Chemistry 9701/33 — May/June 2019

Background Concept

In a rate-of-reaction experiment using the 'disappearing cross' method, the time taken for a fixed amount of precipitate (sulfur) to obscure a mark is inversely proportional to the initial rate of reaction. By varying the concentration of one reactant while keeping the total volume and the concentration of the other reactant constant, the order of reaction with respect to the varied reactant can be determined.

Understanding the Question

The question asks you to design and record a series of experiments (Experiments 3–5) to investigate the effect of thiosulfate concentration on the reaction rate. You are given the method for Experiments 1 and 2 (using 45.00 cm$^3$ and 20.00 cm$^3$ of FA 1 respectively). You must choose three additional volumes of FA 1, add distilled water to keep the combined volume at 45.00 cm$^3$, record the times, and calculate the rates using the formula $\text{rate} = 500 / \text{time}$.

Approach

1. Table Structure: Create a table with four columns: Volume of FA 1, Volume of water, Time, and Rate. Include correct units.
2. Precision: Burette readings (volumes of FA 1 and water) must be recorded to 2 decimal places (nearest 0.05 cm$^3$). Stopwatch times must be recorded to the nearest second.
3. Volume Selection: Choose three additional volumes for FA 1. The mark scheme requires intervals of at least 5.00 cm$^3$ and all volumes must be $\geqslant 25.00$ cm$^3$. A logical choice is 35.00, 30.00, and 25.00 cm$^3$.
4. Data Consistency: Ensure the total volume (FA 1 + water) is exactly 45.00 cm$^3$ for every experiment. Ensure reaction times decrease as the volume (and thus concentration) of FA 1 increases.
5. Calculations: Calculate the rate for each experiment and verify the ratio $t_{20}/t_{45}$ to check for first-order kinetics (expected ratio $\approx 2.25$–$2.50$).

Step-by-Step Reasoning

• Table headings and units: The table needs columns for Volume of FA 1 (cm$^3$), Volume of distilled water (cm$^3$), Time (s), and Rate (s$^{-1}$). Note that 'ml' is rejected for volume and '1/s' is rejected for rate.
• Choosing volumes: We already have 45.00 and 20.00 cm$^3$. We choose 35.00, 30.00, and 25.00 cm$^3$. The corresponding water volumes are $45.00 - 35.00 = 10.00$, $45.00 - 30.00 = 15.00$, and $45.00 - 25.00 = 20.00$ cm$^3$.
• Representative times: For a first-order reaction, halving the concentration roughly doubles the time. From 45.00 to 20.00 cm$^3$ (factor of $\approx 2.25$), the time should increase by a similar factor. If $t_{45} = 20$ s, then $t_{20} \approx 45 \times 1.1 = 49.5$ s. Let's use 48 s. The intermediate times could be 28 s, 35 s, and 40 s.
• Ratio check: $t_{20} / t_{45} = 48 / 20 = 2.40$. This is within the 2.20–2.50 range, confirming the data is realistic for a first-order reaction.
• Rate calculation: $\text{Rate} = 500 / \text{time}$. For 20 s, rate $= 500/20 = 25$ s$^{-1}$. For 48 s, rate $= 500/48 = 10.4$ s$^{-1}$. Round to appropriate significant figures (minimum 2 sf).

Key Takeaways

• Always include correct units in table headings.
• Burette readings must be to 2 decimal places; stopwatch times to the nearest integer.
• Keep the total volume constant when varying concentrations.
• The ratio of times for a first-order reaction should be roughly inversely proportional to the ratio of concentrations.

Common Mistakes

• Using 'ml' instead of 'cm$^3$' in table headings.
• Recording burette volumes to 1 decimal place (e.g., 10.0 cm$^3$) instead of 2 (10.00 cm$^3$).
• Forgetting to add water to make up the total volume of 45.00 cm$^3$.
• Calculating rates with incorrect significant figures.

Things to Be Careful About

• The mark scheme explicitly rejects 'ml' and '1/s'. Use 'cm$^3$' and 's$^{-1}$'.
• Ensure all five experiments (including the two given) are present in the table.
• The ratio $t_{20}/t_{45}$ must be between 2.10 and 2.60 to earn the mark; between 2.20 and 2.50 for full marks.

Background Concept

Plotting a graph correctly is essential for interpreting experimental data. The axes must be clearly labelled with quantities and units, and the scales must be chosen to utilise the available space effectively. A line of best fit should pass as close as possible to the majority of the data points, with roughly equal numbers of points above and below the line.

Understanding the Question

You are asked to plot a graph of the reaction rate (calculated in part a) against the volume of FA 1 (which is proportional to the concentration of thiosulfate ions, since the total volume is constant). You must label any anomalous points and draw a line of best fit.

