How to Choose a Fair Disappearing-Cross Experiment Endpoint
Learn how to choose a consistent endpoint, control liquid depth and other variables, and use disappearance times to compare reaction rates fairly. Includes worked calculations, common experimental errors, and challenging HSC Chemistry questions with solutions.
Two conical flasks sit over identical black crosses. In one, the cross disappears after 30 seconds. In the other, it takes 45 seconds. Would you conclude that the first reaction is faster?
Not necessarily. If the first flask contains a deeper layer of liquid, the cross might disappear sooner simply because you’re looking through more cloudy solution.
To choose a fair endpoint in a disappearing-cross experiment, stop the timer at the first moment the cross is no longer distinguishable through the reacting mixture. Keep the flask, total liquid volume, cross, lighting, viewing position, and endpoint rule the same in every trial.
The aim is to measure how long each reaction takes to produce the same observable level of cloudiness. You’re not measuring when the reaction finishes.
01Why does the cross disappear?
The experiment usually uses sodium thiosulfate solution and dilute hydrochloric acid.
Both solutions are initially clear. When they’re mixed, they react to produce solid sulfur, which forms fine particles suspended throughout the liquid.
The balanced reaction is:
\[
\ce{Na2S2O3 + 2HCl -> 2NaCl + S + SO2 + H2O}
\]
The sulfur is the solid product, \(\ce{S(s)}\). The sulfur dioxide, \(\ce{SO2(g)}\), is a gaseous product.
As more sulfur particles form, they scatter more light. The mixture becomes increasingly cloudy, making the black cross beneath the flask harder to distinguish.
This cloudiness is called turbidity.
Imagine trying to read a road sign through fog. The sign hasn’t changed, but tiny water droplets scatter light and make its outline difficult to see. The suspended sulfur particles play a similar optical role in the experiment.
The analogy has a limit: fog contains water droplets in air, whereas the reaction creates solid sulfur particles in a liquid. The analogy explains why visibility changes, not how quickly the chemical reaction occurs.
A disappearing cross is not a completed reaction
Suppose the cross becomes invisible after 35 seconds.
Has all the sodium thiosulfate reacted?
No. The mixture may continue producing sulfur after the cross disappears. The stopwatch records when the mixture becomes sufficiently opaque, not when the limiting reactant is exhausted.
That makes the disappearing cross an operational endpoint: an event defined by an observable rule.
It is useful because we can apply approximately the same rule to different trials. But we must be careful about what that observation actually measures.
02How to set up a fair endpoint
Picture the apparatus from bottom to top.
A bold black cross is printed on white paper. A clean, transparent conical flask stands directly over it. The flask contains sodium thiosulfate solution, with distilled water added if dilution is required.
Hydrochloric acid is measured separately. When it is added to the flask, the reaction starts.
The mixture is swirled consistently, and an observer watches the cross through the liquid from the same position above the flask.
A reliable timing procedure is:
- Place the same clean flask over the same black cross.
- Measure the required volumes of sodium thiosulfate and water.
- Keep the total reaction volume constant, including the hydrochloric acid that will be added.
- Add the acid and start the stopwatch as it first contacts the solution.
- Swirl the mixture in the same way for every trial.
- Observe from a fixed position under consistent lighting.
- Stop timing at the first moment the cross can no longer be distinguished.
- Repeat the trial and record the times separately.
Agree on the disappearance rule before beginning. Don’t switch from “the cross is becoming faint” in one trial to “absolutely no black is visible” in another.
These criteria can produce noticeably different times.
Which variables must stay constant?
Suppose you’re investigating how sodium thiosulfate concentration affects reaction rate.
| Variable | Why it matters |
|---|---|
| Flask shape and size | Determines how deep the liquid is for a given volume |
| Total liquid volume | Controls the depth of liquid above the cross |
| Cross thickness and contrast | Changes how much cloudiness is needed to hide it |
| Lighting and viewing position | Affect when the observer loses sight of the cross |
| Temperature | Changes reaction rate |
| Hydrochloric acid concentration and volume | Affect the reacting conditions |
| Mixing procedure | Affects how quickly the reactants become uniformly distributed |
| Endpoint rule | Determines the event being timed |
Only the chosen independent variable should deliberately change.
If you’re testing temperature instead of concentration, keep both reactant concentrations and volumes constant while varying temperature.
Why constant liquid depth matters so much
Imagine two equally cloudy samples.
