How to Tell if a Catalyst Changed Reaction Rate or Yield

Two flasks contain the same amount of hydrogen peroxide. After 60 seconds, the flask with a catalyst has produced 54 mL of oxygen gas. The flask without a catalyst has produced only 27 mL.

Has the catalyst doubled the final yield of oxygen?

No. It has produced twice as much so far. That is not the same as producing twice as much when the reaction finishes.

To tell whether a catalyst changed the rate or final yield from a graph, examine two different features:

• Rate: Look at the gradient (slope). A steeper concentration-time or gas-volume-time curve generally indicates a faster change in the measured quantity.
• Final yield: Look at the genuine final plateau, once both reactions have reached their endpoints. If both curves finish at the same gas volume, they have produced the same collected amount under matching conditions, even if one gets there much faster.

Here’s a prediction worth making before looking at the full data. If both flasks started with identical amounts of hydrogen peroxide and eventually decomposed it completely, should they produce different total amounts of oxygen?

They should not. The catalyst changes how quickly oxygen forms, not how many oxygen molecules can form from a fixed amount of reactant.

01How an experiment separates rate from final yield

Consider the decomposition of hydrogen peroxide:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

Two moles of hydrogen peroxide produce one mole of oxygen gas.

An iodide-containing solution can catalyse this decomposition. The iodide provides an alternative reaction pathway, allowing oxygen to form more quickly.

We want to test whether the catalyst changes the speed of oxygen production, the final amount collected, or both.

Set up the experiment

Picture the apparatus from left to right.

A conical flask contains a measured volume and concentration of hydrogen peroxide. The flask sits in a temperature-controlled water bath.

An airtight bung seals the flask. A delivery tube passes through the bung and connects to a graduated gas syringe. A separate injection port allows a solution to enter without opening the flask.

At the start of the experiment, a measured volume of catalyst solution is injected. Oxygen forms in the flask, travels through the delivery tube, and pushes the syringe plunger outwards.

The distance moved by the plunger gives the volume of oxygen collected.

A control experiment uses the same amount of hydrogen peroxide and the same volume of added solvent, but no catalyst.

The starting reactant amounts, total solution volume, temperature, gas pressure, and gas-collection conditions should be matched.

What would the first design show?

Suppose you stopped both experiments after exactly 60 seconds.

The catalysed reaction would have produced 54 mL of oxygen, while the uncatalysed reaction would have produced 27 mL.

You could conclude that more oxygen was collected from the catalysed reaction during that minute.

But you could not conclude that its final yield was greater.

The missing information is what happens after 60 seconds.

The control might still be producing oxygen, while the catalysed reaction is almost finished.

The improved design: follow both reactions to their endpoints

Instead of stopping at an arbitrary time, continue recording the oxygen volume until each reading becomes stable over repeated measurements.

Here is a complete set of illustrative data. These are constructed values for learning how to interpret graphs, not measurements from an actual hydrogen peroxide experiment.

Further readings remain at 60 mL for both runs.

What does this tell us?

The catalyst allows the reaction to reach its observed endpoint much sooner. But both reactions eventually produce the same collected oxygen volume: 60 mL.

The apparent advantage at 60 seconds disappears when the control is allowed to finish.

This is the central pattern to recognise:

A catalyst can change how quickly a curve reaches its plateau without changing the plateau itself.

For an effectively complete reaction, the amount of limiting reactant determines the theoretical product yield. For a reversible reaction at equilibrium, the equilibrium composition is determined by the starting conditions and equilibrium constant, not by the presence of a catalyst.

02Read the gradient and plateau separately

A graph contains information about both how much reaction has occurred and how quickly it is occurring.

These are different quantities.

The gradient tells you the rate

For a graph of gas volume \(V\) against time \(t\), the average rate of gas collection over a time interval is:

\[ \bar r_V=\frac{\Delta V}{\Delta t} \]

where:

• \(\bar r_V\) is the average gas-collection rate, usually in mL/s.
• \(\Delta V\) is the change in collected gas volume, in mL.
• \(\Delta t\) is the elapsed time, in seconds.

