Why Doubling Concentration Doesn’t Always Double Reaction Rate

Two conical flasks contain identical calcium carbonate chips. One receives \(0.50\ \mathrm{mol\,L^{-1}}\) hydrochloric acid. The other receives \(1.00\ \mathrm{mol\,L^{-1}}\) acid. Both produce carbon dioxide, which is collected in gas syringes.

The acid concentration has doubled. Should the second syringe fill exactly twice as quickly?

Not necessarily. Doubling concentration generally increases the opportunities for successful particle collisions, but it does not guarantee that the reaction rate doubles. The exact change depends on the reaction mechanism, the concentrations of other reactants, and the stage of the reaction being measured.

There’s also a measurement problem. Producing twice as much gas over a particular time interval is not necessarily the same as having twice the instantaneous reaction rate.

To see why, we’ll investigate a reaction in which doubling the starting acid concentration produces neither double the average rate nor double the final gas volume.

01The experiment: what does the gas syringe actually measure?

Consider calcium carbonate reacting with hydrochloric acid:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]

Carbon dioxide forms as acid reacts with the exposed calcium carbonate surface.

A suitable gas-collection apparatus has a conical flask containing the reactants, an airtight bung, a delivery tube, and a gas syringe.

The sequence is straightforward:

1. Place calcium carbonate chips in the flask.
2. Introduce the hydrochloric acid and start the stopwatch when it contacts the chips.
3. Carbon dioxide bubbles form and travel through the delivery tube.
4. The gas pushes the syringe plunger outwards.
5. Record the accumulated gas volume at regular intervals.

For an accurate initial-rate investigation, the acid should be introduced without allowing gas to escape while the bung is fitted. A sealed addition arrangement can help.

The syringe measures the volume of carbon dioxide collected, not the concentration of hydrochloric acid or the reaction rate directly.

The distinction is similar to an odometer and a speedometer. An odometer records the total distance travelled. Speed describes how quickly that distance is changing.

Likewise, gas volume is an accumulated quantity. Its gradient against time represents the gas-production rate.

The analogy has a limit: unlike distance, gas volume also depends on temperature and pressure. We must keep those conditions constant for meaningful comparisons.

Predict the initial result

Suppose both flasks contain:

• \(0.400\ \mathrm{g}\) of calcium carbonate chips with comparable exposed surface areas
• \(10.0\ \mathrm{mL}\) of hydrochloric acid
• The same starting temperature and mixing conditions

Trial L uses \(0.500\ \mathrm{mol\,L^{-1}}\) HCl.

Trial H uses \(1.00\ \mathrm{mol\,L^{-1}}\) HCl.

The higher acid concentration means more reacting acid particles per unit volume. At the start, this should generally increase the frequency of successful interactions at the calcium carbonate surface.

But how much faster?

Collision theory alone does not supply an exact multiplier. We need measurements, and we need to be careful about which rate those measurements represent.

Illustrative experimental data

The following values are constructed for teaching. They are not measurements from a reported experiment.

Assume complete gas collection, constant gas temperature and pressure, and negligible dissolution of carbon dioxide.

Both reactions initially produce gas quickly. Their gas-production rates then decrease, and their collected volumes eventually level off.

The more concentrated acid produces more gas in the first 20 seconds. But it does not produce twice as much.

02Why this first experiment cannot establish an exact concentration-rate relationship

The first experiment produces an awkward result for a simple reason: the two reactions don’t remain in equivalent chemical conditions as they progress.

The acid is consumed. The calcium carbonate chips become smaller. The exposed reacting surface can change, and the reactions approach different limiting conditions.

More importantly, the two trials don’t even have the same limiting reactant.

03Revise the experiment: keep the same limiting reactant

How could we improve the investigation so that doubling acid concentration gives a more meaningful rate comparison?

The first design allowed acid to limit one reaction and calcium carbonate to limit the other. The revision makes calcium carbonate the limiting reactant in both.

Use:

• \(0.100\ \mathrm{g}\) of calcium carbonate in each flask
• Comparable initial chip sizes and exposed surface areas
• \(50.0\ \mathrm{mL}\) of acid solution in each flask
• Starting acid concentrations of 0.500 and \(1.00\ \mathrm{mol\,L^{-1}}\)
• Identical temperature, mixing, and gas-collection conditions

The acid is now in substantial excess.