One is 2 cm deep. The other is 4 cm deep.
Which is more likely to hide the cross?
The deeper sample. Light travelling from the cross passes through a longer path containing suspended sulfur particles, so there are more opportunities for scattering.
A deeper mixture may therefore reach the visibility endpoint with a lower concentration of sulfur.
This creates an important problem when changing reactant concentration.
Suppose you use less sodium thiosulfate solution in your second trial. If you simply remove some liquid, you’ve changed both its concentration and the depth of liquid.
Instead, replace the removed sodium thiosulfate solution with distilled water so that the total volume stays constant.
The final volume must include the acid added afterwards.
The limitations of a visual endpoint
Even with an identical flask and liquid depth, two observers may disagree about when the cross has disappeared.
Other problems include a dirty flask, reflections, inconsistent mixing, different cross thicknesses, and changes in sulfur particle size.
The last point is subtle. Turbidity depends not just on the amount of sulfur present, but also on how its particles scatter light.
A fixed visual endpoint is therefore best treated as approximately the same cloudiness, rather than proof that exactly the same number of moles of sulfur has formed.
There is also a safety issue: this reaction releases sulfur dioxide, a harmful gas. Use eye protection, follow your teacher’s risk assessment, work with appropriate ventilation, and avoid leaning over the flask or inhaling the gas. Follow the laboratory’s approved waste-disposal procedure.
03How does disappearance time relate to reaction rate?
Suppose two trials use the same optical setup and reach the same sulfur-production threshold.
One takes 40 seconds and the other takes 80 seconds.
The first trial has produced approximately the same amount of sulfur in half the time. Its average rate of sulfur production up to the endpoint is approximately twice as large.
That is why we often calculate the relative rate indicator:
\[
q=\frac{1}{t}
\]
where:
- \(q\) is the relative rate indicator, measured in \(\mathrm{s^{-1}}\).
- \(t\) is the disappearance time, measured in seconds (s).
A larger \(q\) means a shorter time to reach the selected endpoint.
But \(q\) isn’t an absolute chemical reaction rate.
An actual rate of sulfur formation might have units of \(\mathrm{mol\,L^{-1}\,s^{-1}}\), whereas \(1/t\) has units of \(\mathrm{s^{-1}}\).
The missing information is how much sulfur was produced.
Why \(1/t\) is an average-rate indicator
Imagine that the endpoint always corresponds to producing a fixed amount of sulfur per litre.
Call that amount \(c_*\), measured in \(\mathrm{mol\,L^{-1}}\).
The average sulfur-production rate from the start to the endpoint would be:
\[
\bar r=\frac{c_*}{t}
\]
Here, \(\bar r\) is the average rate and \(t\) is the endpoint time.
If \(c_*\) is constant across trials, comparing \(1/t\) is equivalent to comparing their average rates up to that fixed amount of product.
This reasoning depends on two assumptions:
- The visibility endpoint corresponds to the same amount of sulfur produced per litre in every trial.
- The sulfur remains in the mixture, and the product amount is measured consistently.
Neither assumption is guaranteed simply because the same cross is used.
Also, a reaction usually changes speed as it proceeds. The average rate up to the endpoint is not necessarily the initial rate or the instantaneous rate when the cross disappears.
04Worked example: Comparing two fair trials
A student investigates how sodium thiosulfate concentration affects disappearance time.
The student uses \(0.20\ \mathrm{mol\,L^{-1}}\) sodium thiosulfate stock solution and \(1.0\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
Each trial has a final volume of \(60\ \mathrm{cm^3}\).
The following results are illustrative, not measurements from a real experiment.
| Measurement | Trial A | Trial B |
|---|---|---|
| Sodium thiosulfate stock | 20 cm³ | 30 cm³ |
| Distilled water | 30 cm³ | 20 cm³ |
| Hydrochloric acid | 10 cm³ | 10 cm³ |
| Total volume | 60 cm³ | 60 cm³ |
| First time | 54 s | 35 s |
| Second time | 55 s | 36 s |
| Third time | 56 s | 37 s |
All other experimental conditions are kept the same.
Determine the initial sodium thiosulfate concentration in each mixture, the relative rate indicators, and how much faster Trial B reaches the endpoint.
Step 1
Both trials use the same flask, final liquid volume, acid conditions, temperature, and visual endpoint.
Only the starting sodium thiosulfate concentration changes deliberately.
