Why a Faster Initial Reaction Rate Can Give Less Product
A faster initial reaction doesn't always produce more product. Learn how to compare reaction rates and final yields using gas-volume graphs, worked calculations, catalyst experiments, and HSC Chemistry practice questions.
Two flasks contain exactly the same amount of hydrogen peroxide. One produces oxygen bubbles rapidly, while the other reacts much more slowly. Which flask will produce more oxygen in the end?
If both reactions go to completion by the same chemical pathway, they will produce the same amount. The faster reaction simply gets there sooner.
But suppose the faster flask starts with less hydrogen peroxide. It could produce oxygen more quickly at first and still finish with less gas.
That’s the key distinction: initial reaction rate measures how quickly product forms, while final yield depends on how much product the available reactants can produce, and whether the reaction actually reaches that amount.
01Testing the difference with a catalyst
Consider the decomposition of hydrogen peroxide:
\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]
Two moles of hydrogen peroxide produce one mole of oxygen gas.
Manganese dioxide, \(\ce{MnO2}\), acts as a catalyst. It speeds up the decomposition without being consumed in the overall reaction.
Imagine testing whether using more manganese dioxide increases the amount of oxygen produced.
Why the first experimental design could mislead us
An obvious method is to pour hydrogen peroxide into a conical flask, add manganese dioxide, quickly fit a rubber stopper, and collect the oxygen in a gas syringe.
There’s a problem.
The faster the reaction, the more oxygen might escape before the stopper is fitted. The experiment could then record a smaller final gas volume for the faster reaction, even though more oxygen was not necessarily produced or lost chemically.
The apparatus would be measuring how much oxygen was collected, not necessarily how much formed.
A better arrangement is to place the hydrogen peroxide in the flask and keep the manganese dioxide in a small upright vial inside it, separate from the liquid.
Fit the stopper first. Connect the flask to a horizontal gas syringe using airtight tubing. Then tilt the sealed flask so the catalyst spills into the peroxide, and start the timer.
Oxygen forms in the flask, travels through the delivery tube, and pushes the syringe plunger outwards. The syringe reading gives the collected gas volume.
This revision removes the delay between adding the catalyst and sealing the apparatus.
For a fair catalyst comparison, also keep the hydrogen peroxide concentration and volume, temperature, manganese dioxide particle size, mixing conditions, and gas collection conditions consistent.
We will use 25°C and 100 kPa. Assuming ideal, dry oxygen, its molar volume is:
\[ V_m=24.8\ \mathrm{L\,mol^{-1}} \]
Here, \(V_m\) is the volume occupied by one mole of gas at the stated temperature and pressure.
All experimental readings below are illustrative, not results from a reported experiment. We assume no gas leaks, negligible oxygen dissolution, and complete conversion where stated.
02Experiment 1: Faster reaction, same final amount
Consider two runs with identical amounts of hydrogen peroxide but different amounts of manganese dioxide.
| Condition | Run A | Run B |
|---|---|---|
| Hydrogen peroxide volume | 40.0 mL | 40.0 mL |
| Hydrogen peroxide concentration | 0.250 mol L\(^{-1}\) | 0.250 mol L\(^{-1}\) |
| Manganese dioxide mass | 0.050 g | 0.200 g |
What should we predict?
Run B has more catalyst available. With otherwise comparable conditions, we expect it to produce oxygen faster.
But both flasks contain the same number of hydrogen peroxide molecules. The catalyst cannot create additional oxygen atoms.
Here are illustrative gas-volume readings.
| Time after mixing (s) | Run A: oxygen (mL) | Run B: oxygen (mL) |
|---|---|---|
| 0 | 0 | 0 |
| 10 | 12 | 38 |
| 20 | 23 | 65 |
| 40 | 41 | 96 |
| 80 | 69 | 118 |
| 160 | 99 | 124 |
| After complete decomposition | 124 | 124 |
Run B is much faster initially. By 40 seconds, it has already collected 96 mL, compared with only 41 mL from Run A.
