How to Separate Concentration and Volume Effects in Rate Practicals

Two flasks contain identical strips of magnesium. Each receives 10.0 mL of 1.00 mol/L hydrochloric acid, but one also receives 10.0 mL of water, while the other receives 30.0 mL.

Both flasks contain exactly the same amount of acid. Should they produce hydrogen at the same rate?

Not necessarily. The first mixture has an acid concentration of 0.500 mol/L, while the second has a concentration of 0.250 mol/L. The acid particles are spread through different volumes of solution.

To separate concentration effects from volume effects in a reaction-rate practical, keep the total reaction volume constant when changing concentration. Replace some reactant solution with water rather than simply changing the amount of liquid. Then run a separate comparison in which you change the total liquid volume while keeping the starting concentration constant.

Measure gas production over the same time interval, and check that your comparisons aren’t accidentally changing the amount of limiting reactant.

01Why the original comparison cannot separate the effects

The problem begins with three quantities that sound similar but mean different things.

Only the third quantity is the direct measurement from a gas syringe.

The first two determine the starting concentration.

Equal moles do not mean equal concentrations

Concentration tells us how many moles of a dissolved substance are present in each litre of solution.

The relationship is:

\[ c=\frac{n}{V} \]

Here:

• \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\).
• \(n\) is the amount of dissolved substance in mol.
• \(V\) is the total solution volume in L.

Suppose both flasks receive 10.0 mL of 1.00 mol/L HCl.

The amount of HCl added to either flask is:

\[ \begin{aligned} n&=cV\\ &=1.00\times0.0100\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Now compare the concentrations after dilution.

In the 20.0 mL mixture:

\[ c=\frac{0.0100}{0.0200} =0.500\ \mathrm{mol\,L^{-1}} \]

In the 40.0 mL mixture:

\[ c=\frac{0.0100}{0.0400} =0.250\ \mathrm{mol\,L^{-1}} \]

The smaller mixture has twice the starting concentration.

It’s like mixing the same amount of cordial with two different amounts of water. The total amount of cordial hasn’t changed, but its concentration has.

The analogy stops there: hydrochloric acid actually reacts and is consumed. Its concentration changes during the experiment, even if no extra water is added.

This is the first important limitation of the original comparison. If the more concentrated mixture reacts faster, we cannot conclude that a smaller liquid volume directly caused the faster reaction. Concentration and volume changed together.

02What the apparatus actually measures

Consider the reaction:

\[ \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \]

One mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.

The experiment uses an airtight conical flask and a gas syringe.

A magnesium strip sits at the bottom of the flask. A premixed solution of hydrochloric acid and water is added, and the flask is immediately sealed with a bung.

A delivery tube runs from the bung to the gas syringe. As hydrogen forms, the increase in gas quantity pushes the syringe plunger outwards. The experimenter records its displacement at regular time intervals.

The stopwatch begins when the acid first contacts the magnesium.

The gas syringe does not measure hydrochloric acid concentration. It measures a change in gas volume.

At constant gas temperature and pressure, and assuming effective collection, more hydrogen produced means a greater increase in syringe volume.

What should we predict?

If the magnesium strips have the same exposed surface area and everything else is controlled, the mixture with the higher acid concentration would generally be expected to produce hydrogen faster at the beginning.

Why?

There are more acid particles per unit volume available to react at the magnesium surface. This can increase the frequency of successful reacting collisions.

But predicting the final gas volume requires a different argument.

Suppose each flask contains 0.0243 g of magnesium.

Using a molar mass of \(24.3\ \mathrm{g\,mol^{-1}}\):

\[ \begin{aligned} n(\ce{Mg}) &=\frac{0.0243}{24.3}\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

The balanced equation gives a 1:1 mole ratio between magnesium and hydrogen.

Assume complete reaction and an ideal-gas molar volume of approximately \(24.8\ \mathrm{L\,mol^{-1}}\) at \(25^\circ\mathrm{C}\) and 100 kPa.

The theoretical hydrogen volume is:

\[ \begin{aligned} V(\ce{H2}) &=0.00100\times24.8\\ &=0.0248\ \mathrm{L}\\ &=24.8\ \mathrm{mL} \end{aligned} \]

The magnesium requires:

\[ n(\ce{HCl})=2(0.00100) =0.00200\ \mathrm{mol} \]

Both original mixtures contain 0.0100 mol of HCl, so acid is in excess.

