Rates of Reaction: Question 6

Worked solution

Part (a): Identifying the factor investigated

Everything about the two experiments is kept the same (the mass of calcium carbonate ($0.40\text{ g}$), the volume of acid ($50\text{ cm}^3$) and its concentration) except for the size of the calcium carbonate pieces: large lumps in Experiment 1, small chips in Experiment 2. The factor being investigated is therefore the surface area (particle size) of the solid reactant.

Part (b): Calculating the average rate in Experiment 2

Between $t=0$ and $t=40$ seconds, the volume of gas collected in Experiment 2 rises from $0$ to $68\text{ cm}^3$:

$\text{average rate} = \frac{\text{change in volume}}{\text{change in time}} = \frac{68-0}{40-0} = \frac{68}{40}$

$\text{average rate} = 1.7\text{ cm}^3\text{/s}$

Part (c): Other ways to increase the rate in Experiment 1

Without changing the size of the calcium carbonate pieces, the student could still speed up Experiment 1 by changing one of the other factors that affect rate:

• Increasing the temperature of the hydrochloric acid (and the calcium carbonate) before mixing them, this demonstrates the temperature factor.
• Using a more concentrated dilute hydrochloric acid, while keeping its volume the same, this demonstrates the concentration factor.
• Adding a suitable catalyst to the reaction mixture, this demonstrates the catalyst factor.

Any two of these, each correctly paired with the factor it demonstrates, are acceptable.

Part (d): Predicting the final volume for Experiment 1

Experiment 1 will eventually also reach a maximum of $96\text{ cm}^3$ of gas (exactly the same as Experiment 2. This can be checked using moles: $0.40\text{ g}$ of calcium carbonate is $\dfrac{0.40}{100} = 0.004\text{ mol}$ (using $M_r(\text{CaCO}_3) = 100$), which produces $0.004\text{ mol}$ of $\text{CO}_2$ gas. At room temperature and pressure, this is $0.004 \times 24\,000 = 96\text{ cm}^3$) matching the final reading already reached by Experiment 2.

Both experiments use the same mass, $0.40\text{ g}$, of calcium carbonate, so they contain the same number of moles of the limiting reactant, and the hydrochloric acid is in excess in both cases. Since the total amount of a product formed depends only on the amount of the limiting reactant (not on how quickly it reacts), the same total volume of carbon dioxide gas must eventually be produced in both experiments. Changing the size of the pieces changes only how quickly this volume is reached, not the total volume itself.

Final answers

• (a) The size (surface area) of the calcium carbonate pieces.
• (b) $\boxed{1.7}$ cm3/s.
• (c) Any two of: raise the temperature (temperature); use more concentrated acid (concentration); add a catalyst (catalyst).
• (d) $\boxed{96}$ cm3, the same as Experiment 2, since both use the same mass ($0.004\text{ mol}$) of calcium carbonate with acid in excess, and $0.004\text{ mol} \times 24\,000\text{ cm}^3\text{/mol} = 96\text{ cm}^3$.