Each sample contains approximately 0.00100 mol of calcium carbonate, requiring only 0.00200 mol of HCl to react completely.

Even the less concentrated flask starts with 0.0250 mol HCl.

Therefore, both reactions have the same theoretical carbon dioxide yield:

\[ V_{\max}\approx24.0\ \mathrm{mL} \]

What would the revised results tell us?

Consider another constructed dataset, again not actual measurements.

Both trials eventually produce 24 mL of carbon dioxide because they begin with the same limiting amount of calcium carbonate.

From 0 to 5 seconds:

• Trial L averages \(5/5=1.0\ \mathrm{mL\,s^{-1}}\).
• Trial H averages \(8/5=1.6\ \mathrm{mL\,s^{-1}}\).

The early average rate is 1.6 times larger in the more concentrated trial.

Now compare 20 to 40 seconds.

Trial L:

\[ \bar v_L=\frac{21-15}{20} =0.30\ \mathrm{mL\,s^{-1}} \]

Trial H:

\[ \bar v_H=\frac{23-19}{20} =0.20\ \mathrm{mL\,s^{-1}} \]

The higher-concentration trial now has the smaller average gas-production rate.

That does not mean concentrated hydrochloric acid inherently slows this reaction. Trial H has already consumed more calcium carbonate and is closer to completion. The remaining solid and its exposed surface are different from those in Trial L.

The revision removes the different-final-yield problem, but it still does not establish an exact initial-rate multiplier.

The first five seconds provide an early average, not a measurement at precisely zero seconds. The solid’s surface begins changing as soon as reaction starts.

To estimate the initial instantaneous rate more reliably, we’d collect closely spaced readings near the start, repeat the trials, and estimate a tangent to each smooth gas-volume curve at \(t=0\).

A real experiment also needs checks for leaks, gas dissolving in water, delays in starting the stopwatch, and temperature changes. Shorter intervals reduce changes in reaction conditions but can increase the relative effect of measurement uncertainty.

04Average rate and instantaneous rate answer different questions

Before interpreting any concentration comparison, identify what the data can actually tell you.

Average rate: what happened across an interval?

For a gas-volume graph:

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

This is the gradient of a secant: a straight line connecting two points on the curve.

It describes the average production of gas throughout the chosen interval.

Instantaneous rate: what is happening at one moment?

The instantaneous gas-production rate is the gradient of the tangent to the gas-volume curve at a particular time.

A tangent at 20 seconds describes the rate at 20 seconds.

A secant from 0 to 20 seconds describes the average across the entire interval.

They will generally differ when the curve is not a straight line.

For a detailed tangent calculation, see How to Find Instantaneous Rate Using a Tangent in HSC Chemistry.

What if the data are concentrations or mass measurements?

Different measurements need different signs and units.

The minus sign makes a disappearance or mass-loss rate positive when the measured quantity decreases.

These are rates of change of the measured quantities. When expressing a formal reaction rate from concentration data, stoichiometric coefficients must also be considered.

For example, in:

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

two moles of A disappear for every mole of B formed. At constant solution volume:

\[ r=-\frac12\frac{\Delta[A]}{\Delta t} \]

where \(r\) is the average reaction rate in \(\mathrm{mol\,L^{-1}\,s^{-1}}\), and \(\Delta[A]\) is the concentration change over the chosen interval.

There is no universal numerical equality between a gas-volume rate in mL/s, a mass-loss rate in g/s, and a concentration rate in mol/L/s. Converting between them requires the appropriate physical and stoichiometric information.

05When would doubling concentration actually double the rate?

We’ve seen why one gas-collection experiment can give a rate ratio other than two.

But there’s a deeper chemical question: even if we measured perfect initial instantaneous rates, would doubling one reactant concentration always double the rate?

No. Exact proportionality depends on the reaction’s kinetic behaviour.

The simple collision model

Picture a fixed area of calcium carbonate surface surrounded by acid.

If you put more reacting acid particles into the same solution volume, more particles can encounter the exposed surface each second.

That generally increases the number of successful reaction events.

But successful reaction events may involve several stages. The supply of particles to a surface, the availability of reactive sites, and the chemical steps occurring there can all affect the observed rate.