Under these assumptions, both flasks should eventually produce the same 24.8 mL of hydrogen, even though their rates can differ.

That gives us two separate predictions:

• Starting acid concentration can affect how quickly hydrogen is produced.
• The amount of limiting magnesium determines the final hydrogen yield.

A larger final gas volume is not, by itself, evidence of a faster reaction.

03Revise the practical: make two controlled comparisons

The original design changed both acid concentration and total liquid volume. We can fix that by constructing a set of four trials.

Each trial uses the same:

• 0.0243 g clean magnesium strip, with the same exposed surface area and shape
• 1.00 mol/L HCl stock solution
• Conical flask, bung, delivery tube, and gas syringe
• Starting temperature, mixing procedure, and gas collection conditions

The stock acid and water are premixed before being added to the magnesium.

Four-trial comparison

The square brackets in \([\ce{HCl}]\) mean the molar concentration of hydrochloric acid.

All four trials have enough acid to consume the magnesium completely. Even trial D contains 0.00500 mol of HCl, more than the required 0.00200 mol.

Which trials answer which question?

Compare A with B.

Both contain 40.0 mL of liquid. Their starting HCl concentrations differ.

This is a controlled comparison of concentration, assuming the other experimental conditions are maintained.

Now compare A with C.

Both begin at 0.500 mol/L HCl, but one contains 40.0 mL of liquid and the other contains 20.0 mL.

This comparison tests whether changing the total liquid volume affects the observed gas-production behaviour when the starting acid concentration is held constant.

You can repeat the comparisons using C against D for concentration and B against D for volume.

There is an important subtlety here.

Keeping concentration constant while increasing solution volume also increases the total moles of acid. You cannot independently fix concentration, total volume, and acid amount because \(n=cV\).

That is why keeping magnesium as the limiting reactant is useful. It prevents the different excess acid amounts from automatically producing different final hydrogen yields.

The larger liquid volume can still influence mixing, temperature changes, the gas headspace, and how reactant concentration changes later in the experiment. A matched starting concentration does not eliminate every possible volume-related effect.

For the wider method of controlling practical variables and experimental error, see How to Design a Fair Reaction-Rate Chemistry Practical.

04Read the results without mixing up rate and yield

Suppose the revised practical produced the following data.

Illustrative data only: These values are constructed for teaching. They are not measurements from a reported experiment.

All gas volumes are in mL. Assume complete gas collection at constant temperature and pressure, with negligible dissolution of hydrogen.

Look first at trials A and B.

After 10 seconds, A has produced 12.0 mL of hydrogen, while B has produced 7.0 mL. Both mixtures started with the same total liquid volume.

That supports the prediction that the higher starting HCl concentration gives a faster early reaction under these conditions.

Now compare A and C.

Their starting concentrations are identical. Their gas volumes after 10 seconds are also quite close: 12.0 mL and 11.7 mL.

This suggests that, in these illustrative conditions, changing liquid volume has much less effect on early hydrogen production than changing the starting HCl concentration.

It does not prove that liquid volume never matters.

Average rate is not instantaneous rate

The average rate between two times is the gradient of the straight line joining their points on a gas-volume-time graph. This is a secant.

The instantaneous rate at one particular time is the gradient of the tangent to the smooth curve at that time.

For example, trial A has an average rate of \(1.20\ \mathrm{mL\,s^{-1}}\) between 0 and 10 seconds.

Suppose an illustrative smooth curve fitted through its readings has a tangent at 10 seconds passing through the construction points:

• \(t=5\ \mathrm{s}\), \(V=7.8\ \mathrm{mL}\)
• \(t=15\ \mathrm{s}\), \(V=16.2\ \mathrm{mL}\)

These are points on the tangent line, not additional experimental readings.

The tangent gradient is:

\[ \begin{aligned} r_{10}&=\frac{16.2-7.8}{15-5}\\ &=0.84\ \mathrm{mL\,s^{-1}} \end{aligned} \]

This is an estimated instantaneous gas-production rate at 10 seconds for that constructed curve.

The average rate over the first 10 seconds and the instantaneous rate at 10 seconds answer different questions. The table alone cannot determine an exact instantaneous rate because many smooth curves could pass through the same measured points.

For a fuller treatment of secants, tangents, and changing gradients, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

05A second volume effect: concentration can fall at different rates

There is another trap hiding in the four-trial design.