Yet both eventually reach 124 mL.
The graph contains two different kinds of information.
- The slope tells us how quickly gas is being collected.
- The final plateau tells us the total gas collected once the reaction has finished.
A useful comparison is two taps filling containers. A powerful tap can fill a small container quickly, while a slower tap takes longer to fill a larger one. Flow rate doesn’t determine the container’s capacity.
In the reaction, the rate is like the flow of water. The available reactant determines how much product can potentially form.
The analogy isn’t exact: chemical reactions are governed by mole ratios, and reversible reactions may stop at equilibrium rather than complete conversion.
Worked example: Does more catalyst produce more oxygen?
Run A contains 40.0 mL of 0.250 mol L\(^{-1}\) hydrogen peroxide and 0.050 g manganese dioxide. Run B contains the same hydrogen peroxide solution and 0.200 g manganese dioxide.
In the first 10.0 seconds, A produces 12.0 mL of oxygen, while B produces 38.0 mL. Calculate their average gas-collection rates and the theoretical oxygen volume from each flask at 25°C and 100 kPa.
Step 1
Average rate is the change in gas volume divided by the elapsed time:
\[ \bar r_V=\frac{\Delta V_g}{\Delta t} \]
Here, \(\bar r_V\) is the average gas-collection rate, \(\Delta V_g\) is the increase in gas volume, and \(\Delta t\) is the time interval.
For Run A:
\[ \begin{aligned} \bar r_{V,A}&=\frac{12.0}{10.0}\\ &=1.20\ \mathrm{mL\,s^{-1}} \end{aligned} \]
For Run B:
\[ \begin{aligned} \bar r_{V,B}&=\frac{38.0}{10.0}\\ &=3.80\ \mathrm{mL\,s^{-1}} \end{aligned} \]
Run B’s average oxygen-collection rate over this interval is more than three times that of A.
Step 2
Use:
\[ n=CV \]
Here, \(n\) is amount in moles, \(C\) is concentration in mol L\(^{-1}\), and \(V\) is solution volume in litres.
Both flasks contain:
\[ \begin{aligned} n(\ce{H2O2})&=0.250\times0.0400\\ &=0.0100\ \mathrm{mol} \end{aligned} \]
Step 3
Two moles of hydrogen peroxide produce one mole of oxygen.
\[ \begin{aligned} n(\ce{O2})&=\frac{0.0100}{2}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]
The theoretical gas volume, \(V_g\), is:
\[ \begin{aligned} V_g&=n(\ce{O2})V_m\\ &=0.00500\times24.8\\ &=0.124\ \mathrm{L}\\ &=124\ \mathrm{mL} \end{aligned} \]
Step 4
Run B produces oxygen much faster at first, but both flasks can produce 124 mL of oxygen from the starting hydrogen peroxide.
At 160 seconds, Run A has collected only 99 mL because it is still reacting. That does not mean its theoretical yield is lower.
Average rate is not instantaneous rate
There is one important detail in that calculation.
The values 1.20 and 3.80 mL s\(^{-1}\) are average rates from 0 to 10 seconds. They are not exact instantaneous rates at \(t=0\).
To find the instantaneous rate at a particular moment, draw a tangent to the gas-volume curve at that time and calculate its slope.
A tangent describes the rate at one instant. A line joining two measured points gives an average rate across an interval.
You cannot determine an exact instantaneous rate at \(t=0\) from only the first two table readings. You can calculate the average rate over that first interval.
For more on the distinction, see How to Find Instantaneous Rate Using a Tangent in HSC Chemistry.
If a reactant concentration is being measured instead, it usually decreases with time. We write its disappearance rate as \(-\Delta[\ce{H2O2}]/\Delta t\) so that the reported rate is positive. The gas-volume rate is positive because the collected volume increases.