Step 2
Use the dilution relationship:
\[
c_0=c_{\text{stock}}
\frac{V_{\text{stock}}}{V_{\text{total}}}
\]
Here, \(c_0\) is the concentration after mixing, \(c_{\text{stock}}\) is the stock concentration, and the two \(V\) values are stock and final solution volumes in the same units.
For Trial A:
\[
\begin{aligned}
c_A&=0.20\times\frac{20}{60}\\
&=0.0667\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
For Trial B:
\[
\begin{aligned}
c_B&=0.20\times\frac{30}{60}\\
&=0.100\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
Step 3
\[
\begin{aligned}
\bar t_A&=\frac{54+55+56}{3}\\
&=55\ \mathrm{s}\\[4pt]
\bar t_B&=\frac{35+36+37}{3}\\
&=36\ \mathrm{s}
\end{aligned}
\]
The bar over \(t\) indicates the mean of the repeated times.
Step 4
Using \(q=1/\bar t\):
\[
\begin{aligned}
q_A&=\frac{1}{55}\\
&=0.0182\ \mathrm{s^{-1}}\\[4pt]
q_B&=\frac{1}{36}\\
&=0.0278\ \mathrm{s^{-1}}
\end{aligned}
\]
Step 5
\[
\frac{q_B}{q_A}=\frac{55}{36}=1.53
\]
Trial B reaches the matched visibility endpoint approximately 1.53 times as quickly by the relative-rate measure.
Its higher sodium thiosulfate concentration is consistent with more frequent effective collisions under otherwise comparable conditions.
But the result doesn’t prove that the instantaneous reaction rate is always 1.53 times greater. It compares average progress to an operational endpoint.
05When a shorter disappearance time gives the wrong impression
Now consider a contrasting experiment.
Two students use the same reaction and identical crosses, but their mixtures have different depths.
Independent measurements give the amount of sulfur formed per litre of mixture, \(c_S\), at different times.
For clarity, \(c_S\) is the molar amount of solid sulfur produced per litre of mixture. It does not mean that the sulfur is dissolved.
Worked example: Did the faster-looking trial actually react faster?
The independently determined sulfur-production data are:
| Time (s) | Trial A \(c_S\) (mol L⁻¹) | Trial B \(c_S\) (mol L⁻¹) |
|---|---|---|
| 0 | 0.000 | 0.000 |
| 10 | 0.008 | 0.007 |
| 20 | 0.015 | 0.012 |
| 30 | 0.020 | 0.015 |
| 40 | 0.024 | 0.017 |
Because Trial A has a shallower liquid layer, its cross becomes invisible at \(0.024\ \mathrm{mol\,L^{-1}}\) sulfur.
Trial B has a deeper layer. Its cross becomes invisible at only \(0.015\ \mathrm{mol\,L^{-1}}\).
These thresholds are assumed to come from an independent optical calibration.
Determine which trial has the shorter disappearance time, which produces sulfur faster on average over the first 30 seconds, and how to estimate the rate at 20 seconds.
Step 1
From the supplied thresholds:
- Trial A reaches its endpoint at 40 s.
- Trial B reaches its endpoint at 30 s.
Using only these times, someone might conclude that Trial B reacts faster.
But the endpoints correspond to different amounts of sulfur.
Step 2
The average product-formation rate is:
\[
\bar r=\frac{\Delta c_S}{\Delta t}
\]
Here, \(\Delta c_S\) is the increase in sulfur amount per litre and \(\Delta t\) is the elapsed time.
For the first 30 seconds:
\[
\begin{aligned}
\bar r_A&=\frac{0.020-0}{30}\\
&=0.000667\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
\[
\begin{aligned}
\bar r_B&=\frac{0.015-0}{30}\\
&=0.000500\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
Trial A has the greater average sulfur-production rate over this common interval, even though its cross disappears later.
The shorter visibility time in Trial B is caused by its lower optical threshold.
Step 3
For a concentration-time graph, the instantaneous rate is the gradient of the tangent at the chosen time.
With only tabulated measurements, we can estimate that gradient using points on either side of 20 seconds.
For Trial A:
\[
\begin{aligned}
r_A(20)&\approx\frac{0.020-0.008}{30-10}\\
&=0.000600\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
For Trial B:
\[
\begin{aligned}
r_B(20)&\approx\frac{0.015-0.007}{30-10}\\
&=0.000400\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
These are estimates of the instantaneous rates near 20 seconds, not exact measurements at that instant.