Trials A and C begin at the same HCl concentration, and their early gas-production rates are similar.

Does that mean their HCl concentrations remain the same throughout the reaction?

No.

Imagine removing the same number of acid particles from two different volumes of solution. The smaller volume loses a greater concentration of particles.

This happens even if the chemical reaction consumes nearly the same number of moles per second.

Why matching starting concentrations is not always enough

At the beginning of trials A and C, both acid concentrations are 0.500 mol/L.

After some magnesium has reacted, the smaller volume generally experiences a larger reduction in HCl concentration for the same amount of reaction.

So even if the reactions initially behave similarly, their later gas-production rates may diverge.

There is a further physical complication. Changing the liquid volume in the same flask changes the gas headspace, and potentially the mixing and temperature behaviour.

That means an observed difference between A and C cannot automatically be described as a direct chemical effect of liquid volume alone.

A strong practical conclusion would distinguish three possibilities:

The purpose of a controlled experiment is not to make every effect disappear. It is to make the interpretation of the measured differences as clear as possible.

06Decide what comparison you are actually making

Before calculating a rate, identify which volume is changing and what the measured quantity represents.

One especially useful decision rule is to calculate the starting concentration from the final mixed liquid volume, not from the stock bottle label.

Another is to distinguish a gas-production rate in \(\mathrm{mL\,s^{-1}}\) from a concentration disappearance rate in \(\mathrm{mol\,L^{-1}s^{-1}}\). They are related by stoichiometry, but they are not numerically interchangeable.

Finally, remember that changing the volume of liquid can also affect the gas syringe mechanically if liquid is introduced into an already sealed system. A change in syringe position is only a valid measure of chemical gas production when other causes of gas displacement have been excluded.

That last possibility becomes important in the final question.

07Questions and solutions

Question 1

A student investigates the effect of hydrochloric acid concentration on the reaction with identical magnesium strips.

Both trials use 30.0 mL of total liquid.

• Trial A: 8.0 mL of 1.20 mol/L HCl stock and 22.0 mL of water.
• Trial B: 16.0 mL of 1.20 mol/L HCl stock and 14.0 mL of water.

The gas syringe reads zero at the start. After 12.0 seconds, trial A has produced 6.0 mL of hydrogen, while trial B has produced 9.6 mL.

Calculate the starting HCl concentration and average gas-production rate for each trial. Explain which comparison is fair and whether the results establish the instantaneous rates at 12.0 seconds.

Solution 1

Trial B has the higher starting HCl concentration, \(0.640\ \mathrm{mol\,L^{-1}}\), and the higher average gas-production rate, \(0.800\ \mathrm{mL\,s^{-1}}\).

First, use the dilution relationship:

\[ c_{\mathrm{mix}}= \frac{c_{\mathrm{stock}}V_{\mathrm{stock}}} {V_{\mathrm{total}}} \]

All volumes must use the same units in this ratio.

For trial A:

\[ \begin{aligned} c_A&=\frac{1.20\times8.0}{30.0}\\ &=0.320\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For trial B:

\[ \begin{aligned} c_B&=\frac{1.20\times16.0}{30.0}\\ &=0.640\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The average gas-production rates from 0 to 12.0 seconds are:

\[ \begin{aligned} r_A&=\frac{6.0}{12.0}\\ &=0.500\ \mathrm{mL\,s^{-1}}\\[4pt] r_B&=\frac{9.6}{12.0}\\ &=0.800\ \mathrm{mL\,s^{-1}} \end{aligned} \]

The comparison is fair for investigating starting concentration if temperature, magnesium surface area, mixing, gas collection, and other conditions are also controlled.

Both trials contain the same total liquid volume, so changes in liquid depth and headspace are not introduced by the mixing recipe.

However, neither calculated rate is the instantaneous rate at 12.0 seconds. They are averages over the preceding 12 seconds.

The common mistake would be to identify trial B’s larger accumulated gas volume with its exact rate at that final moment. An instantaneous rate requires a tangent to a suitable gas-volume-time curve.

Question 2

Two students compare reactions involving identical 0.0243 g magnesium strips.

Both use 10.0 mL of 1.00 mol/L HCl stock solution.

• Trial X adds 10.0 mL of water.
• Trial Y adds 30.0 mL of water.

The following illustrative gas volumes are measured under identical conditions.

A student concludes:

“The reaction in X is faster because there is less liquid. The results prove that reducing liquid volume increases reaction rate.”