Doubling the number of acid particles per litre does not establish that the number of completed reaction events per second must double.

The familiar collision explanation predicts a direction of change under suitable conditions. It does not automatically predict an exact numerical factor.

For a fuller explanation of the factors affecting successful collisions, see How to Explain Reaction-Rate Changes in HSC Chemistry.

Extension: reaction order gives the exact multiplier in a kinetic model

The following rate-law notation is a useful extension beyond the basic collision-theory explanation. NSW Chemistry Stage 6 (2017) remains the syllabus in use during 2026; do not assume this formal relationship is required in every HSC rate question.

For a hypothetical reaction where other relevant concentrations and conditions are held constant, suppose the experimentally determined rate law is:

\[ r=k[A]^m \]

Here:

• \(r\) is the instantaneous reaction rate, usually in \(\mathrm{mol\,L^{-1}\,s^{-1}}\).
• \(k\) is the rate constant for the specified conditions, with units depending on reaction order.
• \([A]\) is the molar concentration of reactant A.
• \(m\) is the reaction order with respect to A, determined from experimental evidence.

If \([A]\) doubles:

\[ \frac{r_{\text{new}}}{r_{\text{old}}}=2^m \]

Different values of \(m\) produce different predictions.

A first-order dependence is the special case in which doubling that reactant’s concentration doubles its instantaneous rate, provided the other relevant conditions remain unchanged.

The reaction order is not normally obtained by copying the coefficients from the balanced chemical equation. It depends on the reaction mechanism and must be established experimentally or supplied in the question.

Also notice what this formula does not say. It does not guarantee that doubling the initial concentration doubles the average rate over the same later time interval. The concentrations change during the reaction.

A practical decision rule

When a question says that concentration doubles, work through these distinctions:

1. What changed? Was one reactant concentration doubled while temperature, other reactants, and relevant physical conditions were controlled?
2. What was measured? An average rate, an instantaneous rate, or simply the total amount of product?
3. What does the evidence establish? Collision theory can explain a general increase. An exact multiplier requires suitable kinetic evidence or an explicitly supplied rate law.
4. Has the reaction progressed? Two mixtures compared at the same clock time may have consumed very different amounts of their reactants.

The fourth question becomes particularly interesting when a reaction has more than one reactant. A mixture that reacts faster initially can later have a smaller instantaneous rate because it has depleted another reactant more severely.

06Questions and solutions

These questions move from ordinary HSC rate calculations to more demanding experimental interpretation. Questions 4 and 5 use a supplied kinetic model; you do not need to derive the rate law.

Question 1

A gas-producing reaction is monitored using a gas syringe.

At 10 seconds, the syringe contains 12.0 mL of gas. At 25 seconds, it contains 30.0 mL.

Calculate the average gas-production rate between these times. Explain whether your answer is necessarily the instantaneous rate at 25 seconds.

Solution 1

The average gas-production rate is \(1.20\ \mathrm{mL\,s^{-1}}\). This is not necessarily the instantaneous rate at 25 seconds.

Use the change in gas volume divided by elapsed time:

\[ \begin{aligned} \bar v&=\frac{\Delta V}{\Delta t}\\ &=\frac{30.0-12.0}{25-10}\\ &=\frac{18.0}{15}\\ &=1.20\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The reaction produced an average of 1.20 mL of gas each second between 10 and 25 seconds.

This calculation uses two points, so it gives a secant gradient. The instantaneous rate at 25 seconds requires the tangent gradient there.

The tempting mistake is to treat an interval average as the rate at its final time. If the reaction is slowing, the instantaneous rate at 25 seconds would be lower than this interval average.

Question 2

A hypothetical aqueous reaction follows:

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

A well-mixed flask initially contains \(50.0\ \mathrm{mL}\) of solution with \([A]=0.200\ \mathrm{mol\,L^{-1}}\).

After 10.0 seconds, the concentration of A has decreased to \(0.160\ \mathrm{mol\,L^{-1}}\).

At exactly 10.0 seconds, another 50.0 mL of pure water is added. Assume the water mixes instantly, with no reaction during the mixing step. The total solution volume becomes 100.0 mL.

Immediately after dilution, \([A]=0.0800\ \mathrm{mol\,L^{-1}}\).