The calculation uses a 20-second interval, so it gives the average gradient across that interval as an approximation to the tangent.
Step 4
Trial A produces sulfur faster over the supplied intervals, but Trial B reaches its less demanding visibility threshold sooner.
This is why an endpoint comparison must control the optical conditions, not merely use the same chemical reaction.
The appropriate correction is to repeat the comparison with the same flask and liquid depth, rather than invent a corrected disappearance time from the original readings.
06Average and instantaneous rates: which one do you actually know?
It’s worth keeping three measurements separate.
| Measurement | What it tells you |
|---|---|
| Disappearance time, \(t\) | How long it takes to reach the visibility endpoint |
| Relative indicator, \(1/t\) | A comparison of average progress towards a matched endpoint |
| Instantaneous rate | How quickly product is forming or reactant is disappearing at a particular moment |
If you have a product concentration-time graph, the gradient between two points gives an average rate.
A tangent gives the instantaneous rate at a particular time.
For a reactant that is being consumed, concentration decreases, so the gradient is negative. We usually report the positive rate of disappearance using a minus sign:
\[
r=-\frac{\Delta c_R}{\Delta t}
\]
Here, \(c_R\) is reactant concentration, and the expression gives the average disappearance rate over the chosen interval. The negative sign makes the reported rate positive.
For an instantaneous rate, use the negative tangent gradient instead.
The distinction is the same one used for gas-volume-time graphs. For a more detailed treatment of interpreting graph gradients, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.
A disappearing-cross experiment gives you one selected point in the reaction’s progress. Continuous measurements give you information about how the rate changes throughout the reaction.
07How to improve the endpoint when visual judgement is unreliable
Suppose three repeated measurements give disappearance times of 39, 41, and 40 seconds.
Those results are reasonably consistent.
But suppose another observer obtains 46, 45, and 47 seconds using exactly the same mixtures. Both observers are consistent with themselves, yet they disagree about when the cross has disappeared.
Repeating measurements alone hasn’t removed this systematic difference in judgement.
A useful revision is to replace the observer’s judgement with a light sensor.
From a visual cross to a light sensor
The modified apparatus has a light source on one side of a transparent reaction container and a light sensor on the opposite side. The sensor connects to a data logger.
The light follows this path:
Light source → reacting mixture → light sensor.
As sulfur forms, more light is scattered away from the direct path. The sensor typically detects less transmitted light.
A fixed sensor reading can then define the endpoint.
This revision addresses one weakness of the original method: observers no longer need to decide whether a cross looks sufficiently faint.
But it doesn’t automatically solve every problem.
The optical path length must remain constant. The sensor and light source must be positioned consistently. Particle size and light scattering can still affect the relationship between the signal and the amount of sulfur.
A fixed optical signal remains an operational endpoint unless it has been calibrated against chemical progress.
A useful diagnostic before trusting your results
| Observation | Likely issue | Better decision |
|---|---|---|
| The cross is unclear before mixing | Poor starting visibility | Improve the initial setup before timing |
| One observer stops much earlier | Subjective endpoint | Agree on a fixed criterion or use optical measurement |
| Total volume changes between trials | Different optical path lengths | Keep the final volume constant |
| The cross never disappears | Threshold may be unreachable | Check possible sulfur production before assuming a slow reaction |
| \(1/t\) is reported as an instantaneous rate | Wrong interpretation of the measurement | Use concentration-time data and a tangent for an instantaneous estimate |
The strongest experimental conclusion doesn’t just identify a trend. It explains what was held constant, what was actually measured, and which interpretations the data support.
08Questions and solutions
The questions below progress from an ordinary rate comparison to situations where endpoint choice, chemical stoichiometry, and changing reaction rates interact. All numerical data are illustrative.
Question 1
Two disappearing-cross trials use identical apparatus, liquid depth, lighting, temperature, and endpoint criteria.
Trial X takes 48 seconds to hide the cross. Trial Y takes 72 seconds.
Calculate the relative rate indicator for each trial and the ratio between them. Explain what the ratio means.
Solution 1
Trial X reaches the matched endpoint 1.50 times as quickly by the relative-rate measure.