These quantities have different units. Converting between peroxide consumption and oxygen production also requires the 2:1 mole ratio.
03Experiment 2: Faster initially, but less oxygen overall
Now change the amount of hydrogen peroxide as well as its concentration.
Keep Run A from the previous experiment and introduce Run C.
| Condition | Run A | Run C |
|---|---|---|
| Hydrogen peroxide concentration | 0.250 mol L\(^{-1}\) | 0.500 mol L\(^{-1}\) |
| Hydrogen peroxide volume | 40.0 mL | 10.0 mL |
| Manganese dioxide mass | 0.050 g | 0.050 g |
Run C has twice the hydrogen peroxide concentration.
A student might immediately predict that C will react faster and produce more oxygen.
The first prediction is plausible: a higher peroxide concentration means more peroxide molecules are available in a given volume to interact with the catalyst surface.
But the second prediction ignores how much solution is present.
Run C has only one-quarter of Run A’s solution volume.
Consider these illustrative observations.
| Time (s) | Run A: oxygen (mL) | Run C: oxygen (mL) |
|---|---|---|
| 0 | 0 | 0 |
| 10 | 12 | 20 |
| 40 | 41 | 49 |
| 80 | 69 | 59 |
| 160 | 99 | 62 |
| After complete decomposition | 124 | 62 |
Run C produces more oxygen than A during the first 10 seconds: 20 mL compared with 12 mL.
It remains ahead at 40 seconds.
By 80 seconds, however, Run A has overtaken it.
Run C has nearly exhausted its smaller supply of hydrogen peroxide, while A still has peroxide available to decompose.
This is the clearest counterexample to the idea that faster initial rate means greater final yield.
Worked example: The faster flask has less reactant
Run C contains 10.0 mL of 0.500 mol L\(^{-1}\) hydrogen peroxide and collects 20.0 mL of oxygen in its first 10.0 seconds.
Run A collects 12.0 mL in the same interval and has the conditions given previously.
Calculate Run C’s average initial gas-collection rate, its theoretical oxygen volume, and the point at which the gas-volume data show Run A overtaking C.
Step 1
\[ \begin{aligned} \bar r_{V,C}&=\frac{20.0}{10.0}\\ &=2.00\ \mathrm{mL\,s^{-1}} \end{aligned} \]
Run C has the higher average gas-collection rate over the first 10 seconds:
\[ 2.00>1.20\ \mathrm{mL\,s^{-1}} \]
Step 2
\[ \begin{aligned} n(\ce{H2O2})&=CV\\ &=0.500\times0.0100\\ &=0.00500\ \mathrm{mol} \end{aligned} \]
That’s only half the amount in Run A.
Step 3
\[ \begin{aligned} n(\ce{O2})&=\frac{0.00500}{2}\\ &=0.00250\ \mathrm{mol} \end{aligned} \]
Then:
\[ \begin{aligned} V_g&=0.00250\times24.8\\ &=0.0620\ \mathrm{L}\\ &=62.0\ \mathrm{mL} \end{aligned} \]
Step 4
At 40 seconds:
- Run A has produced 41 mL.
- Run C has produced 49 mL.
At 80 seconds:
- Run A has produced 69 mL.
- Run C has produced 59 mL.
Therefore, A overtakes C sometime between 40 and 80 seconds.
The table cannot give the exact crossing time without additional information or an assumed model for the curves.
Step 5
Run C begins with a higher gas-collection rate but only half the number of peroxide moles.
The measured early rate determines how quickly the available peroxide is being converted into oxygen. The balanced equation and available peroxide determine the theoretical total.
There is also an experimental limitation: A and C have different liquid volumes, which may change mixing and contact with the catalyst. The observations establish which run was faster initially, but they do not prove that concentration alone caused the rate difference.