At 20.0 seconds, \([A]=0.0700\ \mathrm{mol\,L^{-1}}\).

Assume no side reactions and constant solution volume between these events.

(a) Calculate the amount of A consumed during 0-10 seconds and 10-20 seconds.

(b) Calculate the average rate of formation of B, in mol/s, over each interval.

(c) A student subtracts \(0.0700\) from \(0.160\) and divides by 10 to calculate the disappearance rate of A during the second interval. Explain why this is incorrect.

(d) Can these results determine the instantaneous reaction rate immediately before or after dilution?

Solution 2

The reaction consumes 0.00200 mol of A in the first interval and 0.00100 mol in the second. The average B-formation rate halves, even though the concentration of A shows a much larger apparent drop across the dilution event.

(a) Calculate the amounts consumed.

Initially:

\[ \begin{aligned} n_0(A)&=0.200(0.0500)\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

At 10 seconds, before dilution:

\[ \begin{aligned} n_{10}(A)&=0.160(0.0500)\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

Therefore, during the first interval:

\[ \begin{aligned} n_{\text{used},1} &=0.0100-0.00800\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

Adding water changes concentration, but does not consume A.

Immediately after dilution, the amount of A remains 0.00800 mol.

At 20 seconds:

\[ \begin{aligned} n_{20}(A)&=0.0700(0.1000)\\ &=0.00700\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} n_{\text{used},2} &=0.00800-0.00700\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

(b) Calculate the B-formation rates.

The equation shows that two moles of A form one mole of B.

During the first interval:

\[ \begin{aligned} n_1(B)&=\frac{0.00200}{2}\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

The average formation rate is:

\[ \begin{aligned} \bar v_{B,1} &=\frac{0.00100}{10.0}\\ &=1.00\times10^{-4}\ \mathrm{mol\,s^{-1}} \end{aligned} \]

During the second interval:

\[ \begin{aligned} n_2(B)&=\frac{0.00100}{2}\\ &=0.000500\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} \bar v_{B,2} &=\frac{0.000500}{10.0}\\ &=5.00\times10^{-5}\ \mathrm{mol\,s^{-1}} \end{aligned} \]

The average B-formation rate in mol/s is half as large after dilution.

(c) Explain the faulty calculation.

The student’s subtraction includes a concentration change caused by adding water.

The concentration falls from 0.160 to \(0.0800\ \mathrm{mol\,L^{-1}}\) immediately upon dilution, even though no A reacts during that step.

The actual concentration decrease caused by reaction during the second interval is:

\[ 0.0800-0.0700 =0.0100\ \mathrm{mol\,L^{-1}} \]

That is much smaller than the student’s apparent decrease of \(0.0900\ \mathrm{mol\,L^{-1}}\).

To calculate amounts in mol, the different solution volumes must also be included.

(d) Interpret the instantaneous rates.

The instantaneous rates immediately before and after dilution cannot be determined from these two ten-second averages.

Dilution changes the concentration abruptly and may change the instantaneous reaction rate. However, determining its numerical effect requires a kinetic relationship or suitable measurements close to the dilution time.

The central trap is treating every concentration decrease as chemical consumption. Concentration can fall because particles are reacting, because the solution is being diluted, or both.

Question 3

A calcium carbonate-hydrochloric acid reaction is monitored using two measurements from the same apparatus.

The reaction flask and its contents sit on an electronic balance. The balance does not support the external delivery tube or gas syringe.

As carbon dioxide leaves the flask through the delivery tube, the recorded flask mass decreases. A known leak in the tube allows some carbon dioxide to escape before the remainder reaches the gas syringe.

During the first 20.0 seconds:

• The recorded flask mass decreases by 0.132 g.
• The gas syringe collects 48.0 mL of carbon dioxide.

Assume carbon dioxide is the only substance leaving the weighed flask, the gas is dry, no significant amount dissolves, and all uncollected gas escapes through the tube leak. The molar mass of \(\ce{CO2}\) is \(44.0\ \mathrm{g\,mol^{-1}}\), and the molar gas volume is \(24.0\ \mathrm{L\,mol^{-1}}\).

(a) Calculate the total amount and volume of carbon dioxide that left the reaction flask during the first 20.0 seconds.

(b) Determine the gas volume lost through the leak.