A large positive gradient indicates rapid gas collection.

A small positive gradient indicates slower gas collection.

A horizontal section has a gradient of approximately zero: the measured volume is no longer increasing.

Notice that the volume can be 60 mL while the gradient is zero. The syringe still contains oxygen, even though it is no longer filling.

Average rate is not instantaneous rate

There is another trap hidden in the word rate.

The calculation over the first 20 seconds gives an average. It does not tell you exactly how quickly oxygen was being produced at 20 seconds.

To estimate the instantaneous rate at a particular time, draw a tangent to the curve at that time.

A tangent is a straight line matching the curve’s slope at the chosen point. Its gradient gives the instantaneous rate.

For a continuous gas-volume curve:

\[ r_V(t)=\frac{dV}{dt} \]

Here \(r_V(t)\) means the instantaneous gas-collection rate at time \(t\), and \(dV/dt\) is the gradient of the tangent.

Both average and instantaneous rates can be expressed in mL/s. The difference is the time interval they describe.

A table with readings every 20 seconds allows you to calculate average rates over those intervals. It does not normally provide enough information to determine an exact instantaneous rate at a single moment.

If you are given only tabulated data, a short-interval gradient may estimate an instantaneous rate, but you should identify it as an approximation.

For a more detailed treatment of secants, tangents, and gas-volume calculations, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

03A surprising result: the catalysed curve can be flatter later

You might expect the catalysed curve to be steeper at every moment.

But that is not necessarily what you will observe.

Imagine two runners approaching the same finish line. One runs towards it very quickly and has almost stopped by the time the other is still moving steadily.

The analogy helps with the graph, but it has a limit. At chemical equilibrium, particles do not stop reacting. Forward and reverse reactions continue at equal rates, so the overall concentrations remain constant.

In a reversible reaction, the catalysed mixture may already be close to equilibrium while the uncatalysed mixture is still moving towards it.

At the same clock time, the uncatalysed curve can therefore have a larger net gradient.

This does not mean the catalyst has stopped working.

It means the two reaction mixtures are at different stages.

04What if the graph shows concentration or mass instead?

The method stays the same, but you must interpret the direction of the curve correctly.

Reactant concentration decreases

Suppose the concentration of reactant A falls from 0.80 mol/L to 0.50 mol/L in 2.0 minutes.

The average rate of disappearance of A is:

\[ \begin{aligned} \bar r_A&=-\frac{\Delta[A]}{\Delta t}\\ &=-\frac{0.50-0.80}{2.0}\\ &=0.15\ \mathrm{mol\,L^{-1}\,min^{-1}} \end{aligned} \]

The negative sign makes the disappearance rate positive.

The graph itself has a negative gradient because concentration is falling.

If the balanced equation contains different stoichiometric coefficients, remember that the formation rate of one species may differ from the disappearance rate of another. For example, in \(\ce{A -> 2B}\), B forms twice as quickly as A disappears when both rates are expressed as changes in molar concentration per unit time at constant volume.

Mass can decrease while product is forming

Suppose a reaction releases gas into the surrounding air from an open flask sitting on a balance.

The mass reading falls because gas escapes.

A faster reaction may produce a steeper downward mass-time curve.

However, the mass remaining is not itself the mass of gas produced.

You must calculate:

\[ m_{\mathrm{gas}}=m_{\mathrm{initial}}-m_{\mathrm{final}} \]

This assumes that the gas is the only substance leaving the apparatus.

If the entire apparatus is sealed and weighed, its total mass remains constant even while gas forms inside.

05When different plateaus do not prove that a catalyst changed yield

Suppose the catalysed curve finishes at 48 mL while the uncatalysed curve finishes at 60 mL.

Can you conclude that the catalyst reduced the final yield?

Not from those two numbers alone.

Before interpreting a difference in plateau heights, ask these questions.

1. Have both curves actually finished?

A line can look almost horizontal because the reaction has become very slow.