04How to avoid confusing rate, amount, and yield
These are three questions that may look similar but require different evidence.
| What are you trying to determine? | What should you examine? |
|---|---|
| Which reaction is faster initially? | The initial tangent slope, or an early average rate if only discrete data are available |
| Which has produced more at a particular time? | The actual gas volume or product amount at that time |
| Which can produce more in total? | Initial reactant moles, the balanced equation, limiting reagents, and any equilibrium or competing reactions |
For a gas-volume graph at constant temperature and pressure, a steeper curve means a greater rate of gas collection.
A higher final plateau means more gas was collected.
But a plateau doesn’t automatically prove that all reactant has been consumed. A catalyst may become inactive, mixing may stop, or the reaction may be limited by equilibrium or another condition.
Also distinguish theoretical yield from actual yield.
Theoretical yield is the amount predicted from the available reactants and reaction stoichiometry under the assumed pathway. Actual yield is the amount of product obtained. If actual yield is smaller, the difference needs an explanation.
A catalyst does not change the theoretical amount given by a particular balanced reaction and fixed initial reactant quantities. However, it can affect how much product is obtained within a limited time. In a reversible reaction at fixed conditions, a catalyst speeds up both directions without changing the equilibrium composition.
If the gas-volume graph is unfamiliar, How to Calculate Reaction Rate from a Gas-Volume-Time Graph develops the slope calculations in more detail.
05Questions and solutions
Question 1
Two reactions produce oxygen, which is collected in gas syringes at the same temperature and pressure.
Both syringes initially read zero.
After 15.0 seconds, Run D has collected 18.0 mL and Run E has collected 30.0 mL.
After both reactions finish, the final volumes are 80.0 mL for D and 65.0 mL for E.
Calculate the average gas-collection rate over the first 15 seconds for each run. Identify which reaction is initially faster and which produces more oxygen overall.
Solution 1
Run E has the higher average gas-collection rate during the first 15 seconds, but Run D produces more oxygen overall.
For D:
\[ \begin{aligned} \bar r_{V,D}&=\frac{18.0}{15.0}\\ &=1.20\ \mathrm{mL\,s^{-1}} \end{aligned} \]
For E:
\[ \begin{aligned} \bar r_{V,E}&=\frac{30.0}{15.0}\\ &=2.00\ \mathrm{mL\,s^{-1}} \end{aligned} \]
Thus, E is faster over the measured initial interval.
However, D finishes at 80.0 mL, compared with E’s 65.0 mL.
D collects 15.0 mL more oxygen overall.
The tempting mistake is to assume that E’s faster start means it must finish with more product. The early rate describes how quickly gas accumulates, not the total amount available for production.
Question 2
Two runs each begin with 0.00800 mol of hydrogen peroxide and different amounts of manganese dioxide.
The reaction is:
\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]
Both runs use airtight gas syringes, and oxygen is measured at 25°C and 100 kPa.
The results are:
| Observation | Run F | Run S |
|---|---|---|
| Oxygen after 10.0 s | 30.0 mL | 15.0 mL |
| Oxygen after 90.0 s | 78.0 mL | 69.0 mL |
| Behaviour after 90 s | Remains at 78.0 mL | Still increasing |
A chemical test confirms that unreacted hydrogen peroxide remains in Run F. Mixing is maintained, and no leak is detected.
Fresh manganese dioxide is then introduced into F through a sealed port. A further 21.2 mL of oxygen is collected before the reaction finishes.
Run S eventually produces 99.2 mL.
Explain whether F has a lower theoretical yield, calculate the theoretical oxygen volume, and identify what the additional catalyst experiment establishes.
Solution 2
Run F does not have a lower theoretical oxygen yield. Both runs can theoretically produce 99.2 mL because they begin with the same amount of hydrogen peroxide.