(c) Calculate the actual average carbon dioxide production rate in mL/s.

(d) Explain why comparing this syringe’s recorded rate with a second trial using a different leak size would not be a valid test of concentration dependence.

Solution 3

The flask produced 72.0 mL of carbon dioxide, of which 24.0 mL escaped through the leak. The actual average gas-production rate was \(3.60\ \mathrm{mL\,s^{-1}}\), not the \(2.40\ \mathrm{mL\,s^{-1}}\) suggested by the syringe alone.

(a) Calculate total carbon dioxide produced.

Under the supplied assumptions, the entire 0.132 g mass loss represents carbon dioxide leaving the flask.

\[ \begin{aligned} n(\ce{CO2}) &=\frac{m}{M}\\ &=\frac{0.132}{44.0}\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

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

Convert moles into gas volume:

\[ \begin{aligned} V_{\text{total}} &=0.00300(24.0)\\ &=0.0720\ \mathrm{L}\\ &=72.0\ \mathrm{mL} \end{aligned} \]

(b) Calculate the leaked gas volume.

\[ \begin{aligned} V_{\text{leaked}} &=72.0-48.0\\ &=24.0\ \mathrm{mL} \end{aligned} \]

One-third of the produced carbon dioxide failed to reach the syringe.

(c) Calculate the true average gas-production rate.

\[ \begin{aligned} \bar v_{\text{actual}} &=\frac{72.0}{20.0}\\ &=3.60\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The syringe alone suggests:

\[ \begin{aligned} \bar v_{\text{syringe}} &=\frac{48.0}{20.0}\\ &=2.40\ \mathrm{mL\,s^{-1}} \end{aligned} \]

Both quantities are interval averages. Neither determines the instantaneous rate at 20 seconds.

(d) Evaluate the comparison.

Different leaks could cause different fractions of the produced gas to be collected.

A smaller syringe reading could therefore result from slower gas production, greater gas leakage, or a combination of both.

The tempting assumption is that the syringe measures all the carbon dioxide produced. Under the given conditions, it measures only the portion successfully collected.

The experiment must control gas collection efficiency before the readings can be used to infer a concentration effect.

Question 4

A hypothetical aqueous reaction produces a gas:

\[ \ce{A(aq) + B(aq) -> G(g)} \]

Two trials use identical apparatus, constant temperature, and the same solution volume of 100.0 mL.

Their initial concentrations are:

Assume the experimentally established instantaneous rate law is:

\[ r=k[A][B] \]

Here \(r\) is in \(\mathrm{mol\,L^{-1}\,s^{-1}}\), and \(k\) is the same rate constant in both trials.

The molar gas volume is \(24.0\ \mathrm{L\,mol^{-1}}\), and all G is collected.

At some point, each gas syringe records exactly 120.0 mL of G. These readings do not necessarily occur at the same time.

(a) Calculate the remaining concentrations of A and B in each trial when 120.0 mL of gas has been produced.

(b) Determine the ratio of their instantaneous reaction rates at those points.

(c) Explain why this ratio differs from the initial rate ratio.

Solution 4

At the point where each trial has produced 120.0 mL of gas, Trial H has three times the instantaneous reaction rate of Trial L. Initially, its rate was only twice as large.

(a) Find the remaining concentrations.

Convert the gas volume to moles:

\[ \begin{aligned} n(G)&=\frac{0.1200}{24.0}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

The balanced equation has a 1:1:1 mole ratio.

Therefore, forming 0.00500 mol G consumes 0.00500 mol of both A and B.

Each flask contains 0.1000 L of solution, so the concentration decrease of each reactant is:

\[ \begin{aligned} \Delta c&=\frac{0.00500}{0.1000}\\ &=0.0500\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The remaining concentrations are:

(b) Compare instantaneous rates.

Using \(r=k[A][B]\):

\[ r_L=k(0.0500)(0.0500) \]

\[ r_H=k(0.150)(0.0500) \]

The common rate constant and B concentration cancel in the ratio:

\[ \begin{aligned} \frac{r_H}{r_L} &=\frac{0.150}{0.0500}\\ &=3.00 \end{aligned} \]

Trial H has three times the instantaneous rate when both have produced 120.0 mL of gas.

(c) Explain the difference from the initial rate ratio.