A short observation period does not necessarily establish the final amount produced.

You need sufficiently long measurements, ideally supported by another check such as the amount of reactant remaining.

2. Did both reactions start with the same limiting reactant amount?

If one flask contained less hydrogen peroxide, it could produce less oxygen even if its reaction was faster.

A smaller final gas volume might reflect the starting amounts rather than any catalytic effect.

3. Was the product collected completely?

Gas might escape through a loose connection or dissolve in the reaction mixture.

A syringe could reach its maximum capacity while gas production continues.

A flat line caused by an instrument limit is not a chemical endpoint.

4. Were the conditions genuinely the same?

Gas volume depends on temperature and pressure, as well as the amount of gas present.

For a fair volume comparison, both runs must use the same measurement conditions.

In a reversible reaction, temperature can also change the equilibrium position. A catalyst cannot be blamed for a difference produced by changing temperature.

5. Does ‘final yield’ mean equilibrium yield or product collected at a deadline?

This is especially important.

A catalyst does not change the equilibrium constant of a particular reversible reaction at fixed temperature. It helps the system approach equilibrium sooner.

But suppose the products are collected after exactly five minutes and the reaction is then stopped.

The catalysed experiment may deliver more product within that five-minute window. That is a genuine difference in the amount isolated at the chosen deadline, not a change in the equilibrium composition.

In a system with competing reactions, a catalyst may also affect which products form more quickly and therefore alter the product mixture collected. This is a different question from whether it shifts the equilibrium of one reversible reaction.

For the equilibrium reasoning behind this distinction, see Why a Catalyst Does Not Change Equilibrium Yield.

06A reliable decision process for HSC graph questions

When you are given two reaction graphs, use this order.

1. Identify the quantity on the vertical axis. Is it product concentration, reactant concentration, gas volume, or mass? This determines what an increasing or decreasing gradient means.

2. Compare the correct gradients. For average rates, use the same specified time interval. For instantaneous rates, use tangents at the requested times. Do not mix the two.

3. Check whether both endpoints are genuine. Look for stable readings, check the measurement range, and consider whether equilibrium or reaction completion has occurred.

4. Compare final amounts under equivalent conditions. Equal genuine plateaus indicate equal measured final amounts. Different plateaus require investigation of starting quantities, equilibrium conditions, gas losses, or competing processes.

5. State what the evidence actually supports. A difference in gradient is evidence of a difference in measured rate. A difference in final product amount is not, by itself, proof that the catalyst caused it.

The strongest exam response usually gives both the observation and the reason: the catalysed reaction has a steeper early gradient, while both curves approach the same final value, showing faster progress but no change in final collected yield.

07Questions and solutions

The questions below use original, illustrative data. Assume measurements are accurate except where a question explicitly introduces a limitation.

Question 1

Two identical gas-producing reactions are compared. One contains a catalyst.

Calculate the average gas-collection rate over the first 20 seconds for each reaction. State whether the catalyst changed the final collected gas volume and explain your answer.

Solution 1

The catalyst increased the average gas-collection rate over the first 20 seconds from 0.50 mL/s to 1.20 mL/s, but both reactions eventually collected 40 mL of gas.

With the catalyst:

\[ \begin{aligned} \bar r_V&=\frac{\Delta V}{\Delta t}\\ &=\frac{24-0}{20}\\ &=1.20\ \mathrm{mL\,s^{-1}} \end{aligned} \]

Without the catalyst:

\[ \begin{aligned} \bar r_V&=\frac{10-0}{20}\\ &=0.50\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The catalysed reaction therefore has 2.4 times the average collection rate during this interval.

However, both curves stabilise at 40 mL.

It would be incorrect to compare their volumes at 60 seconds and call the difference a change in final yield. The uncatalysed reaction is still progressing at that time.

Question 2

Two open flasks contain identical initial amounts of hydrogen peroxide: 0.0375 mol each.

One flask contains an additional 0.20 g of a solid catalyst, which remains in the flask throughout the experiment.

The flasks are placed on separate balances.