Using \(q=1/t\):
\[
\begin{aligned}
q_X&=\frac{1}{48}\\
&=0.0208\ \mathrm{s^{-1}}\\[4pt]
q_Y&=\frac{1}{72}\\
&=0.0139\ \mathrm{s^{-1}}
\end{aligned}
\]
The ratio is:
\[
\frac{q_X}{q_Y}=\frac{72}{48}=1.50
\]
Because the optical conditions are matched, Trial X reaches approximately the same sulfur-production endpoint in less time.
The tempting mistake is to say the instantaneous rate of Trial X is exactly 1.50 times that of Trial Y at every moment.
The calculation establishes a relative measure of average progress to the chosen endpoint. It doesn’t describe the full reaction-rate curve.
Question 2
A student tries to investigate the effect of sodium thiosulfate concentration.
The stock solutions are \(0.20\ \mathrm{mol\,L^{-1}}\) sodium thiosulfate and \(1.0\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
The student runs two trials:
| Volume | Trial 1 | Trial 2 |
|---|---|---|
| Sodium thiosulfate stock | 15 cm³ | 30 cm³ |
| Distilled water | 35 cm³ | 0 cm³ |
| Hydrochloric acid | 10 cm³ | 10 cm³ |
| Disappearance time | 52 s | 37 s |
The same conical flask, cross, temperature, and viewing position are used.
The student concludes:
“Doubling the sodium thiosulfate volume made the reaction faster. Everything else stayed the same.”
Assess this conclusion by calculating the initial concentrations of both reactants after mixing. Identify the confounding variables, and redesign Trial 2 to test a doubling of sodium thiosulfate concentration fairly.
Solution 2
The conclusion is not valid. Trial 2 changes sodium thiosulfate concentration, hydrochloric acid concentration, and the depth of liquid, so the shorter time cannot be attributed to sodium thiosulfate concentration alone.
First, find the total volumes:
\[
\begin{aligned}
V_1&=15+35+10=60\ \mathrm{cm^3}\\
V_2&=30+0+10=40\ \mathrm{cm^3}
\end{aligned}
\]
The sodium thiosulfate concentrations are:
\[
\begin{aligned}
c_{T,1}&=0.20\times\frac{15}{60}\\
&=0.050\ \mathrm{mol\,L^{-1}}\\[4pt]
c_{T,2}&=0.20\times\frac{30}{40}\\
&=0.150\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
Here, \(c_T\) represents sodium thiosulfate concentration immediately after mixing.
The concentration triples rather than doubles.
The hydrochloric acid concentrations also differ:
\[
\begin{aligned}
c_{H,1}&=1.0\times\frac{10}{60}\\
&=0.167\ \mathrm{mol\,L^{-1}}\\[4pt]
c_{H,2}&=1.0\times\frac{10}{40}\\
&=0.250\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
Here, \(c_H\) represents hydrochloric acid concentration immediately after mixing.
So Trial 2 has more concentrated hydrochloric acid as well as more concentrated sodium thiosulfate.
The liquid is also shallower, which changes the optical endpoint. A shallower layer generally needs more sulfur per litre to hide the cross, making the timing comparison even less straightforward.
To redesign Trial 2, use:
- 30 cm³ sodium thiosulfate stock
- 20 cm³ distilled water
- 10 cm³ hydrochloric acid
The final volume is now \(60\ \mathrm{cm^3}\).
Its sodium thiosulfate concentration becomes:
\[
\begin{aligned}
c_{T,2}&=0.20\times\frac{30}{60}\\
&=0.100\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
This is exactly twice Trial 1’s concentration.
The hydrochloric acid concentration remains \(0.167\ \mathrm{mol\,L^{-1}}\), and the liquid depth is matched.
The experiment must be repeated with this corrected mixture. The original 37-second measurement cannot simply be reused or mathematically adjusted.
Question 3
A separate analysis measures sodium thiosulfate concentration during a disappearing-cross reaction.
The results are:
| Time (s) | Sodium thiosulfate concentration (mol L⁻¹) |
|---|---|
| 0 | 0.120 |
| 10 | 0.100 |
| 20 | 0.085 |
| 30 | 0.074 |
| 40 | 0.066 |
The cross disappears at 30 seconds.
Calculate:
(a) The average rate of sodium thiosulfate disappearance from 0 to 30 seconds.
(b) An estimate of the instantaneous disappearance rate at 30 seconds, using the concentration measurements on either side of that time.
(c) The relative rate indicator from the disappearance time.
(d) Explain why these three quantities are not interchangeable, and identify what the data suggest about the reaction rate as time passes.