Initially, Trial H has twice the A concentration and the same B concentration:

\[ \frac{r_{H,0}}{r_{L,0}}=2 \]

But after both have produced the same amount of product, they have consumed the same absolute concentration of A and B.

Subtracting \(0.0500\ \mathrm{mol\,L^{-1}}\) from the two starting A concentrations changes their ratio from \(0.200/0.100=2\) to \(0.150/0.0500=3\).

The important constraint is equal reaction progress, not equal elapsed time.

A common mistake is to assume that because one mixture began with twice as much A per litre, it must always have exactly twice the rate. That ignores how the reactant concentrations evolve.

Question 5

Two closed, well-mixed reaction flasks produce gas G according to:

\[ \ce{A(aq) + B(aq) -> G(g)} \]

Both use a constant solution volume of 100.0 mL, identical temperature and pressure, and perfect gas collection.

The rate law is:

\[ r=k[A][B] \]

The rate constant is:

\[ k=0.200\ \mathrm{L\,mol^{-1}\,s^{-1}} \]

The initial concentrations are:

The molar gas volume is \(24.0\ \mathrm{L\,mol^{-1}}\).

After 80.0 seconds, the concentrations are approximately:

These values are consistent, to rounding, with the supplied rate law and starting conditions.

(a) Calculate the initial instantaneous reaction rates. Confirm which trial initially produces gas faster.

(b) At 80.0 seconds, calculate the instantaneous reaction rate in each trial. Which gas syringe is moving faster at that moment?

(c) Determine which trial has collected more gas at 80.0 seconds.

(d) A student argues:

“Trial L is producing gas faster at 80 seconds, so its total collected gas volume must eventually overtake Trial H.”

Is this possible under the supplied conditions? Justify your conclusion without solving the concentration-time equations.

(e) Determine the theoretical final gas volume in each trial.

Solution 5

Trial H initially produces gas twice as quickly as Trial L. At 80 seconds, however, Trial L has the greater instantaneous rate, while Trial H still has more accumulated gas. Trial L can never overtake Trial H in total collected gas under the stated kinetic model. Both approach the same theoretical final gas volume of 240 mL.

The apparent contradiction disappears when we separate reaction speed from accumulated progress.

(a) Calculate the initial rates.

The rate law is:

\[ r=k[A][B] \]

For Trial L:

\[ \begin{aligned} r_{L,0} &=0.200(0.100)(0.100)\\ &=0.00200\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

For Trial H:

\[ \begin{aligned} r_{H,0} &=0.200(0.200)(0.100)\\ &=0.00400\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

Trial H initially has twice the reaction rate.

Because both use the same solution volume and gas-collection conditions, Trial H also initially produces gas at twice the volume rate.

(b) Compare the rates at 80 seconds.

For Trial L:

\[ \begin{aligned} r_{L,80} &=0.200(0.0385)(0.0385)\\ &=2.96\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

For Trial H:

\[ \begin{aligned} r_{H,80} &=0.200(0.1112)(0.0112)\\ &=2.49\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

Trial L now has the higher instantaneous rate.

For the gas syringes, convert reaction rate into mol/s by multiplying by the solution volume of 0.1000 L, then into mL/s by multiplying by \(24\,000\ \mathrm{mL\,mol^{-1}}\).

Trial L:

\[ \begin{aligned} v_{L,80} &=(2.96\times10^{-4})(0.1000)\\ &\quad\times24\,000\\ &\approx0.711\ \mathrm{mL\,s^{-1}} \end{aligned} \]

Trial H:

\[ \begin{aligned} v_{H,80} &=(2.49\times10^{-4})(0.1000)\\ &\quad\times24\,000\\ &\approx0.598\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The syringe in Trial L is moving faster at 80 seconds.

Why?

Trial H still has more A, but it has consumed much more B.

Its remaining B concentration is only \(0.0112\ \mathrm{mol\,L^{-1}}\), compared with \(0.0385\ \mathrm{mol\,L^{-1}}\) in Trial L.

Because the rate depends on the product \([A][B]\), the severe depletion of B in Trial H outweighs its greater remaining A concentration.

The student who considers only A concentration will predict the wrong rate ordering.

(c) Calculate the collected gas volumes.

Every mole of B consumed produces one mole of G.

