Why Powdered Calcium Carbonate Reacts Faster Than Marble Chips

Drop powdered calcium carbonate into hydrochloric acid, and it fizzes rapidly. Drop the same mass of calcium carbonate in as a few large marble chips, and the bubbling is usually slower. Does that mean the powder will produce more carbon dioxide overall? No. Under identical conditions, powdered calcium carbonate usually reacts faster because its smaller particles expose more surface area to the acid, allowing more successful collisions at the solid’s surface each second. If the acid is in excess and all the calcium carbonate reacts, both samples produce the same total amount of carbon dioxide. The difference is how quickly they produce it.

01Why breaking calcium carbonate into smaller pieces increases its reaction rate

Calcium carbonate is a solid. Hydrochloric acid is an aqueous solution, so the acid can only react with calcium carbonate where the solution reaches the solid.

Imagine a single large marble chip sitting at the bottom of a flask. Acid can reach its outside, but not the carbonate buried inside.

Now imagine crushing that chip into hundreds of smaller grains. Much of the material that was previously inside the solid becomes exposed.

The mass hasn’t increased. The amount of calcium carbonate hasn’t increased. But the total surface available for reaction has.

The reaction is:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]

Here, calcium carbonate is the solid, hydrochloric acid and calcium chloride are aqueous, carbon dioxide is the gas, and water is liquid.

The bubbles are carbon dioxide, \(\ce{CO2}\), being produced as carbonate reacts with the acid.

Picture the change in surface area

Consider a simplified solid cube with sides of 1.0 cm.

Its surface area is:

\[ \begin{aligned} A &= 6l^2\\ &= 6(1.0)^2\\ &= 6.0\ \mathrm{cm^2} \end{aligned} \]

Here, \(A\) is surface area in square centimetres, and \(l\) is the cube’s side length in centimetres.

Now divide the cube into eight smaller cubes, each with sides of 0.50 cm.

Each small cube has six exposed faces. If all eight cubes are separated, their combined surface area is:

\[ \begin{aligned} A &= 8\times6(0.50)^2\\ &= 12.0\ \mathrm{cm^2} \end{aligned} \]

The same volume of solid now has twice the surface area.

This is a geometric model, not a claim that real calcium carbonate particles are cubes. Actual powder particles are irregular, and some surfaces may be inaccessible if the powder clumps together.

But the underlying idea remains: smaller particles generally provide more exposed surface per gram.

What collision theory adds

Collision theory explains why that extra surface matters.

For calcium carbonate to react with hydrochloric acid, acid particles must reach reactive sites at the solid surface. In aqueous solution, the reacting acid species can be represented as \(\mathrm{H}^{+}\), shorthand for hydrated hydrogen ions.

A successful encounter at the surface contributes to the chemical reaction.

With a large piece, relatively few surface sites are available at one time.

With powder, many more sites are accessible, so more successful encounters can occur per second.

This increases the reaction rate.

An important detail: crushing calcium carbonate does not automatically give acid particles more kinetic energy or lower the reaction’s activation energy. Those would be different explanations.

The powder reacts faster because more surface is available, not because individual collisions are necessarily more energetic.

02How could we test this properly?

Watching which flask fizzes more vigorously is a useful first observation. But bubbles alone aren’t a precise measurement.

A better investigation measures how quickly carbon dioxide is produced while changing only the particle size.

First design: measure the mass lost from an open flask

One possible method is to place a conical flask containing hydrochloric acid on a digital balance.

Add calcium carbonate and start a stopwatch.

As carbon dioxide forms, bubbles travel upwards through the solution and escape through the open neck. The balance reading decreases because gas is leaving the apparatus.

What should we predict?

The powdered sample should lose mass more quickly at the beginning, because it should produce escaping carbon dioxide faster.

This method can work, but it has a weakness.

A vigorous powdered sample might splash tiny droplets out of the flask. The balance would record that lost liquid as well as the escaping gas.