Evaluate this conclusion by:

1. Calculating both starting concentrations and both average gas-production rates between 5 and 20 seconds.
2. Explaining whether the data establish an independent liquid-volume effect.
3. Designing one additional trial that would help isolate the liquid-volume effect.
4. Predicting the final hydrogen volume in X and Y.

Assume complete reaction, excess HCl, full gas collection, and a gas molar volume of \(24.8\ \mathrm{L\,mol^{-1}}\).

Solution 2

Trial X has a higher average gas-production rate, \(0.740\ \mathrm{mL\,s^{-1}}\), than trial Y, \(0.533\ \mathrm{mL\,s^{-1}}\), between 5 and 20 seconds. But the data do not establish that liquid volume directly caused the difference, because starting concentration also changed.

1. Starting concentrations and average rates

Both trials contain 0.0100 mol of HCl.

For trial X:

\[ c_X=\frac{0.0100}{0.0200} =0.500\ \mathrm{mol\,L^{-1}} \]

For trial Y:

\[ c_Y=\frac{0.0100}{0.0400} =0.250\ \mathrm{mol\,L^{-1}} \]

The average gas-production rates are:

\[ \begin{aligned} r_X&=\frac{17.9-6.8}{20-5}\\ &=\frac{11.1}{15}\\ &=0.740\ \mathrm{mL\,s^{-1}}\\[4pt] r_Y&=\frac{12.0-4.0}{20-5}\\ &=\frac{8.0}{15}\\ &=0.533\ \mathrm{mL\,s^{-1}} \end{aligned} \]

2. Interpretation

Trial X is more concentrated. Its greater average rate could be explained by the difference in HCl concentration.

The student has confused keeping acid moles equal with keeping acid concentration equal.

Because concentration and liquid volume changed together, their individual effects cannot be isolated from X and Y alone.

3. Additional trial

A suitable trial Z would use:

• 20.0 mL of 1.00 mol/L HCl stock
• 20.0 mL of water
• The same magnesium strip and controlled conditions

Trial Z would contain 40.0 mL of liquid at 0.500 mol/L HCl.

Comparing X with Z would hold the starting concentration constant while changing total liquid volume.

Comparing Y with Z would hold the total volume constant while changing starting concentration.

These comparisons would be more informative, although differences in gas headspace, temperature changes, or mixing would still need consideration.

4. Final hydrogen yield

Each strip contains:

\[ n(\ce{Mg})=\frac{0.0243}{24.3} =0.00100\ \mathrm{mol} \]

Both trials contain more than the required 0.00200 mol of HCl.

Therefore, magnesium is limiting in both.

\[ \begin{aligned} V(\ce{H2}) &=0.00100\times24.8\\ &=0.0248\ \mathrm{L}\\ &=24.8\ \mathrm{mL} \end{aligned} \]

Both should produce 24.8 mL of hydrogen at completion under the stated ideal assumptions.

The same final gas yield is compatible with different reaction rates.

Question 3

Two reactions start with 0.500 mol/L hydrochloric acid and identical magnesium strips.

• Flask P contains 20.0 mL of reaction solution.
• Flask Q contains 40.0 mL.

A chemical analysis of the solutions gives the following illustrative concentrations:

A student claims that P must have produced twice as much hydrogen because its HCl concentration decreased twice as much.

Determine the moles of HCl consumed and the predicted hydrogen volume produced in each flask over the 10 seconds.

Then compare the average HCl concentration disappearance rates with the average hydrogen gas-production rates. Explain why these comparisons appear to disagree.

Assume the liquid volumes remain constant, all HCl consumption follows the magnesium reaction, and the gas molar volume is \(24.8\ \mathrm{L\,mol^{-1}}\).

Solution 3

Both flasks consumed 0.00100 mol of HCl and therefore produced 12.4 mL of hydrogen over the 10 seconds.

The student’s claim is incorrect because concentration change must be multiplied by solution volume to determine the amount consumed.

For flask P:

\[ \begin{aligned} n(\ce{HCl\ used}) &=(0.500-0.450)(0.0200)\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

For flask Q:

\[ \begin{aligned} n(\ce{HCl\ used}) &=(0.500-0.475)(0.0400)\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

The balanced equation is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

Therefore, each flask produces:

\[ n(\ce{H2})=\frac{0.00100}{2} =0.000500\ \mathrm{mol} \]

Converting to gas volume:

\[ \begin{aligned} V(\ce{H2}) &=0.000500\times24.8\\ &=0.0124\ \mathrm{L}\\ &=12.4\ \mathrm{mL} \end{aligned} \]

Both average hydrogen gas-production rates are:

\[ r=\frac{12.4}{10} =1.24\ \mathrm{mL\,s^{-1}} \]

But their average HCl concentration disappearance rates are different.