For Trial L, the concentration of B consumed is:

\[ \begin{aligned} \Delta[B]_L &=0.100-0.0385\\ &=0.0615\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The amount of gas produced is:

\[ \begin{aligned} n_L(G) &=0.0615(0.1000)\\ &=0.00615\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} V_L&=0.00615(24.0)\\ &=0.1476\ \mathrm{L}\\ &=147.6\ \mathrm{mL} \end{aligned} \]

For Trial H:

\[ \begin{aligned} \Delta[B]_H &=0.100-0.0112\\ &=0.0888\ \mathrm{mol\,L^{-1}} \end{aligned} \]

\[ \begin{aligned} n_H(G)&=0.0888(0.1000)\\ &=0.00888\ \mathrm{mol} \end{aligned} \]

\[ \begin{aligned} V_H&=0.00888(24.0)\\ &=0.21312\ \mathrm{L}\\ &\approx213.1\ \mathrm{mL} \end{aligned} \]

At 80 seconds, Trial H has collected approximately 213.1 mL, compared with 147.6 mL for Trial L.

So Trial L is currently producing gas faster, but Trial H has already produced substantially more gas.

A faster current rate does not mean a larger accumulated amount.

(d) Can Trial L eventually overtake Trial H?

No. Under the supplied rate law and constant conditions, Trial L cannot overtake Trial H in accumulated gas volume.

This is the hardest part because it requires comparing the reactions at a hypothetical future point where their gas volumes might become equal.

Suppose both had produced exactly the same amount of gas.

Because G forms in a 1:1 ratio with B, both must then have consumed exactly the same amount of B.

They would therefore have equal remaining B concentrations.

But Trial H started with \(0.100\ \mathrm{mol\,L^{-1}}\) more A.

Since both consume equal amounts of A when they produce equal amounts of G, Trial H would still have \(0.100\ \mathrm{mol\,L^{-1}}\) more A.

At equal accumulated gas volumes:

• Both trials have the same remaining B concentration.
• Trial H has more remaining A.
• Their rate law is \(r=k[A][B]\), with the same positive \(k\).

Therefore, as long as B remains available, Trial H must have the higher instantaneous rate whenever their accumulated volumes are equal.

That is the decisive constraint.

Trial H begins producing gas faster and moves ahead. Although Trial L may later have the steeper gas-volume curve, Trial L cannot pass Trial H: at any hypothetical point of equality before completion, Trial H would again be producing gas faster.

The curves can approach one another as they approach their shared final volume, but they cannot cross.

(e) Determine the final gas yields.

Both trials initially contain:

\[ \begin{aligned} n_0(B)&=0.100(0.1000)\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Trial L has the same initial amount of A and B, while Trial H has excess A.

In either case, at most 0.0100 mol of B can react, producing 0.0100 mol G.

Thus:

\[ \begin{aligned} V_{\max}&=0.0100(24.0)\\ &=0.240\ \mathrm{L}\\ &=240\ \mathrm{mL} \end{aligned} \]

Both approach a theoretical final gas volume of 240 mL.

In this idealised rate-law model, the final yield is approached asymptotically rather than reached exactly at a finite mathematical time.

The result reveals three different comparisons:

• At the start, Trial H has twice the instantaneous rate.
• At 80 seconds, Trial L has the greater instantaneous rate, but Trial H has more accumulated gas.
• At eventual completion, both have the same total gas yield.

The reaction with the larger initial concentration can finish no further ahead in total product, even though it remains ahead throughout the reaction.

07What to carry into the next reaction-rate question

When concentration doubles, do not immediately write that the rate doubles.

First identify the measurement. A gas volume tells you how much product has accumulated. A gradient over an interval gives an average rate. A tangent gives an instantaneous rate.

Then ask whether the two reactions are being compared at the same starting conditions, the same elapsed time, or the same extent of reaction. Those are different comparisons.

Finally, separate the chemical prediction from the measurement. Collision theory can explain why increased concentration generally increases reaction rate, but an exact multiplier requires additional kinetic evidence or a supplied rate law.

This leads to a useful next step in experimental chemistry: using carefully controlled initial-rate measurements to determine how a reaction actually depends on concentration, rather than assuming that its rate follows directly from the balanced chemical equation.