Some carbon dioxide can also remain dissolved in the solution rather than escape immediately. Evaporation and differences in mixing can introduce further uncertainty.

Mass loss therefore measures material leaving the flask, not carbon dioxide production perfectly.

Revised design: collect the gas in a syringe

A gas syringe gives us a more direct measurement of the gas collected.

Imagine the apparatus from left to right:

1. A conical flask contains a measured volume of hydrochloric acid.
2. A small dry inner vial holds the calcium carbonate above the acid.
3. A rubber bung seals the flask, with a delivery tube passing through it.
4. The tube connects to a horizontal gas syringe with a graduated barrel and movable plunger.

The flask is sealed before the reactants touch. At the start, the flask is tipped gently so the calcium carbonate falls into the acid.

Carbon dioxide forms at the solid’s surface, bubbles rise through the liquid, and gas travels through the delivery tube into the syringe. As gas enters, the plunger moves outwards.

Record the gas volume at regular time intervals.

Pre-sealing the apparatus matters particularly for powder. If the solid is added before the bung is fitted, a fast reaction may release substantial carbon dioxide before collection begins.

The syringe method still isn’t perfect: leaks, gas remaining dissolved, and delays in gas transfer can affect readings.

Keep the comparison fair

For two trials, use the following conditions:

Use fresh acid for each trial. Keep the purity of the calcium carbonate, the starting procedure, and the gas-collection conditions the same.

The independent variable is particle size. The dependent variable is the volume of carbon dioxide collected over time.

The main controlled variables are the amounts of reactants, acid concentration, temperature, and measurement conditions.

We can now interpret the results as a test of the effect of particle size.

03What the gas-volume data should show

The following data are illustrative, constructed to demonstrate a plausible result. They are not measurements from a reported experiment.

Both samples begin producing carbon dioxide at the start.

The powder produces much more gas during the first 30 seconds. It then slows and reaches its final collected volume of 98 mL at around 90 seconds.

The large pieces produce gas more slowly. Their volume continues increasing until about 180 seconds, when it reaches the same 98 mL.

Three observations matter.

First, the powder produces carbon dioxide more rapidly initially.

Second, both reactions slow as they proceed.

Third, both produce the same final gas volume under these conditions.

The first observation supports the exposed-surface explanation because the other important starting conditions were controlled.

The third observation is consistent with both samples containing the same number of moles of calcium carbonate.

But equal final gas volumes don’t automatically prove complete reaction. If the acid were limiting in both trials, both could stop while calcium carbonate remained. We need a mole calculation to establish why 98 mL is reasonable here.

04Calculate the rate without confusing it with the gas volume

The gas volume tells us how much carbon dioxide has been collected.

The rate tells us how quickly that volume is changing.

For these trials, we define the volumetric gas-collection rate as positive when collected gas volume increases.

Over a time interval:

\[ r_{\mathrm{avg}}=\frac{\Delta V}{\Delta t} \]

Here:

• \(r_{\mathrm{avg}}\) is the average gas-collection rate, in mL/s.
• \(\Delta V\) is the change in collected gas volume, in mL.
• \(\Delta t\) is the elapsed time, in seconds.

This gives an average over an interval, not the exact rate at one instant.

Average rate is not instantaneous rate

Suppose you want the rate at exactly 20 seconds.

Dividing the first 20 seconds’ gas volume by 20 gives the average rate over that interval. It does not give the exact rate at 20 seconds.

To estimate the instantaneous rate from a smooth gas-volume-time curve, draw a tangent at the required time.

A tangent is a straight line that follows the curve’s direction at one point. Its gradient represents the rate at that instant.

For a gas-volume graph:

• A steeper upward tangent means a greater instantaneous gas-collection rate.
• A flatter tangent means a smaller instantaneous rate.
• A horizontal tangent means the collected gas volume is no longer increasing at that instant.