For P:

\[ r_{c,P}=\frac{0.050}{10} =0.00500\ \mathrm{mol\,L^{-1}s^{-1}} \]

For Q:

\[ r_{c,Q}=\frac{0.025}{10} =0.00250\ \mathrm{mol\,L^{-1}s^{-1}} \]

The concentration disappearance rate is twice as large in P because its liquid volume is half as large.

The gas-production rates agree because both reactions consumed the same amount of HCl and produced the same number of hydrogen molecules during the interval.

The key distinction is between moles consumed per second and moles consumed per litre per second.

Neither pair of concentration readings determines the instantaneous rate at exactly 10 seconds. They describe average changes over the interval.

Question 4

Two identical 0.0243 g magnesium strips react separately with 0.500 mol/L HCl.

• Trial A uses 40.0 mL of acid solution.
• Trial C uses 20.0 mL.

At the moment each gas syringe has collected 12.4 mL of hydrogen, a tangent to its gas-volume-time curve gives the following instantaneous gas-production rates:

• Trial A: \(0.82\ \mathrm{mL\,s^{-1}}\)
• Trial C: \(0.70\ \mathrm{mL\,s^{-1}}\)

A student argues:

“The starting concentrations were equal. Trial C now has a lower rate, so this proves that smaller liquid volume directly slows the reaction.”

Evaluate this argument.

Calculate the HCl concentration in each flask when 12.4 mL of hydrogen has been produced.

Then determine the starting HCl concentration trial C would need in order for its concentration to match trial A’s at this same amount of hydrogen production.

Explain why it is impossible to keep both the starting concentrations and these later concentrations equal while using the two different fixed liquid volumes.

Assume complete gas collection at constant temperature and pressure, negligible changes in liquid volume, and a gas molar volume of \(24.8\ \mathrm{L\,mol^{-1}}\).

Solution 4

Trial C has a lower HCl concentration at the comparison point: 0.450 mol/L compared with 0.475 mol/L in trial A. Therefore, the observed difference in instantaneous rates does not prove an independent liquid-volume effect.

To match trial A’s concentration after producing 12.4 mL of hydrogen, trial C would need to start at 0.525 mol/L HCl.

First, calculate the amount of hydrogen produced.

\[ \begin{aligned} n(\ce{H2})&=\frac{0.0124}{24.8}\\ &=0.000500\ \mathrm{mol} \end{aligned} \]

The acid consumed is:

\[ n(\ce{HCl\ used}) =2(0.000500) =0.00100\ \mathrm{mol} \]

Next, calculate the remaining HCl concentrations.

In trial A, the concentration decrease is:

\[ \Delta c_A=\frac{0.00100}{0.0400} =0.0250\ \mathrm{mol\,L^{-1}} \]

So:

\[ c_A=0.500-0.0250 =0.475\ \mathrm{mol\,L^{-1}} \]

In trial C:

\[ \Delta c_C=\frac{0.00100}{0.0200} =0.0500\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ c_C=0.500-0.0500 =0.450\ \mathrm{mol\,L^{-1}} \]

Although the trials started at the same concentration, their acid concentrations are no longer equal after the same amount of magnesium has reacted.

The smaller mixture has lost a greater concentration of HCl.

Now adjust the starting concentration for trial C.

To finish at 0.475 mol/L after losing 0.0500 mol/L:

\[ \begin{aligned} c_{C,\mathrm{initial}} &=0.475+0.0500\\ &=0.525\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This would match the acid concentrations at the chosen reaction extent.

But trial C would then start at 0.525 mol/L, while trial A would still start at 0.500 mol/L.

It is impossible to match both concentrations at the start and after consuming the same non-zero amount of HCl when the fixed solution volumes differ.

The reason follows from:

\[ c_{\mathrm{later}} =c_{\mathrm{initial}}-\frac{n_{\mathrm{used}}}{V} \]

The same number of moles consumed produces different concentration decreases when \(V\) differs.

What can we conclude from the instantaneous rates?

