Factors Affecting the Rate of a Chemical Reaction Lab Report Answers

Answers

1. Enter the observations before calculating

The report sheet requires the stock molarities of oxalic acid and potassium permanganate, three elapsed-time trials for each concentration experiment, and three trials at each temperature. These values are not printed on the sheet. Record them from the reagent labels and your experiment.

For every row, calculate the mean time:

$$\bar t=\frac{t_1+t_2+t_3}{3}$$

2. Initial permanganate concentration

All mixtures contain 12 drops, so the dilution relation is:

$$[KMnO_4]_{initial}=[KMnO_4]_{stock}\times\frac{\text{drops of }KMnO_4}{12}$$

The report's operational average rate is:

$$\text{average rate}=\frac{[KMnO_4]_{initial}}{\bar t}$$

Its units are $\mathrm{mol\,L^{-1}\,s^{-1}}$. Substitute the labeled stock concentration and your measured mean time into each row.

3. Discussion: effect of concentration

Experiments 1 and 2: oxalic acid increases from 5 to 10 drops in the same 12-drop total volume, so its initial concentration doubles. KMnO4 and temperature are held constant. The expected trend is a higher rate and shorter endpoint time in experiment 2. Whether the rate exactly doubles must be decided from the calculated rates, not assumed.

Experiments 1 and 3: KMnO4 increases from 1 to 2 drops in the same total volume, so its initial concentration doubles. Oxalic acid and temperature are held constant. The calculated operational rate is expected to increase, but the endpoint time is not guaranteed to halve because both the initial amount of purple permanganate and the reaction rate change. Use the measured times.

The 1.5-drop prediction: keep oxalic acid at 5 drops and use 5.5 drops of water so the total remains 12 drops. Then:

$$[KMnO_4]_{initial}=\frac{1.5}{12}[KMnO_4]_{stock}=\frac{1}{8}[KMnO_4]_{stock}$$

Estimate the elapsed time by interpolating between experiments 1 and 3 only if their observed trend is smooth. The protocol alone does not supply a unique number.

4. Discussion: effect of temperature

The reactant proportions stay at 6 drops water, 5 drops H2C2O4, and 1 drop KMnO4. As temperature rises, the expected rate increases and the elapsed time decreases because a larger fraction of collisions have enough energy to cross the activation barrier.

The handout's “rate doubles for each 10 °C increase” is a rule of thumb, not a universal law. Test it by comparing measured rate ratios for temperature pairs about 10 °C apart:

$$\text{rate ratio}=\frac{rate_{hot}}{rate_{cool}}$$

If the rate at an ice-bath temperature 10 °C below room temperature is estimated with that rule, use approximately half the room-temperature rate. With the same initial permanganate concentration, the estimated endpoint time would be about twice as long. Label this as a model-based estimate rather than an observation.

Conclusion

The sheet fixes the dilution factors and comparison design, but the numerical rates and conclusions about how closely the data follow a doubling model depend on the student's stock concentrations and elapsed times.