This distinction becomes especially important when the powder has already reacted substantially, while the large pieces still have plenty of unreacted solid.

05When powder might not appear to react faster

Here’s a more demanding prediction.

Imagine powdered calcium carbonate in very dilute hydrochloric acid and a large marble chip in concentrated hydrochloric acid.

Must the powder produce carbon dioxide faster?

No. The concentration difference could outweigh the surface-area advantage.

Higher acid concentration generally means more acid particles are available to encounter the calcium carbonate surface each second.

If you change particle size and acid concentration together, you cannot confidently attribute the observed difference to either variable alone.

More surface area only helps if the acid can reach it

There is another limitation.

Imagine a sample of very fine powder compacted into a dense lump. Acid may initially contact only its exterior and any accessible pores.

Now imagine a large, porous piece of calcium carbonate with cracks and channels that the acid can enter.

Despite its larger overall size, the porous piece might offer more accessible surface than the tightly packed powder.

This doesn’t contradict collision theory.

The important quantity is not simply the number of particles or the surface area calculated from their ideal shapes. It is the surface that is actually accessible to the reacting solution.

Also, accessible surface can change during reaction. Particles shrink, pieces fracture, and newly exposed surfaces may become available.

Don’t confuse reaction rate with the amount that reacts

Suppose both flasks begin with equal amounts of acid, but the acid runs out before all the calcium carbonate reacts.

The powder may reach the final gas volume much sooner. Large pieces may take longer.

Neither can produce more carbon dioxide than the limiting amount of acid permits.

Likewise, faster bubbling does not establish that the reaction is more exothermic. Rate describes how quickly a reaction proceeds; enthalpy change describes the energy change associated with the reaction. These are separate ideas, explored in Why a Faster Reaction Is Not Necessarily More Exothermic.

For this calcium carbonate investigation, the strongest conclusion comes from three linked checks: control the starting conditions, compare gas production over equivalent time intervals, and use stoichiometry to check the final gas amount.

06Questions and solutions

Question 1

Two students react equal 0.500 g samples of pure calcium carbonate with identical excess quantities of hydrochloric acid.

Sample A is a fine powder. Sample B consists of several large marble chips.

Both reactions occur at the same temperature, with identical mixing and gas-collection conditions.

(a) Predict which sample should initially produce carbon dioxide faster.

(b) Predict whether the total amount of carbon dioxide produced will be different once both reactions are complete.

(c) Explain your answers using collision theory and the balanced chemical equation.

Solution 1

Sample A should initially produce carbon dioxide faster, but both samples should produce the same total amount of carbon dioxide after complete reaction.

(a) The powder has a greater exposed surface area for the same mass of calcium carbonate. More reactive surface sites are available for collisions with acid particles, increasing the number of successful encounters per second.

(b) Both samples contain 0.500 g of the same pure substance. Because the acid is in excess, all the calcium carbonate can react.

(c) The equation

\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]

shows that one mole of calcium carbonate produces one mole of carbon dioxide.

Changing particle size does not change this mole ratio or the starting number of moles.

The tempting mistake is to assume that more vigorous bubbling means more gas in total. It means gas is being produced faster, not necessarily that more gas can be produced.

Question 2

A student compares powdered calcium carbonate with large marble chips using the following conditions.

Both samples contain 0.400 g of pure calcium carbonate.

The temperature is 25°C. Assume a molar gas volume of 24.5 L/mol and complete collection of the carbon dioxide.

(a) Calculate the average gas-collection rate during the first 20 seconds for both trials.

(b) Calculate the maximum carbon dioxide volume that each trial can produce.

(c) A student concludes, “Large marble chips react faster than powder.” Evaluate this conclusion and propose a fairer investigation.

Solution 2

Large pieces have the higher measured average gas-collection rate in these trials, but this does not establish that large pieces are intrinsically faster. The trials use different acid concentrations, and the powder trial is limited by its smaller amount of acid.