Assume oxygen is the only substance escaping, both reactions eventually stop because all hydrogen peroxide is consumed, and the molar mass of oxygen gas is 32.0 g/mol.

(a) Determine which reaction has the greater average rate of mass loss over the first 60 seconds.

(b) Calculate the final amount of oxygen produced by each reaction, in moles.

(c) Explain why the identical balance readings at 60 seconds and the different final balance readings cannot be used directly to compare oxygen yields.

(d) Predict how the mass-time graph would differ if each entire reaction system, including its gas, were sealed and weighed.

Solution 2

The catalysed reaction loses mass twice as quickly over the first 60 seconds, but both reactions ultimately produce 0.01875 mol of oxygen. The different total balance readings arise from the additional catalyst mass.

(a) Compare the early mass-loss rates.

With the catalyst:

\[ \begin{aligned} \bar r_m&=-\frac{\Delta m}{\Delta t}\\ &=-\frac{151.90-152.30}{60}\\ &=0.00667\ \mathrm{g\,s^{-1}} \end{aligned} \]

Without the catalyst:

\[ \begin{aligned} \bar r_m&=-\frac{151.90-152.10}{60}\\ &=0.00333\ \mathrm{g\,s^{-1}} \end{aligned} \]

Here \(\bar r_m\) is the positive rate of mass loss. The balance readings themselves have negative gradients.

The catalysed flask loses mass at twice the average rate.

(b) Calculate the final oxygen yield.

With the catalyst:

\[ \begin{aligned} m_{\ce{O2}}&=152.30-151.70\\ &=0.60\ \mathrm{g} \end{aligned} \]

Without the catalyst:

\[ \begin{aligned} m_{\ce{O2}}&=152.10-151.50\\ &=0.60\ \mathrm{g} \end{aligned} \]

For either reaction:

\[ \begin{aligned} n_{\ce{O2}}&=\frac{m}{M}\\ &=\frac{0.60}{32.0}\\ &=0.01875\ \mathrm{mol} \end{aligned} \]

Here \(n\) is the amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in g/mol.

The balanced equation is:

\[ \ce{2H2O2 -> 2H2O + O2} \]

Therefore, 0.0375 mol of hydrogen peroxide produces a theoretical maximum of:

\[ n_{\ce{O2}}=\frac{0.0375}{2} =0.01875\ \mathrm{mol} \]

The measured mass losses are consistent with complete decomposition in both flasks.

(c) Interpret the balance readings.

At 60 seconds, both balances show 151.90 g. But the catalysed flask started 0.20 g heavier.

It has already lost 0.40 g, whereas the control has lost only 0.20 g.

Equal absolute mass readings do not mean equal amounts of gas have escaped.

Similarly, the final readings differ by 0.20 g because the catalyst remains in the flask.

The useful comparison is the change in mass from each experiment’s own starting value.

(d) Predict the sealed-system graph.

If each complete apparatus were sealed and all produced oxygen remained inside the weighed system, its total mass would stay constant.

Each graph would be horizontal from beginning to end, assuming nothing entered or left either system.

Oxygen could still be forming rapidly inside. A horizontal total-mass graph in a closed system does not establish a zero chemical reaction rate.

Question 3

A hypothetical reversible reaction is represented by:

\[ \ce{A(aq) <=> 2B(aq)} \]

Two experiments have the same initial concentrations, constant solution volumes, and temperature of 25°C.

Initially:

\[ [A]_0=0.600\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0 \]

The measured concentrations of A are:

Further readings confirm that both concentrations remain at 0.200 mol/L.

(a) Calculate the average rate of formation of B over the first four minutes in each experiment.

(b) Calculate the final equilibrium concentration of B in each experiment.

(c) A student uses the rates from part (a) as the instantaneous rates at exactly four minutes. Explain whether this is justified.

(d) Explain why the different concentrations at four minutes do not imply different equilibrium positions.

Solution 3

The catalysed experiment has the greater average rate of B formation over the first four minutes, but both experiments reach the same equilibrium B concentration of 0.800 mol/L. The given readings do not determine exact instantaneous rates at four minutes.