First, calculate the oxygen moles:
\[ \begin{aligned} n(\ce{O2})&=\frac{0.00800}{2}\\ &=0.00400\ \mathrm{mol} \end{aligned} \]
Convert to gas volume:
\[ \begin{aligned} V_g&=0.00400\times24.8\\ &=0.0992\ \mathrm{L}\\ &=99.2\ \mathrm{mL} \end{aligned} \]
F has a faster early average gas-collection rate:
\[ \begin{aligned} \bar r_{V,F}&=\frac{30.0}{10.0}\\ &=3.00\ \mathrm{mL\,s^{-1}},\\ \bar r_{V,S}&=\frac{15.0}{10.0}\\ &=1.50\ \mathrm{mL\,s^{-1}} \end{aligned} \]
But F’s plateau at 78.0 mL does not establish complete decomposition.
The positive peroxide test shows reactant remains. Adding fresh manganese dioxide restores gas production, so the earlier stoppage is consistent with the original catalyst becoming inactive or unavailable for effective contact.
The new total is:
\[ 78.0+21.2=99.2\ \mathrm{mL} \]
This matches the theoretical oxygen volume.
The tempting interpretation is that a flat graph must mean the reaction has exhausted its limiting reactant. Here, the independent peroxide test and renewed gas production rule that out.
The fresh-catalyst result does not identify the exact reason the original catalyst stopped working. Surface blockage, loss of activity, or poor accessibility would require further investigation.
Question 3
Two open conical flasks containing calcium carbonate react with hydrochloric acid. The flasks sit on electronic balances, and the carbon dioxide escapes.
The reaction is:
\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]
The experiment uses the following conditions.
| Condition | Run P | Run Q |
|---|---|---|
| Calcium carbonate mass | 0.200 g | 0.150 g |
| Calcium carbonate form | Fine powder | Large chips |
| Hydrochloric acid volume | 25.0 mL | 25.0 mL |
| Hydrochloric acid concentration | 0.100 mol L\(^{-1}\) | 0.200 mol L\(^{-1}\) |
| Mass lost in first 15.0 s | 0.020 g | 0.012 g |
Assume all measured mass loss is due to escaping carbon dioxide, evaporation is negligible, and each reaction continues until its limiting reactant is consumed.
Use \(M(\ce{CaCO3})=100.0\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\), where \(M\) represents molar mass.
(a) Which run has the higher average initial mass-loss rate?
(b) Identify the limiting reactant in each run.
(c) Calculate the final mass loss in each run and explain why the initial-rate comparison does not determine the final result.
Solution 3
Run P has the faster initial mass-loss rate, but Run Q produces more carbon dioxide overall. Hydrochloric acid limits P, while calcium carbonate limits Q.
(a) Compare the early average rates.
For P:
\[ \begin{aligned} \bar r_{m,P}&=\frac{0.020}{15.0}\\ &=0.00133\ \mathrm{g\,s^{-1}} \end{aligned} \]
For Q:
\[ \begin{aligned} \bar r_{m,Q}&=\frac{0.012}{15.0}\\ &=0.000800\ \mathrm{g\,s^{-1}} \end{aligned} \]
Here, \(\bar r_m\) is the average rate of mass loss. It is positive when defined as mass lost per unit time.
P loses mass faster during the first 15 seconds, so carbon dioxide escapes more rapidly during that interval.
(b) Find the limiting reactants.
Run P contains:
\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.200}{100.0}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]
Its acid amount is:
\[ \begin{aligned} n(\ce{HCl})&=0.100\times0.0250\\ &=0.00250\ \mathrm{mol} \end{aligned} \]
The balanced equation requires two moles of HCl for each mole of calcium carbonate.
Therefore, P’s acid can react with only:
\[ \frac{0.00250}{2}=0.00125\ \mathrm{mol} \]
of calcium carbonate.
HCl is limiting in P.
For Q:
\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.150}{100.0}\\ &=0.00150\ \mathrm{mol} \end{aligned} \]
The acid amount is:
\[ \begin{aligned} n(\ce{HCl})&=0.200\times0.0250\\ &=0.00500\ \mathrm{mol} \end{aligned} \]
Reacting all the carbonate would require \(2\times0.00150=0.00300\) mol of acid.