(a) For powder:

\[ \begin{aligned} r_{\mathrm{avg,P}} &=\frac{24.0}{20}\\ &=1.20\ \mathrm{mL/s} \end{aligned} \]

For large pieces:

\[ \begin{aligned} r_{\mathrm{avg,L}} &=\frac{30.0}{20}\\ &=1.50\ \mathrm{mL/s} \end{aligned} \]

Large pieces have the higher average gas-collection rate over this interval. These values are not the instantaneous rates at exactly 20 seconds.

(b) The powder trial begins with:

\[ \begin{aligned} n(\ce{HCl}) &=0.100\times0.0400\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

The 2:1 acid-to-carbonate mole ratio means this can produce at most:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.00400}{2}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

Thus:

\[ \begin{aligned} V(\ce{CO2}) &=0.00200\times24.5\\ &=0.0490\ \mathrm{L}\\ &=49.0\ \mathrm{mL} \end{aligned} \]

For large pieces, the starting amount of acid is:

\[ \begin{aligned} n(\ce{HCl}) &=0.500\times0.0400\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

The 0.400 g of calcium carbonate contains approximately 0.00400 mol, which needs 0.00800 mol of HCl to react completely.

The acid is in excess in this trial.

Consequently:

\[ \begin{aligned} V(\ce{CO2}) &=0.00400\times24.5\\ &=0.0980\ \mathrm{L}\\ &=98.0\ \mathrm{mL} \end{aligned} \]

(c) The student’s conclusion is not justified.

The marble chips received acid that was five times as concentrated. This provides more acid particles per unit volume and can increase the reaction rate enough to overcome the powder’s surface-area advantage.

Furthermore, the powder trial cannot produce as much total carbon dioxide because its acid is limiting.

A fair repeat should use equal masses of calcium carbonate, identical volumes and concentrations of excess hydrochloric acid, and identical temperatures, mixing procedures, and gas-collection apparatus.

Only particle size should change.

Question 3

Two new trials use 0.400 g of pure calcium carbonate with 50.0 mL of 0.500 mol/L hydrochloric acid.

All carbon dioxide is assumed to be collected at 25°C, with a molar gas volume of 24.5 L/mol.

The following readings are illustrative.

Both trials eventually reach approximately 98 mL.

(a) Calculate each trial’s average gas-collection rate from 0 to 40 seconds and from 30 to 40 seconds.

(b) Is it valid to conclude that the large pieces have a higher instantaneous reaction rate at exactly 40 seconds?

(c) Calculate the approximate mass of unreacted calcium carbonate remaining in the powder trial at 40 seconds.

(d) Explain how the ordering of average rates can reverse over a later interval without contradicting the original surface-area explanation.

Solution 3

Powder has the greater average rate over the full 0-40 s interval, but large pieces have the slightly greater average rate over 30-40 s. This does not establish their exact instantaneous rate ordering at 40 seconds.

(a) Over 0-40 seconds:

For powder:

\[ \begin{aligned} r_{\mathrm{avg,P}} &=\frac{91}{40}\\ &=2.275\ \mathrm{mL/s} \end{aligned} \]

For large pieces:

\[ \begin{aligned} r_{\mathrm{avg,L}} &=\frac{36}{40}\\ &=0.900\ \mathrm{mL/s} \end{aligned} \]

Over 30-40 seconds:

For powder:

\[ \begin{aligned} r_{\mathrm{avg,P}} &=\frac{91-83}{40-30}\\ &=0.800\ \mathrm{mL/s} \end{aligned} \]

For large pieces:

\[ \begin{aligned} r_{\mathrm{avg,L}} &=\frac{36-27}{40-30}\\ &=0.900\ \mathrm{mL/s} \end{aligned} \]

The average rate ranking has reversed over this later interval.

(b) The instantaneous rate at exactly 40 seconds cannot be determined from these readings alone.

The 30-40 s calculation gives a mean gradient over ten seconds. It does not provide the gradient of a tangent at exactly 40 seconds.