Solution 3
The average disappearance rate from 0 to 30 seconds is \(0.00153\ \mathrm{mol\,L^{-1}\,s^{-1}}\). The estimated instantaneous rate at 30 seconds is smaller, approximately \(0.000950\ \mathrm{mol\,L^{-1}\,s^{-1}}\), while the cross-time indicator is \(0.0333\ \mathrm{s^{-1}}\).
Part (a): Average disappearance rate
Because sodium thiosulfate is being consumed, use the negative concentration gradient:
\[
\bar r=-\frac{\Delta c_T}{\Delta t}
\]
Substituting:
\[
\begin{aligned}
\bar r&=-\frac{0.074-0.120}{30-0}\\
&=\frac{0.046}{30}\\
&=0.00153\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
This represents the mean rate of sodium thiosulfate consumption over the first 30 seconds.
Part (b): Estimated instantaneous rate
To estimate the gradient at 30 seconds, use the points at 20 and 40 seconds.
\[
\begin{aligned}
r(30)&\approx-\frac{0.066-0.085}{40-20}\\
&=\frac{0.019}{20}\\
&=0.000950\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
This is a centred secant estimate, not an exact instantaneous measurement.
An actual tangent to a sufficiently detailed concentration-time curve would give a better instantaneous estimate.
Part (c): Relative rate indicator
\[
\begin{aligned}
q&=\frac{1}{30}\\
&=0.0333\ \mathrm{s^{-1}}
\end{aligned}
\]
This value is an indicator based on the visibility endpoint.
Part (d): Interpretation
The three values describe different things.
The first is an average chemical disappearance rate over 30 seconds.
The second estimates the chemical disappearance rate near one particular time.
The third is simply the reciprocal of the time needed to hide the cross.
They also have different units, so treating them as interchangeable would be incorrect.
The first 10-second interval has a disappearance rate of:
\[
\begin{aligned}
\bar r_{0-10}&=\frac{0.120-0.100}{10}\\
&=0.00200\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
This exceeds the later estimated rate of \(0.000950\ \mathrm{mol\,L^{-1}\,s^{-1}}\).
The data are consistent with the reaction slowing as reactants are consumed.
The tempting mistake is to use the average rate from 0 to 30 seconds as the exact instantaneous rate when the cross disappears. The declining concentration gradient shows why that assumption is unreliable.
Question 4
A teacher investigates the disappearing-cross endpoint using the same flask, optical conditions, and liquid volume in two reactions.
An independent calibration shows that the cross becomes invisible only when the mixture contains at least \(0.010\ \mathrm{mol\,L^{-1}}\) of suspended sulfur.
Both trials begin with \(0.020\ \mathrm{mol\,L^{-1}}\) sodium thiosulfate, but have different initial hydrochloric acid concentrations:
| Condition | Trial A | Trial B |
|---|---|---|
| Initial sodium thiosulfate | 0.020 mol L⁻¹ | 0.020 mol L⁻¹ |
| Initial hydrochloric acid | 0.060 mol L⁻¹ | 0.012 mol L⁻¹ |
Trial A hides the cross after 55 seconds.
Trial B develops some cloudiness, but the cross remains visible even after the reaction has stopped.
A student suggests that Trial B must have a reaction rate of zero.
Determine whether the student’s conclusion is correct. Calculate the maximum possible amount of sulfur formed per litre in each trial, and propose a way to compare early reaction rates without changing the starting reactant concentrations.
Assume the stated reaction is the only reaction occurring, volumes remain constant, sulfur remains in the mixture, and the calibrated optical threshold applies to both trials.
Solution 4
The student is incorrect. Trial B can react and produce sulfur, but it cannot produce enough sulfur to reach the cross-disappearance threshold.
The key constraint is the amount of hydrochloric acid available.
From the balanced reaction:
\[
\ce{Na2S2O3 + 2HCl -> 2NaCl + S + SO2 + H2O}
\]
One mole of sodium thiosulfate can produce one mole of sulfur, but it requires two moles of hydrochloric acid.
For a constant-volume system, the maximum amount of sulfur produced per litre is determined by the limiting reactant:
\[
c_{S,\max}=\min\left(c_T,\frac{c_H}{2}\right)
\]
Here, \(c_T\) and \(c_H\) are the initial molar concentrations of sodium thiosulfate and hydrochloric acid.