Q has more than enough acid, so calcium carbonate is limiting.
(c) Calculate the total carbon dioxide produced.
For P, the acid determines the carbon dioxide moles:
\[ \begin{aligned} n(\ce{CO2})&=\frac{0.00250}{2}\\ &=0.00125\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ \begin{aligned} m(\ce{CO2})&=nM\\ &=0.00125\times44.0\\ &=0.0550\ \mathrm{g} \end{aligned} \]
For Q, the calcium carbonate determines the carbon dioxide moles:
\[ n(\ce{CO2})=0.00150\ \mathrm{mol} \]
So:
\[ \begin{aligned} m(\ce{CO2})&=0.00150\times44.0\\ &=0.0660\ \mathrm{g} \end{aligned} \]
Q eventually loses more mass, despite having less calcium carbonate and a slower early mass-loss rate.
There are two traps here. First, the carbonate mass alone doesn’t determine the yield because acid is limiting in P. Second, fine powder and acid concentration both differ, so the observed rate difference cannot be attributed to particle size alone.
The limiting-reagent calculation settles the final amount independently of the early rate measurement.
Question 4
A student compares oxygen production from two reactions using gas syringes.
The gas conditions are different. Each syringe contains dry oxygen at the stated constant temperature and pressure throughout its run.
| Measurement | Run A | Run B |
|---|---|---|
| Temperature | 300 K | 330 K |
| Pressure | 100 kPa | 100 kPa |
| Oxygen volume after 20.0 s | 20.0 mL | 21.0 mL |
| Final oxygen volume | 100.0 mL | 105.0 mL |
Both readings begin at zero. The runs may have different initial reactant amounts.
Assume ideal-gas behaviour, no leaks, and negligible oxygen dissolution.
Use:
\[ PV=nRT \]
where \(P\) is pressure in kPa, \(V\) is gas volume in litres, \(n\) is gas amount in moles, \(T\) is absolute temperature in kelvin, and \(R=8.31\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).
(a) Which run has the larger average gas-volume increase per second over the first 20 seconds?
(b) Which run actually forms oxygen faster in mol s\(^{-1}\) during that interval?
(c) Which run produces more moles of oxygen overall?
Explain why the raw syringe-volume comparison can be misleading.
Solution 4
Run B has the larger early gas-volume increase, but Run A actually produces oxygen faster in moles per second and produces more oxygen moles overall.
The crucial issue is that gas volume also depends on temperature.
(a) Compare the volume-based average rates.
For A:
\[ \bar r_{V,A}=\frac{20.0}{20.0} =1.00\ \mathrm{mL\,s^{-1}} \]
For B:
\[ \bar r_{V,B}=\frac{21.0}{20.0} =1.05\ \mathrm{mL\,s^{-1}} \]
B appears faster if we compare only syringe volume per second.
(b) Convert each early gas volume into moles.
Rearrange the ideal gas equation:
\[ n=\frac{PV}{RT} \]
For A:
\[ \begin{aligned} n_A&=\frac{100\times0.0200}{8.31\times300}\\ &=0.000802\ \mathrm{mol} \end{aligned} \]
Its average molar production rate is:
\[ \begin{aligned} \bar r_{n,A}&=\frac{0.000802}{20.0}\\ &=4.01\times10^{-5}\ \mathrm{mol\,s^{-1}} \end{aligned} \]
For B:
\[ \begin{aligned} n_B&=\frac{100\times0.0210}{8.31\times330}\\ &=0.000766\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ \begin{aligned} \bar r_{n,B}&=\frac{0.000766}{20.0}\\ &=3.83\times10^{-5}\ \mathrm{mol\,s^{-1}} \end{aligned} \]
Here, \(\bar r_n\) is the average oxygen production rate in moles per second.