More closely spaced readings or a sufficiently accurate continuous curve would be required to estimate the instantaneous rate.

(c) At 40 seconds, the powder has produced 91 mL, or 0.091 L, of carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.091}{24.5}\\ &=0.003714\ \mathrm{mol} \end{aligned} \]

The mole ratio between calcium carbonate consumed and carbon dioxide formed is 1:1.

The mass of calcium carbonate consumed is:

\[ \begin{aligned} m&=nM\\ &=0.003714\times100.1\\ &\approx0.372\ \mathrm{g} \end{aligned} \]

The unreacted mass is therefore:

\[ \begin{aligned} m_{\mathrm{remaining}} &=0.400-0.372\\ &=0.028\ \mathrm{g} \end{aligned} \]

Only about 0.028 g of calcium carbonate remains in the powder trial.

(d) Powder initially has a greater exposed surface area, so it reacts rapidly.

But by 40 seconds, nearly all of its calcium carbonate has reacted. The remaining powder may offer much less total reactive surface than it did initially.

The large pieces have reacted less extensively and may still offer substantial reactive surface.

The acid concentration has also fallen further in the powder trial because it has consumed more acid.

These changing conditions can allow the large pieces to have a higher rate over a later interval.

The tempting mistake is to treat the original surface-area advantage as fixed throughout the reaction. It isn’t: the solids and solutions change as the reaction proceeds.

Question 4

A student attempts to compare powdered calcium carbonate with large marble chips.

Both samples contain 0.400 g of pure calcium carbonate and receive identical excess hydrochloric acid.

For the powder trial, the student drops the powder into the flask and takes six seconds to fit the bung connected to the gas syringe.

For the large-piece trial, the student uses an inner vial to release the chips only after sealing the apparatus.

After both reactions have finished, all the original calcium carbonate has disappeared.

The gas syringe readings are:

Take the molar gas volume as 24.5 L/mol. Assume pure calcium carbonate reacts only with the acid and temperature and pressure remain constant.

(a) Does the result establish that the large pieces produced more carbon dioxide chemically?

(b) Calculate the theoretical gas volume corresponding to 0.400 g of calcium carbonate and the volume not accounted for by the powder syringe reading.

(c) If all of the powder trial’s uncollected gas had escaped during the six-second delay, calculate its average escape rate during that delay. Explain why this estimate cannot be treated as a measured reaction rate.

(d) Explain how the experiment should be redesigned.

Solution 4

No. The large pieces have a higher recorded gas volume, but this does not mean they chemically produced more carbon dioxide. The powder trial’s collection method could have missed gas produced before sealing.

(a) Both trials use 0.400 g of pure calcium carbonate and excess acid.

They therefore have the same theoretical chemical yield. The unequal syringe readings reveal a gas-collection problem or another measurement limitation.

(b) The amount of calcium carbonate is:

\[ \begin{aligned} n(\ce{CaCO3}) &=\frac{0.400}{100.1}\\ &\approx0.00400\ \mathrm{mol} \end{aligned} \]

This produces approximately 0.00400 mol of carbon dioxide.

Its equivalent gas volume is:

\[ \begin{aligned} V(\ce{CO2}) &=0.00400\times24.5\\ &=0.0980\ \mathrm{L}\\ &=98.0\ \mathrm{mL} \end{aligned} \]

The powder syringe recorded 71 mL.

\[ V_{\mathrm{unaccounted}} =98-71 =27\ \mathrm{mL} \]

Approximately 27 mL of the theoretical carbon dioxide volume is not accounted for by the syringe reading.

(c) If every millilitre of that difference escaped during the six-second delay, the estimated average escape rate would be:

\[ \begin{aligned} r_{\mathrm{avg}} &=\frac{27}{6}\\ &=4.5\ \mathrm{mL/s} \end{aligned} \]

However, this result depends on an unverified assumption.

