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Rate evidence

Which runs a rate claim can be tested on, how to treat repeats and anomalies, and when a set of data cannot tell two explanations apart.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to name the independent, dependent and control variables of a rate investigation, choose the runs that fairly test a claim, and decide whether a set of rate data supports a claim, contradicts it, or cannot tell. You will be able to leave out an anomalous repeat and report a mean, and to say which extra run would separate two explanations that the existing data cannot.

2. What you already have

You have four levers on a rate — concentration, temperature, surface area and a catalyst — and the collision model that explains each one. A model predicts; it does not prove itself. This lesson is about the other half of science: deciding what a set of measurements actually shows.

3. Words for this lesson

The independent variable is the quantity an investigator changes on purpose. The dependent variable is the quantity measured to see the effect. Control variables are everything else that could affect the result and is kept the same. A fair test changes only the independent variable between runs. Repeats are runs done again under the same conditions; readings that agree closely are concordant, and a reading far from its repeats is an anomaly. Two quantities that change together in some data are confounded when the data cannot separate their effects.

4. Two explanations, one pair of results

Two learners each run marble chips in hydrochloric acid and time how long it takes to collect 50 mL of carbon dioxide.

One learner says: the higher temperature made Q eight times faster — two 10 °C steps is only four times, so the reaction must be extra sensitive to temperature. The other says: Q was faster because its chips were smaller — temperature hardly mattered.

Before reading on, decide which of them the data supports.

The honest answer is neither. Q differs from P in two ways at once, and both are levers on the rate. Every result the first learner can explain with temperature, the second can explain with surface area. The data cannot choose between them, however striking the eight-fold difference looks.

Now ask what each explanation predicts for a run the learners have not done. Suppose run R uses large chips at 40 °C. If temperature did most of the work, R should be nearly as fast as Q. If surface area did, R should be nearly as slow as P. Those are different predictions, so run R can tell the explanations apart. Suppose R takes 24 s. Then going from P to R, changing only the temperature, made the reaction four times faster, as the doubling rule suggests; and going from R to Q, changing only the chip size, made it twice as fast again. Both levers mattered, and the data now says how much.

5. Change one thing, and know in advance what each answer would mean

A rate can be changed by several things at once, so a difference between two runs is evidence about a particular cause only when that cause is the only difference. That is all a fair test is:

1. One independent variable. Decide the single quantity the claim is about, and change only that between runs.
2. Control everything else that is a lever. For a rate, that list is known from the collision model: concentrations, temperature, particle size and mass of any solid, catalyst, and the total volume.
3. A measure of the rate. The dependent variable must be something that tracks the rate: a volume in a fixed time, a time to a fixed volume, a fall in mass, a clock time.
4. Predictions before data. For each explanation on the table, say what result it predicts. An experiment is useful exactly when the explanations predict different results.

The same rule tells you which rows of someone else's data may be compared. In a table of runs, two rows test a claim about a quantity only if they differ in that quantity and in nothing else. If every pair of rows that differs in temperature also differs in particle size, the data is confounded for temperature and cannot say anything about it alone — however many rows there are.

6. Repeats, anomalies and the mean

One run is not enough, because every measurement has some random scatter: a stopwatch pressed a fraction late, a cross judged to vanish a moment early, a chip slightly larger than the rest. Repeating a run and averaging reduces that scatter.

Suppose four repeats of one thiosulfate run give 52, 54, 71 and 53 s. Three agree to within two seconds; 71 s disagrees with all of them. That is an anomaly — perhaps the flask was not swirled, or the timer was started late — and it is left out before averaging: $(52 + 54 + 53) \div 3 = 53$ s. Including it would give 57.5 s, a value no careful run produced.

An anomaly is left out because it disagrees with its own repeats, never because it disagrees with what someone expected. Leaving out every result that does not fit a prediction is not tidying the data; it is deciding the answer first.

The spread of the concordant repeats is also information. Two conditions whose means differ by less than the spread of their repeats have not been shown to differ at all.

7. Reading a table of runs

Here is data from four runs of one reaction, with the initial rate in arbitrary units.

Concentration. Runs 1 and 2 differ only in concentration: doubling it doubled the rate. Runs 3 and 4 agree. The claim that concentration increases the rate is supported.

Temperature. Every pair that differs in temperature (1 and 4, 2 and 3) also differs in particle size. The rise in temperature might have doubled the rate and the lumps halved it again, or neither might matter. The data is confounded for temperature and cannot tell. A fifth run — 30 °C, 0.5 mol/L, powder — would settle it by comparison with run 1.

This habit of comparing only rows that differ in one quantity is exactly what the next two lessons use to find how a rate depends on each concentration.

8. Where this goes wrong

One dramatic result settles it. An eight-fold difference between two runs that differ in two ways is evidence that something matters, not which thing.

Any two rows of a table can be compared. Only rows that differ in the quantity of interest and nothing else test a claim about it.