A actually forms oxygen faster over the first 20 seconds.
(c) Compare the final amounts.
For A:
\[ \begin{aligned} n_{A,\text{final}} &=\frac{100\times0.1000}{8.31\times300}\\ &=0.00401\ \mathrm{mol} \end{aligned} \]
For B:
\[ \begin{aligned} n_{B,\text{final}} &=\frac{100\times0.1050}{8.31\times330}\\ &=0.00383\ \mathrm{mol} \end{aligned} \]
A produces more moles of oxygen overall.
The plausible but incorrect approach is to treat millilitres as a direct comparison of moles when the temperatures differ.
A given amount of gas occupies more volume at a higher temperature and the same pressure. We must convert to moles before comparing actual chemical production rates or amounts.
A higher temperature can increase reaction rate when other relevant conditions are controlled, but these data do not establish such a controlled comparison.
Question 5
Two sealed flasks each start with 0.0100 mol of hydrogen peroxide.
Both contain manganese dioxide to catalyse oxygen production. Run B also contains 0.00500 mol of sulfite ions, \(\mathrm{SO_{3}}^{2-}\), which can react with hydrogen peroxide.
Under the experimental conditions, assume that only these two irreversible reactions occur:
Oxygen-producing pathway:
\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]
Competing oxidation pathway:
\[ \ce{H2O2(aq) + SO3^2-(aq) -> SO4^2-(aq) + H2O(l)} \]
The sulfate produced remains stable, and the reactions do not reverse. There is no gas loss, and collected oxygen is measured at 25°C and 100 kPa.
Run A contains no sulfite. It collects 12.0 mL of oxygen in its first 10.0 seconds and eventually produces 124.0 mL.
Run B collects 20.0 mL of oxygen during its first 10.0 seconds. A separate analysis at 10.0 seconds shows that 0.00100 mol of sulfate has already formed.
At the end of Run B, there is no hydrogen peroxide or sulfite remaining. The final measurements show 62.0 mL of oxygen and 0.00500 mol of sulfate.
(a) Using only the information available at 10.0 seconds, determine the greatest total oxygen volume that Run B could possibly produce, even if none of its remaining peroxide reacts with sulfite.
(b) Show that Run B’s final oxygen and sulfate quantities account for all the initial hydrogen peroxide.
(c) Compare the early average oxygen-production rates and explain why Run B’s faster oxygen production does not contradict its smaller final oxygen yield.
(d) Explain why this result cannot be explained as merely adding more catalyst to an otherwise identical reaction.
Solution 5
Run B cannot reach Run A’s 124.0 mL oxygen total, despite producing oxygen faster during the first 10 seconds. The decisive fact is that some of B’s hydrogen peroxide has already been irreversibly consumed by a pathway that produces no oxygen.
(a) Find the highest possible oxygen yield using the 10-second evidence.
At 10 seconds, B has produced 0.00100 mol of sulfate.
The competing reaction has a 1:1 mole ratio between hydrogen peroxide and sulfate.
Therefore, at least 0.00100 mol of hydrogen peroxide has already been used in the oxygen-free pathway.
That peroxide cannot later produce oxygen, because the reaction is assumed irreversible and the sulfate remains stable.
The greatest amount of peroxide that could ever contribute to oxygen formation is therefore:
\[ \begin{aligned} n_{\max}(\ce{H2O2}) &=0.0100-0.00100\\ &=0.00900\ \mathrm{mol} \end{aligned} \]
Even in the most favourable case, the greatest oxygen amount would be:
\[ \begin{aligned} n_{\max}(\ce{O2}) &=\frac{0.00900}{2}\\ &=0.00450\ \mathrm{mol} \end{aligned} \]
Convert to gas volume:
\[ \begin{aligned} V_{\max}&=0.00450\times24.8\\ &=0.1116\ \mathrm{L}\\ &=111.6\ \mathrm{mL} \end{aligned} \]
This is the greatest total oxygen volume B can reach, including the oxygen already produced.