For Trial A:
\[
\begin{aligned}
c_{S,\max}
&=\min(0.020,0.030)\\
&=0.020\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
Sodium thiosulfate is limiting. The maximum sulfur production exceeds the required \(0.010\ \mathrm{mol\,L^{-1}}\).
Trial A can therefore reach the endpoint.
For Trial B:
\[
\begin{aligned}
c_{S,\max}
&=\min(0.020,0.006)\\
&=0.006\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]
Hydrochloric acid is limiting.
Even if the reaction proceeds as far as possible, Trial B cannot produce the \(0.010\ \mathrm{mol\,L^{-1}}\) of sulfur required to obscure the cross.
This is why its cross never disappears.
The absence of an endpoint does not mean the reaction never occurred. It means the chosen endpoint is chemically unreachable under those conditions.
A useful revised experiment would use a calibrated optical threshold corresponding to a smaller amount of sulfur, such as \(0.004\ \mathrm{mol\,L^{-1}}\), and apply it to both trials.
Both reactions could reach that amount, allowing a comparison of average progress to the same early chemical threshold.
An alternative is to measure sodium thiosulfate concentration over an early common time interval and compare the concentration-time gradients.
That would be more informative about early chemical rates than treating an unreachable disappearance endpoint as an infinite time or a zero reaction rate.
Question 5
Two hypothetical reaction conditions, A and B, produce suspended sulfur according to the independently measured data below.
Both use the same flask geometry, total volume, viewing position, and lighting. The sulfur particles are assumed to have identical light-scattering behaviour, and both reactions have enough reactants to produce all the listed amounts.
The sulfur-production data are assumed to change linearly between the recorded times.
| Time (s) | Condition A \(c_S\) (mol L⁻¹) | Condition B \(c_S\) (mol L⁻¹) |
|---|---|---|
| 0 | 0.000 | 0.000 |
| 10 | 0.012 | 0.008 |
| 20 | 0.019 | 0.015 |
| 30 | 0.023 | 0.022 |
| 40 | 0.025 | 0.028 |
| 50 | 0.026 | 0.033 |
Two different cross designs are available:
- A thin, faint cross becomes invisible when \(c_S=0.008\ \mathrm{mol\,L^{-1}}\).
- A bold, high-contrast cross becomes invisible when \(c_S=0.025\ \mathrm{mol\,L^{-1}}\).
The optical thresholds have been determined independently.
Each comparison uses the same cross design for both reaction conditions.
(a) Determine which reaction hides the faint cross first, and calculate the disappearance times.
(b) Repeat the calculation for the bold cross.
(c) Calculate the average sulfur-production rates over 0-10 seconds and 30-40 seconds for both conditions.
(d) A student argues that a fair cross experiment must always identify the same reaction as faster, regardless of which cross design is chosen. Evaluate this claim.
(e) If the aim is to compare early reaction rates, which measurement would provide the more defensible conclusion?
Solution 5
Both cross designs are internally fair, but they reverse which reaction appears faster. Condition A hides the faint cross first, while Condition B hides the bold cross first.
The apparent contradiction arises because the reactions don’t maintain the same relative rate throughout the experiment. Condition A produces sulfur faster early on, but Condition B later produces sulfur faster and overtakes it.
Part (a): The faint cross
The faint cross disappears at:
\[
c_S=0.008\ \mathrm{mol\,L^{-1}}
\]
For Condition A, sulfur concentration rises from zero to \(0.012\ \mathrm{mol\,L^{-1}}\) in the first 10 seconds.
Under the stated linear assumption:
\[
\begin{aligned}
r_A&=\frac{0.012}{10}\\
&=0.00120\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
The time to produce \(0.008\ \mathrm{mol\,L^{-1}}\) is:
\[
\begin{aligned}
t_A&=\frac{0.008}{0.00120}\\
&=6.67\ \mathrm{s}
\end{aligned}
\]
Condition B produces exactly \(0.008\ \mathrm{mol\,L^{-1}}\) after 10 seconds.
Therefore:
\[
t_B=10.0\ \mathrm{s}
\]
Condition A hides the faint cross first.
Part (b): The bold cross
The bold cross requires:
\[
c_S=0.025\ \mathrm{mol\,L^{-1}}
\]
Condition A reaches this amount at exactly 40 seconds.
