This question is about the rate of the reaction between hydrochloric acid and calcium carbonate - AQA - GCSE Chemistry - Question 8 - 2020 - Paper 2

The mean rate of reaction can be calculated using the change in the number of moles of gas produced between the time intervals of 20 seconds and 105 seconds. Using the values from Figure 4 (or based on the calculated data), the difference in moles must be obtained (for example: if it was 0.0035 moles at 105 seconds and 0.0010 moles at 20 seconds), giving:

The student's results indicate that the large calcium carbonate lumps produced fewer moles of gas over the same time intervals compared to the smaller lumps. This observation suggests that fewer reactions occurred due to the reduced surface area in contact with the hydrochloric acid, supporting the conclusion that larger lumps react more slowly.

The surface area (SA) and volume (V) of a cube with side length (s) of 0.5 cm can be calculated as follows:

• Surface area: $SA = 6 \cdot s^2 = 6 \cdot (0.5)^2 = 6 \cdot 0.25 = 1.5 \, \text{cm}^2$

• Volume: $V = s^3 = (0.5)^3 = 0.125 \, \text{cm}^3$

Thus, the surface area to volume ratio is:

$\text{Ratio} = \frac{SA}{V} = \frac{1.5}{0.125} = 12 : 1$

The simplest whole number ratio is 12:1.

The larger cube of calcium carbonate with a side length of 5 cm has a greater volume compared to the smaller cube. The surface area to volume ratio will decrease as the size of the cube increases. Specifically, the larger cube's surface area will be less significant relative to its volume, leading to a ratio that decreases by a factor of 10 compared to the 12:1 ratio of the smaller cube in Figure 5.