At the comparison point, trial A is producing hydrogen faster: 0.82 rather than \(0.70\ \mathrm{mL\,s^{-1}}\).

But trial A also has a higher current acid concentration.

That concentration difference is a possible explanation for the rate difference. The observation does not isolate a separate physical effect of liquid volume.

Matching the amount of hydrogen already produced is useful because it compares the reactions at the same extent of magnesium consumption. Even then, other effects, such as differing temperature histories or mixing conditions, would need to be considered.

The tempting assumption was that equal starting concentrations guarantee equal concentrations throughout a reaction. They do not.

Question 5

A student investigates magnesium reacting with hydrochloric acid using a sealed conical flask connected to a freely moving gas syringe.

The apparatus has a second, self-sealing inlet through which liquid can be injected without opening the flask.

Initially, the flask contains:

• 0.0243 g of magnesium
• 20.0 mL of 0.500 mol/L HCl
• A gas syringe reading of 0.0 mL

The system is maintained at \(25^\circ\mathrm{C}\) and 100 kPa, and the gas molar volume is taken as \(24.8\ \mathrm{L\,mol^{-1}}\).

At 20 seconds, the gas syringe reads 12.4 mL.

The student then rapidly injects 20.0 mL of pure water through the sealed inlet.

The injection takes negligible time compared with the chemical reaction. No additional hydrogen is produced during the instant of injection.

Assume:

• All hydrogen produced is retained in the gas system.
• The injected water contains no trapped air.
• Gas temperature and pressure remain constant.
• The syringe plunger moves freely.
• Injected liquid displaces an equal volume of gas from the flask into the syringe.
• Solution volumes are additive, and other volume changes are negligible.
• Magnesium remains the limiting reactant.

Determine:

1. The HCl concentration immediately before and immediately after the water injection.
2. The gas syringe reading immediately after injection.
3. The final gas syringe reading after all magnesium has reacted, and the final HCl concentration.
4. Whether the sudden syringe movement proves that the instantaneous chemical reaction rate has increased.
5. A control injection that would help distinguish the concentration effect of dilution from the mechanical effect of injecting extra liquid.

Solution 5

The water injection halves the HCl concentration from 0.450 to 0.225 mol/L at that moment, but also pushes 20.0 mL of existing gas into the gas syringe.

The syringe reading therefore jumps from 12.4 mL to 32.4 mL without any additional hydrogen being produced during the injection.

That is the crucial distinction: not every change in collected gas volume is caused by chemical gas production.

1. Concentration immediately before and after injection

At 20 seconds, the true hydrogen production corresponds to 12.4 mL.

Convert that volume into moles:

\[ \begin{aligned} n(\ce{H2})&=\frac{0.0124}{24.8}\\ &=0.000500\ \mathrm{mol} \end{aligned} \]

The reaction consumes two moles of HCl for each mole of hydrogen.

\[ n(\ce{HCl\ used})=2(0.000500) =0.00100\ \mathrm{mol} \]

Initially, the flask contains:

\[ \begin{aligned} n_0(\ce{HCl})&=0.500(0.0200)\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

After 20 seconds, the HCl remaining is:

\[ n(\ce{HCl})=0.0100-0.00100 =0.00900\ \mathrm{mol} \]

Before injection, this is distributed through 0.0200 L:

\[ c_{\mathrm{before}} =\frac{0.00900}{0.0200} =0.450\ \mathrm{mol\,L^{-1}} \]

Injecting 20.0 mL of pure water increases the total liquid volume to 40.0 mL.

The injection does not add or consume any HCl.

Therefore:

\[ c_{\mathrm{after}} =\frac{0.00900}{0.0400} =0.225\ \mathrm{mol\,L^{-1}} \]

The HCl concentration halves, but the moles of HCl remain unchanged at the instant of dilution.

2. Gas syringe reading immediately after injection

Before injection, the gas syringe reads 12.4 mL.

The added water occupies 20.0 mL of space that previously contained gas inside the sealed flask.

Because the gas is maintained at constant pressure and temperature, that existing gas is displaced into the freely moving syringe.

Thus:

\[ \begin{aligned} V_{\mathrm{reading}} &=12.4+20.0\\ &=32.4\ \mathrm{mL} \end{aligned} \]

No additional hydrogen is needed to explain the syringe movement.

A student who divides this sudden 20.0 mL increase by the very short injection time would obtain an enormous apparent gas-production rate.