Averaging everything is the honest thing to do. A reading that disagrees with its own repeats is left out before averaging; including it produces a mean no run supports. What is not honest is leaving out readings because they disagree with the expected answer.

Controlling one variable makes a test fair. A fair test controls every variable that could affect the result. Using the same chips while changing both the temperature and the acid still leaves two causes in play.

9. Choosing the runs that test a claim

1. Runs: A at 20 °C with 0.1 mol/L acid; B at 30 °C with 0.2 mol/L; C at 30 °C with 0.1 mol/L. Claim: temperature speeds the reaction up.

   List every difference between each pair of runs.

2. A and B differ in temperature and concentration; B and C differ only in concentration; A and C differ only in temperature.

   Only a pair with one difference, in the right quantity, is a fair test.

3. So A and C test the temperature claim. If C is faster than A, the claim is supported for this reaction over this range.

   The conclusion is limited to what the comparison actually covered.

10. A mean with an anomaly

1. Repeats give 31, 30, 44 and 29 mL of gas in the first 30 s.

   Look for a reading far from the rest before averaging.

2. 31, 30 and 29 agree; 44 disagrees with all three and is left out as an anomaly.

   It disagrees with its own repeats, which is the only reason for leaving it out.

3. Mean $= (31 + 30 + 29) \div 3 = 30$ mL, so the mean rate is $30 \div 30 = 1.0$ mL/s.

   Average the concordant repeats, then turn the mean into a rate.

11. Your turn: repeats of a clock reaction give 64, 66, 65 and 48 s. What mean time should be reported?

1. 64, 66 and 65 agree closely; 48 is far from all three.

   Find the reading that disagrees with its repeats.

2. Leave out 48 and average the rest: $(64 + 66 + 65) \div 3 = \ldots$

3. Your turn: work this step out. Its working is at the end of the packet.

   …$65$ s, the mean of the three concordant repeats.

12. Guided practice

A learner investigates how the temperature of the acid affects the time for a cross to vanish under a flask of sodium thiosulfate. Match each role to the quantity that plays it.

13. Guided practice

Four repeats of the same thiosulfate run give times of $57$ s, $59$ s, $81$ s and $58$ s for the cross to vanish. What mean time should be reported, in s?

Answer: s

14. Practice

Three runs of marble chips in hydrochloric acid use the same mass of the same chips. Run A: $20$ °C, acid at $0.1$ mol/L. Run B: $30$ °C, acid at $0.2$ mol/L. Run C: $30$ °C, acid at $0.1$ mol/L. A learner claims that raising the temperature is what speeds the reaction up. Which two runs should be compared to test that claim?

• Runs A and C
• Runs B and C
• Runs A and B
• Any two of them, since all three used the same chips

15. Practice

A learner tests whether temperature affects how fast magnesium reacts with acid. Select the one step that makes the comparison unfair.

This task has no paper form; do it on a device.

16. Practice

Four runs of the same reaction give these initial rates. Run 1: 20 °C, 0.5 mol/L, powder, $5$ units. Run 2: 20 °C, 1.0 mol/L, powder, $10$ units. Run 3: 30 °C, 1.0 mol/L, lumps, $10$ units. Run 4: 30 °C, 0.5 mol/L, lumps, $5$ units. For each claim, decide whether this data supports it, contradicts it, or cannot tell.

17. Practice

A learner wants to test the claim that a catalyst speeds up the decomposition of hydrogen peroxide. Put the stages of a sound test in order.

Number the steps in order (write the number in the box):

• State the claim and the result it predicts
• Choose the one quantity to change and list what to hold fixed
• Take repeated measurements for each condition
• Discard anomalies and find the mean for each condition
• Compare the means with the prediction

18. Somewhere new

A gardener claims a new additive makes garden waste rot faster. He mixes it into a new heap in a sunny corner and compares it with an old heap without the additive in the shade. After $6$ weeks the new heap has rotted much further. What is the strongest criticism of his evidence?

• The heaps differ in position as well as additive, so the extra warmth of the sunny corner could explain the faster rotting.
• He should have waited longer, because $6$ weeks is too short to see any rotting.
• Nothing: the new heap rotted faster, so the additive works.
• He should have used more additive to make the effect larger.

19. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

20. Test question

Four repeats of the same run collect $46$ mL, $44$ mL, $28$ mL and $48$ mL of gas in the first 20 s. What mean rate over the first 20 s should be reported, in mL/s?

Answer: mL/s

21. What you can do now

You can judge a rate claim by finding runs that differ in one quantity only, and you can handle repeats honestly. Say out loud why a run at 40 °C with small chips cannot show that temperature alone speeds a reaction up. Next: rate equations, which turn these controlled comparisons into a law.

Working for the steps left to you

11. Your turn: repeats of a clock reaction give 64, 66, 65 and 48 s. What mean time should be reported?, step 3