of collisions has energy equal to or greater than the activation energy, so the frequency of successful collisions increases and the rate increases. These are separate effects and must be tested separately.
3. Calculate the final volume of carbon dioxide expected in each run and decide whether the classmate is correct that the volumes should differ.
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n(CaCO₃) = 2.00 / 100.1 = 0.0200 mol. n(HCl) in Run 1 = 1.00 × 0.0500 = 0.0500 mol. n(HCl) in Run 2 = 2.00 × 0.0500 = 0.100 mol. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so 0.0200 mol CaCO₃ needs 0.0400 mol HCl. Both runs supply more than 0.0400 mol HCl, so calcium carbonate is limiting in both. Therefore n(CO₂) = 0.0200 mol in both runs. Using the molar volume at room temperature and pressure, V = 0.0200 × 24.0 = 0.480 dm³ (about 480 cm³). The final volume should be the same; the higher acid concentration in Run 2 changes the rate, not the yield, because the acid is in excess in both experiments. The classmate is wrong on that point.