Condition B has already reached \(0.022\ \mathrm{mol\,L^{-1}}\) after 30 seconds and reaches \(0.028\ \mathrm{mol\,L^{-1}}\) after 40 seconds.
The additional sulfur needed after 30 seconds is:
\[
0.025-0.022=0.003\ \mathrm{mol\,L^{-1}}
\]
Condition B produces \(0.006\ \mathrm{mol\,L^{-1}}\) during the 30-40 second interval.
Under the linear assumption, producing half that amount takes half the interval:
\[
\begin{aligned}
t_B&=30+10\times\frac{0.003}{0.006}\\
&=35\ \mathrm{s}
\end{aligned}
\]
Therefore, for the bold cross:
- Condition A: 40 s
- Condition B: 35 s
Condition B hides the bold cross first.
Part (c): Compare the rates over two intervals
For 0-10 seconds:
\[
\begin{aligned}
\bar r_A&=\frac{0.012-0}{10}\\
&=0.00120\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
\[
\begin{aligned}
\bar r_B&=\frac{0.008-0}{10}\\
&=0.000800\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
Condition A produces sulfur faster during the first 10 seconds.
For 30-40 seconds:
\[
\begin{aligned}
\bar r_A&=\frac{0.025-0.023}{10}\\
&=0.000200\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
\[
\begin{aligned}
\bar r_B&=\frac{0.028-0.022}{10}\\
&=0.000600\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
Condition B now produces sulfur three times as quickly as Condition A over this later interval.
Part (d): Why does the ranking reverse?
The student’s claim is false because reaction rate can change with time.
At 30 seconds, Condition A has produced \(0.023\ \mathrm{mol\,L^{-1}}\) sulfur, slightly more than Condition B’s \(0.022\ \mathrm{mol\,L^{-1}}\).
Condition A is still ahead in total sulfur production.
But Condition B is producing sulfur more rapidly by this stage.
During the following 10 seconds, Condition B produces an additional \(0.006\ \mathrm{mol\,L^{-1}}\), compared with only \(0.002\ \mathrm{mol\,L^{-1}}\) for Condition A.
Condition B overtakes A before the 40-second measurement.
This is the crucial distinction: having produced more product so far does not mean a reaction is currently producing product faster.
Both experiments are fair within their respective cross designs because each paired comparison uses one consistent optical threshold.
Yet they answer slightly different questions:
- The faint cross compares progress through an early part of the reaction.
- The bold cross compares progress through a later, larger amount of sulfur production.
The two comparisons need not rank the reactions identically.
A disappearing-cross experiment can establish which reaction reaches a specified endpoint first. It cannot automatically establish which reaction is faster at every moment.
Part (e): A better comparison of early reaction rates
To compare early reaction rates, use concentration-time measurements near the start and calculate their gradients over a short, common time interval.
The supplied 0-10 second data give:
\[
\begin{aligned}
\bar r_A&=0.00120\ \mathrm{mol\,L^{-1}\,s^{-1}}\\
\bar r_B&=0.000800\ \mathrm{mol\,L^{-1}\,s^{-1}}
\end{aligned}
\]
Condition A has the greater average rate over this early interval.
Under the question’s piecewise-linear model, these gradients also represent the rates within that first interval.
For a real reaction, more closely spaced measurements near time zero would provide a better estimate of the initial instantaneous rate.
The tempting conclusion is that there must be one universally correct answer to “Which reaction is faster?”
There isn’t, unless you specify what rate you’re comparing and when.
The endpoint must match the scientific question. Otherwise, even a carefully controlled experiment can measure a different aspect of the reaction from the one you intended.
09What to decide before your next disappearing-cross experiment
Before collecting any times, decide exactly what disappearance means.
Use one repeatable visibility criterion, a constant optical path length, and enough reactant to ensure the cross can actually disappear. Keep the independent variable separate from other factors that affect both reaction speed and what the observer sees.
Then choose your interpretation:
- If you’re comparing times to the same visibility threshold, report disappearance times and, where appropriate, \(1/t\) as a relative average-rate indicator.
- If you’re calculating average chemical rates, measure the change in concentration or another product quantity over a stated time interval.
- If you’re investigating the rate at a particular moment, obtain a sufficiently detailed time-series graph and estimate the tangent gradient.
The next useful question is often not “How quickly did the cross disappear?” but “How did the reaction rate change before it disappeared?”
That is where continuous concentration, mass, or gas-volume measurements become more informative than a single stopwatch reading.