That would be a measurement artefact, not a genuine chemical rate.

3. Final syringe reading and HCl concentration

The initial amount of magnesium was:

\[ n(\ce{Mg})=\frac{0.0243}{24.3} =0.00100\ \mathrm{mol} \]

Because magnesium is limiting, the total hydrogen eventually produced is 0.00100 mol.

Its gas volume is:

\[ V(\ce{H2})=0.00100(24.8) =0.0248\ \mathrm{L} \]

So the reaction produces 24.8 mL of hydrogen in total.

But the syringe also registers the 20.0 mL of gas displaced by the injected water.

The final reading is therefore:

\[ V_{\mathrm{final}}=24.8+20.0 =44.8\ \mathrm{mL} \]

Only 24.8 mL of that net displacement is attributable to hydrogen production.

The total HCl consumed by the magnesium is:

\[ n(\ce{HCl\ used}) =2(0.00100) =0.00200\ \mathrm{mol} \]

The final amount of HCl remaining is:

\[ n(\ce{HCl\ remaining}) =0.0100-0.00200 =0.00800\ \mathrm{mol} \]

Distributed through 0.0400 L:

\[ c_{\mathrm{final}} =\frac{0.00800}{0.0400} =0.200\ \mathrm{mol\,L^{-1}} \]

The final acid concentration is 0.200 mol/L.

4. Does the syringe jump indicate a faster reaction?

No.

The syringe reading contains two contributions:

• Volume displaced by genuine hydrogen production.
• Volume displaced mechanically by injecting liquid into the sealed flask.

During the assumed instantaneous injection, the first contribution is zero, while the second is 20.0 mL.

The apparent surge in gas volume is not a surge in chemical reaction rate.

After the injection, the HCl concentration is lower. If all other relevant conditions remain unchanged, the lower concentration would generally be expected to reduce the hydrogen-production rate at the magnesium surface.

However, without further kinetic information, we cannot calculate the exact new instantaneous rate or claim that it must fall by precisely half.

There is also a second mathematical trap. The sudden fall in HCl concentration from 0.450 to 0.225 mol/L is caused by dilution, not instantaneous chemical consumption.

The simple expression

\[ -\frac{\Delta c}{\Delta t} \]

only describes the chemical disappearance of HCl when concentration changes are being interpreted appropriately, including the condition of constant solution volume.

Across this injection, the correct quantity to track is the amount of HCl:

\[ n=cV \]

That amount does not change during the assumed instantaneous dilution.

5. Design a matched injection control

A useful control would be to repeat the experiment from the same pre-injection state, but inject 20.0 mL of 0.450 mol/L HCl instead of pure water.

The two injections would increase the liquid volume by the same amount.

Both should therefore cause the same immediate 20.0 mL mechanical displacement in the idealised apparatus.

But their concentration effects would differ.

With water, the HCl concentration falls to 0.225 mol/L.

With 0.450 mol/L HCl solution, the concentration remains 0.450 mol/L immediately after mixing, because the added solution matches the concentration already in the flask.

The experimenter could subtract the mechanical syringe displacement and compare the subsequent hydrogen-production rates, while controlling temperature and mixing.

This creates a much stronger comparison between dilution and unchanged concentration at the same liquid volume.

A non-reacting blank using the same injection apparatus would also help verify that the immediate syringe movement is mechanical.

The surprising result is that a gas syringe can jump forwards at exactly the moment the reacting acid becomes less concentrated. Those observations only seem contradictory if every syringe movement is assumed to represent newly formed hydrogen.

07What this means for your next rate practical

Before comparing two reaction-rate curves, reconstruct the starting mixture.

Calculate the actual concentration after dilution, check the total liquid volume, and identify which reactant determines the final amount of product.

Then decide whether you are measuring accumulated product, gas produced per second, or reactant concentration lost per second.

These are related quantities, but they answer different questions.

For a concentration investigation, replacing acid stock with water while holding total volume constant is the clean starting method. For a liquid-volume investigation, preserve the initial concentration by scaling the acid and water together, and recognise that later concentrations and physical measurement conditions may still diverge.

The next important experimental decision is where to compare rates: over an identical early time interval, from tangents at specified times, or at a matched amount of reaction progress.

Choosing that comparison before collecting the data makes the result much easier to interpret. A graph can reveal what happened, but it cannot repair a mixing recipe that changed two uncontrolled variables at once.