(a) Calculate the formation rates of B.

From the equation, every mole of A consumed produces two moles of B.

At constant volume:

\[ \Delta[B]=-2\Delta[A] \]

For the catalysed reaction:

\[ \begin{aligned} \Delta[B]&=2(0.600-0.240)\\ &=0.720\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \bar r_B&=\frac{0.720}{4}\\ &=0.180\ \mathrm{mol\,L^{-1}\,min^{-1}} \end{aligned} \]

For the uncatalysed reaction:

\[ \begin{aligned} \Delta[B]&=2(0.600-0.390)\\ &=0.420\ \mathrm{mol\,L^{-1}}\\ \bar r_B&=\frac{0.420}{4}\\ &=0.105\ \mathrm{mol\,L^{-1}\,min^{-1}} \end{aligned} \]

The catalyst increases the average net rate of B formation over the first four minutes.

A tempting mistake is to treat the disappearance rate of A as equal to the formation rate of B. That ignores the coefficient of 2 in the balanced equation.

(b) Calculate the equilibrium concentration of B.

Both experiments finish with:

\[ [A]_{\mathrm{eq}}=0.200\ \mathrm{mol\,L^{-1}} \]

The concentration of A consumed is:

\[ 0.600-0.200=0.400\ \mathrm{mol\,L^{-1}} \]

So the concentration of B formed is:

\[ \begin{aligned} [B]_{\mathrm{eq}}&=2(0.400)\\ &=0.800\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The equilibrium composition is the same for both reactions.

(c) Distinguish average and instantaneous rates.

The rates in part (a) are averages across the interval from 0 to 4 minutes.

The instantaneous rate at exactly four minutes is the tangent gradient at that point.

Since only three concentration readings are supplied, there is insufficient information to determine the exact tangent gradient.

(d) Explain the different intermediate concentrations.

At four minutes, the catalysed reaction has progressed further towards equilibrium.

The uncatalysed reaction still contains more A and will continue producing a net amount of B until it reaches equilibrium.

The catalyst has changed the rate at which equilibrium is approached, not the final equilibrium composition.

Question 4

Two identical amounts of hydrogen peroxide are decomposed completely.

Each flask initially contains 0.00500 mol of hydrogen peroxide.

The reaction is:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

A catalyst is added to experiment C, while experiment U has no catalyst.

The experimenters accidentally use different gas syringes:

• Experiment C uses a syringe with a maximum capacity of 40 mL. A safety outlet opens when the syringe is full, allowing additional oxygen to escape.
• Experiment U uses a 100 mL syringe.

Both experiments are performed under conditions where the molar gas volume is 24.0 L/mol.

All oxygen formed reaches its syringe, and oxygen that cannot be retained by a full syringe leaves through its safety outlet. Ignore dissolved oxygen and gas remaining in the flask or tubing.

The recorded volumes are:

(a) Calculate the total volume of oxygen that each reaction must produce.

(b) Determine the volume of oxygen that escaped through the safety outlet in experiment C by completion.

(c) Explain what can and cannot be concluded about the average rate of oxygen formation over the first 30 seconds.

(d) Explain how you would improve the apparatus to make a valid comparison of final collected yields.

Solution 4

Both reactions produce 60.0 mL of oxygen, but the catalysed syringe can retain only 40 mL. Its apparent plateau is an instrument limit, not a chemical endpoint. By completion, 20.0 mL of oxygen has escaped through the safety outlet.

(a) Calculate the theoretical oxygen volume.

The balanced equation shows that two moles of hydrogen peroxide produce one mole of oxygen.

Therefore:

\[ \begin{aligned} n_{\ce{O2}}&=\frac{0.00500}{2}\\ &=0.00250\ \mathrm{mol} \end{aligned} \]

Using the given molar gas volume:

\[ \begin{aligned} V&=nV_m\\ &=0.00250(24.0)\\ &=0.0600\ \mathrm{L}\\ &=60.0\ \mathrm{mL} \end{aligned} \]

Here \(V\) is the oxygen volume and \(V_m\) is the molar gas volume.