Run A eventually produces 124.0 mL, so B cannot match A.
Notice the subtlety: B’s early lead in collected oxygen doesn’t remove the constraint imposed by its competing reaction.
The 0.00100 mol of sulfate is evidence of peroxide that has permanently been diverted away from oxygen production.
(b) Account for the final products.
Run B eventually produces 62.0 mL of oxygen.
Convert this to moles:
\[ \begin{aligned} n(\ce{O2})&=\frac{0.0620}{24.8}\\ &=0.00250\ \mathrm{mol} \end{aligned} \]
The oxygen-producing reaction consumes two moles of peroxide per mole of oxygen.
Therefore, peroxide consumed through decomposition is:
\[ \begin{aligned} n_{\text{decomp}} &=2\times0.00250\\ &=0.00500\ \mathrm{mol} \end{aligned} \]
Here, \(n_{\text{decomp}}\) is the amount of peroxide consumed in the oxygen-producing pathway.
The final sulfate amount is 0.00500 mol, requiring another 0.00500 mol of peroxide through the competing reaction.
Total peroxide consumed:
\[ \begin{aligned} n_{\text{total}} &=0.00500+0.00500\\ &=0.0100\ \mathrm{mol} \end{aligned} \]
This exactly equals the initial peroxide amount.
The final measurements are consistent with complete peroxide consumption through the two stated pathways.
Half the peroxide is used in oxygen production, and half is used to oxidise sulfite.
(c) Compare the early oxygen-production rates.
For A:
\[ \begin{aligned} \bar r_{V,A}&=\frac{12.0}{10.0}\\ &=1.20\ \mathrm{mL\,s^{-1}} \end{aligned} \]
For B:
\[ \begin{aligned} \bar r_{V,B}&=\frac{20.0}{10.0}\\ &=2.00\ \mathrm{mL\,s^{-1}} \end{aligned} \]
B has the faster average oxygen-production rate in the first 10 seconds.
This is entirely compatible with B producing less oxygen in total.
Its first 10 seconds tell us how quickly oxygen is forming during that interval. They do not tell us how much peroxide will ultimately be directed into the oxygen-producing pathway.
There is an additional distinction between the rate of oxygen formation and the total rate of peroxide consumption.
For example, by 10 seconds B has produced:
\[ \begin{aligned} n(\ce{O2})&=\frac{0.0200}{24.8}\\ &=0.000806\ \mathrm{mol} \end{aligned} \]
This required approximately \(2\times0.000806=0.00161\) mol of peroxide.
But B also consumed 0.00100 mol of peroxide to form sulfate.
So its total peroxide consumption over those 10 seconds was approximately:
\[ \begin{aligned} n_{\text{consumed}} &=0.00161+0.00100\\ &=0.00261\ \mathrm{mol} \end{aligned} \]
Measuring oxygen alone would not reveal all the peroxide being consumed when a competing pathway exists.
(d) Explain why a catalyst-only model is insufficient.
If both flasks contained only peroxide and manganese dioxide, and all peroxide decomposed completely through the same balanced reaction, both would theoretically produce:
\[ \begin{aligned} V_g&=\frac{0.0100}{2}\times24.8\\ &=0.124\ \mathrm{L}\\ &=124\ \mathrm{mL} \end{aligned} \]
Changing the amount of catalyst alone would not change that theoretical volume.
Run B is fundamentally different because it contains sulfite, an additional reactant that consumes peroxide without producing oxygen.
The deeper lesson is that a product-formation rate is not automatically the same as the disappearance rate of a reactant when several chemical pathways are available.
In a real experiment, an unexpectedly low final gas volume should lead to a new investigation: was the limiting reactant completely consumed, was gas lost, was equilibrium reached, or did some reactant form a different product?
Those measurements, rather than the initial slope alone, determine what the final yield means.