The method and the results are on the reaction mechanism lab. Do the analysis yourself first, then check the answers below.
Quick answer: A reaction mechanism is a sequence of elementary steps whose sum is the overall equation. These answers go with the AP Chemistry lab on reaction mechanism: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model answers follow. The calculated total matches the measured plateau within 1 %, which supports the stoichiometry. Using ln(106 − V) against t gives a straight line, so the decomposition is first order in peroxide, and the rate constant comes from the slope.
The volume rises steeply at first and then levels off at 106 cm³. The slope falls because [H₂O₂] decreases as it is used, so the rate falls; the reaction ends when the peroxide has gone: decreasing slope, plateau at 106 cm³.
Question 3 Average rates
0–20 s: 42/20 = 2.1 cm³/s. 100–120 s: (101 − 97)/20 = 0.20 cm³/s. The rate is about ten times smaller later.
Question 4 Order and rate constant
106 − V = 64, 39, 24, 14, 9, 5 cm³, so ln = 4.16, 3.66, 3.18, 2.64, 2.20, 1.61. The plot is a straight line; slope = (1.61 − 4.16)/100 = −0.0255 s⁻¹, so k ≈ 0.026 s⁻¹ and the reaction is first order.
Question 5 Is the catalyst used up?
The residue still has a mass of 0.20 g, the same as at the start, and the dried residue can catalyse a new sample at the same speed (reuse test). Without it the control gave no gas.
Question 6 Control variables
Same volume and concentration of H₂O₂; same mass and particle size of catalyst; same temperature (water bath) and same stirring. Three of these are enough.
Question 7 Improving the lab
Use a water bath to hold temperature, add the catalyst with a tube to avoid losing gas, and repeat each run. Doubling the mass of powder gives more surface, so the rate is higher (about twice the initial rate) while the final volume stays 106 cm³.