Both reactions produce this amount because both start with the same moles of hydrogen peroxide and decompose it completely.

(b) Calculate the oxygen that escapes.

Experiment C retains a maximum of 40 mL.

\[ \begin{aligned} V_{\mathrm{escaped}}&=V_{\mathrm{total}}-V_{\mathrm{retained}}\\ &=60.0-40.0\\ &=20.0\ \mathrm{mL} \end{aligned} \]

The 40 mL plateau is therefore not evidence that the catalysed reaction stopped at 40 mL of oxygen production.

(c) Interpret the rates.

For the uncatalysed experiment, the first 30-second average rate is:

\[ \begin{aligned} \bar r_U&=\frac{18-0}{30}\\ &=0.600\ \mathrm{mL\,s^{-1}} \end{aligned} \]

For the catalysed experiment, at least 40 mL of oxygen formed in the first 30 seconds, because that much was retained.

Therefore:

\[ \begin{aligned} \bar r_C&\geq\frac{40}{30}\\ &\geq1.33\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The actual rate of oxygen formation could be higher if the safety outlet had already released additional gas before 30 seconds.

So the catalysed reaction certainly produced oxygen faster on average during the first 30 seconds, but its exact average formation rate cannot be calculated from the syringe reading alone.

This is the important constraint: once the syringe is full, its reading stops tracking the actual gas production.

(d) Improve the apparatus.

Use gas syringes of equal capacity, comfortably greater than the expected final volume of 60 mL.

Ensure both systems are airtight, with no gas lost through overflow or leakage, and maintain the same temperature and pressure.

Then compare the entire curves, including their genuine final plateaus.

The surprising result is that the catalysed reaction can appear to have a lower final gas volume simply because the measuring equipment reached its limit.

Question 5

A reversible, exothermic reaction takes place in two sealed, rigid vessels:

\[ \ce{A(aq) <=> B(aq)} \]

Both vessels contain 1.00 L of solution, initially with:

\[ [A]_0=0.90\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0.10\ \mathrm{mol\,L^{-1}} \]

The two mixtures are chemically identical except that vessel C contains a catalyst and vessel U does not.

For the first 10 minutes, both vessels are held at temperature \(T_1\), where the equilibrium concentration of B is 0.70 mol/L.

At exactly 10 minutes, both vessels are heated rapidly to a higher temperature \(T_2\).

At \(T_2\), the equilibrium concentration of B is 0.20 mol/L.

The volumes remain constant. The catalyst affects reaction rates but does not alter either equilibrium constant, and no side reactions occur.

The following concentrations are measured:

At 20 minutes, both reactions are quenched instantaneously, preventing any further chemical change. All B present at that instant is then isolated quantitatively.

(a) Determine the direction of the net reaction in each vessel immediately before and immediately after the temperature increase.

(b) Explain why the two product-concentration curves must cross between 10 and 20 minutes, even though a catalyst does not change equilibrium position.

(c) Calculate which vessel produces the greater isolated yield of B at 20 minutes and by how many moles.

(d) A student states: “The catalyst must have lowered the final equilibrium yield because the catalysed vessel contains less B at 20 minutes.” Evaluate this claim.

Solution 5

Both reactions initially proceed towards B, then reverse direction after heating. Their concentration curves cross because the new equilibrium lies below both mixtures’ concentrations at 10 minutes. At the 20-minute quench, the uncatalysed vessel contains 0.050 mol more B, even though the catalyst has not changed either equilibrium composition.

(a) Determine the directions of net reaction.

Immediately before heating, both solutions contain less B than the equilibrium concentration at \(T_1\):

\[ [B]_{\mathrm{eq},1}=0.70\ \mathrm{mol\,L^{-1}} \]

Vessel C contains 0.65 mol/L, and vessel U contains 0.35 mol/L.