Approach

1. Axis labels: $y$-axis is Rate / s$^{-1}$, $x$-axis is Volume of FA 1 / cm$^3$.
2. Scales: Choose linear scales (e.g., 0–30 s$^{-1}$ on $y$-axis, 0–50 cm$^3$ on $x$-axis) that allow the points to spread over more than half the grid in both directions.
3. Plotting: Plot each (Volume, Rate) pair accurately.
4. Line of best fit: Draw a straight line (or smoothly curved line if the relationship is non-linear, though here it should be linear) that best represents the trend. Ignore anomalous points when drawing the line.

Step-by-Step Reasoning

• Using the example data from part (a): (45.00, 25), (35.00, 18), (30.00, 14), (25.00, 13), (20.00, 10).
• The $x$-axis can range from 15 to 50 cm$^3$, with major divisions of 5 cm$^3$.
• The $y$-axis can range from 0 to 30 s$^{-1}$, with major divisions of 5 s$^{-1}$.
• Plot the points. They should form a roughly straight line passing through or near the origin.
• Draw a straight line of best fit. If one point is significantly off the line (e.g., if the 25.00 cm$^3$ point gave a rate of 10 s$^{-1}$ instead of 13), mark it as anomalous and do not include it in the line of best fit.

Key Takeaways

• Always label axes with quantity and unit.
• Use linear scales that maximise the use of the graph paper.
• A line of best fit should not be forced through the origin unless physically justified; however, for this reaction, it should pass near the origin.

Common Mistakes

• Using non-linear scales (e.g., logarithmic) when a linear scale is required.
• Drawing a 'line of connection' (connecting the dots) instead of a line of best fit.
• Forgetting to label the axes with units.

Things to Be Careful About

• The mark scheme requires the graph to occupy more than half the available length for both axes.
• Points must be plotted accurately (within half a small square).
• Anomalous points must be clearly marked (e.g., with a circle or cross) and excluded from the line of best fit.

Background Concept

The rate law for a reaction can be expressed as $\text{rate} = k[\text{S}_2\text{O}_3^{2-}]^n$. If a graph of rate against concentration (or a variable proportional to concentration, like volume when total volume is constant) is a straight line passing through the origin, then $n = 1$, meaning the reaction is first order with respect to that reactant, and the rate is directly proportional to its concentration.

Understanding the Question

From the graph in part (b), you must state the relationship between the rate of reaction and the concentration of thiosulfate ions. Since the total volume of the mixture is constant (45.00 cm$^3$ FA 1/water + 10.0 cm$^3$ FA 2), the concentration of thiosulfate is directly proportional to the volume of FA 1 used.

Approach

Look at the shape of the graph. If it is a straight line through the origin, the relationship is direct proportionality.

Step-by-Step Reasoning

• The graph of rate against volume of FA 1 is a straight line passing through the origin.
• This indicates that rate $\propto$ volume of FA 1.
• Since volume of FA 1 $\propto$ concentration of thiosulfate ions (constant total volume), it follows that rate $\propto$ concentration of thiosulfate ions.
• Therefore, the rate is proportional to the concentration of the thiosulfate ions.

Key Takeaways

• A straight line through the origin on a rate vs. concentration graph indicates first-order kinetics and direct proportionality.
• When total volume is constant, volume of a stock solution is proportional to its concentration in the mixture.

Common Mistakes

• Saying 'rate increases with concentration' without specifying 'proportional'.
• Confusing the relationship with the time (rate is inversely proportional to time).

Things to Be Careful About

• Use the word 'proportional' or 'directly proportional'. 'Linear' is not sufficient unless 'through the origin' is also stated.

Background Concept

When calculating a value from a measurement with an uncertainty, the maximum and minimum possible values of the result are found by using the maximum and minimum possible values of the input. For a division $\text{rate} = 500 / t$, the minimum rate occurs when the denominator (time $t$) is at its maximum.

Understanding the Question

You are given an error in time of $\pm 2$ s. You must calculate the minimum value for the reaction rate in Experiment 2. To get the minimum rate, you must use the maximum possible time.

Approach

1. Take the recorded time for Experiment 2.
2. Add the error (+2 s) to get the maximum time.
3. Divide 500 by this maximum time.

Step-by-Step Reasoning

• Recorded time for Experiment 2 (20.00 cm$^3$ FA 1) = 48 s.
• Maximum time = $48 + 2 = 50$ s.
• Minimum rate = $500 / 50 = 10.0$ s$^{-1}$.
• If a candidate used a different representative time (e.g., 45 s), their minimum rate would be $500 / (45 + 2) = 500 / 47 = 10.6$ s$^{-1}$. The mark scheme awards method marks for showing $500 / (t + 2)$.

Key Takeaways

• Minimum rate corresponds to maximum time.
• Always add the uncertainty to the measured value when calculating a minimum for an inversely proportional quantity.

Common Mistakes

• Subtracting the error from the time (which would give the maximum rate).
• Forgetting to add the error to the time.
• Not showing the working with the candidate's actual time.

Things to Be Careful About

• The question asks for the minimum value, so use $t + 2$, not $t - 2$.
• Show the substitution clearly: $500 / (\text{candidate time} + 2)$.