Both therefore have a net forward reaction:

\[ \ce{A -> B} \]

The catalysed vessel is much closer to the original equilibrium.

At 10 minutes, the temperature increases.

Because the forward reaction is exothermic, raising the temperature favours the endothermic reverse direction. The new equilibrium concentration of B is lower:

\[ [B]_{\mathrm{eq},2}=0.20\ \mathrm{mol\,L^{-1}} \]

Immediately after heating, both mixtures contain more B than the new equilibrium requires.

The net reaction therefore reverses in both vessels:

\[ \ce{B -> A} \]

The direction changes because the equilibrium conditions changed, not because the catalyst began favouring the reverse reaction.

(b) Explain the crossing curves.

At 10 minutes:

\[ [B]_C=0.65\ \mathrm{mol\,L^{-1}} \]

\[ [B]_U=0.35\ \mathrm{mol\,L^{-1}} \]

The catalysed vessel contains more B.

However, both mixtures must now approach the lower equilibrium concentration of 0.20 mol/L.

The catalyst allows vessel C to respond more quickly to the new conditions. Its B concentration falls rapidly.

By 20 minutes:

\[ [B]_C=0.23\ \mathrm{mol\,L^{-1}} \]

\[ [B]_U=0.28\ \mathrm{mol\,L^{-1}} \]

The order has reversed.

Because concentrations change continuously and the vessels have exchanged their relative positions, the two concentration curves must have crossed at least once between 10 and 20 minutes.

The exact crossing time cannot be determined from the table.

The key insight is that faster movement towards equilibrium does not always mean faster production of the product.

Before the temperature change, approaching equilibrium means producing B.

After the change, approaching equilibrium means consuming B.

The catalyst speeds the approach in whichever direction is required.

(c) Calculate the isolated amounts.

At 20 minutes, vessel C contains:

\[ \begin{aligned} n_C&=[B]_CV\\ &=0.23(1.00)\\ &=0.23\ \mathrm{mol} \end{aligned} \]

Vessel U contains:

\[ \begin{aligned} n_U&=[B]_UV\\ &=0.28(1.00)\\ &=0.28\ \mathrm{mol} \end{aligned} \]

Here \(n\) is the amount of B in moles and \(V\) is the solution volume in litres.

The difference is:

\[ \begin{aligned} \Delta n&=0.28-0.23\\ &=0.050\ \mathrm{mol} \end{aligned} \]

The uncatalysed vessel provides 0.050 mol more isolated B when both reactions are quenched at 20 minutes.

(d) Evaluate the equilibrium-yield claim.

The student’s conclusion is incorrect.

The final equilibrium concentration at \(T_2\) is 0.20 mol/L for both vessels.

If neither reaction were quenched, both would eventually contain:

\[ n_{\mathrm{eq}}=0.20(1.00)=0.20\ \mathrm{mol} \]

of B at the new equilibrium.

The different amounts isolated at 20 minutes arise because the mixtures are stopped before reaching equilibrium.

The catalyst made vessel C approach the lower equilibrium concentration more quickly, so more B had already been converted back into A when the reaction was quenched.

This is the conceptual trap: a catalyst can reduce the amount of a chosen product present at a particular time, even while correctly accelerating the approach to equilibrium.

You must know which direction the net reaction is moving, which equilibrium applies, and whether the graph shows an intermediate state or a true final equilibrium.

08What to decide when you see the next catalyst graph

Don’t start by asking which curve is higher.

Start by identifying what is measured, then separate the gradient from the eventual endpoint.

A steep gas-volume curve tells you how quickly gas is being collected. A horizontal plateau tells you the measured volume has stopped changing. Neither observation alone proves that a catalyst has changed the amount of product chemically available.

When the graphs have different final values, check the starting reactant amounts, temperature, gas collection, apparatus limits, and whether both reactions have genuinely finished.

And if the question involves a reversible reaction, ask one further question: is the mixture moving towards equilibrium from the reactant side or the product side?

That determines whether a catalyst makes the product-concentration curve rise faster, fall faster, or approach a horizontal line sooner.