Some carbon dioxide may have remained dissolved in the solution, escaped through a leak after sealing, or been affected by the collection apparatus.

The data don’t establish how much escaped specifically during the six-second interval.

Even if 4.5 mL/s represented the actual average gas loss during that interval, it would not be the instantaneous reaction rate at a particular moment.

(d) Both trials should use a pre-sealed gas-collection apparatus with an internal tipping vial.

Each solid sample should contact the acid only after its flask is sealed.

The temperature, starting volumes, acid concentration, particle mass, and mixing procedure should be controlled.

The apparatus should also be checked for leaks, and measurements should begin at the moment of contact.

The important experimental principle is that a very fast reaction requires a collection system that begins measuring before significant product can escape.

Question 5

Two sealed flasks initially contain identical quantities of reactants:

• 0.400 g of pure calcium carbonate.
• 20.0 mL of 0.400 mol/L hydrochloric acid.

Flask P contains powdered carbonate. Flask L contains large pieces.

Assume the solution volume remains approximately 20.0 mL during the first 40 seconds, all carbon dioxide is collected, and the temperature remains at 25°C. Take the molar gas volume as 24.5 L/mol.

At exactly 40 seconds, the observations are:

A student argues that the large pieces must now have more exposed surface area because their instantaneous rate is higher.

For this question, use the following simplified model at a particular instant:

\[ r=kAc \]

Here, \(r\) is the instantaneous gas-collection rate in mL/s, \(A\) is the accessible wetted surface area in cm², \(c\) is the hydrochloric acid concentration in mol/L, and \(k\) is the same proportionality constant in both flasks under the stipulated conditions.

This is a supplied local model, not a universal rate law for calcium carbonate.

(a) Calculate the HCl concentration remaining in each flask at 40 seconds.

(b) Determine the ratio of the accessible surface area in P to that in L at that moment. Is the student’s conclusion correct?

(c) At exactly 40 seconds, 20.0 mL of fresh 0.400 mol/L HCl is added to each flask. Assume it mixes instantly, no reaction occurs during the addition, the accessible surface areas do not change during that instant, and the model remains valid. Calculate both new acid concentrations and predict the instantaneous gas-collection rates immediately after mixing.

(d) A second student says, “Adding the same acid to both flasks makes acid concentration a controlled variable again.” Evaluate this claim. Predict the final total carbon dioxide volume for each flask once the reactions have finished.

Solution 5

The student’s surface-area conclusion is wrong. At 40 seconds, the powder still has approximately 1.67 times the accessible surface area of the large pieces, despite its lower instantaneous rate. The reason is that the powder has consumed more acid, leaving a lower acid concentration.

Adding the same fresh acid to both flasks then reverses the instantaneous rate ordering under the supplied model, but it does not equalise their acid concentrations.

(a) First calculate the original amount of hydrochloric acid in each flask:

\[ \begin{aligned} n_0(\ce{HCl}) &=0.400\times0.0200\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

In flask P, the carbon dioxide produced is:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.0735}{24.5}\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

Each mole of carbon dioxide corresponds to two moles of HCl consumed.

Therefore:

\[ n_{\mathrm{used,P}}(\ce{HCl}) =0.00600\ \mathrm{mol} \]

The acid remaining in P is:

\[ \begin{aligned} n_{\mathrm{P}} &=0.00800-0.00600\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

The concentration is:

\[ \begin{aligned} c_{\mathrm{P}} &=\frac{0.00200}{0.0200}\\ &=0.100\ \mathrm{mol/L} \end{aligned} \]

For flask L:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.0490}{24.5}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

The acid consumed is 0.00400 mol, leaving:

\[ \begin{aligned} n_{\mathrm{L}} &=0.00800-0.00400\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

Thus:

\[ \begin{aligned} c_{\mathrm{L}} &=\frac{0.00400}{0.0200}\\ &=0.200\ \mathrm{mol/L} \end{aligned} \]

At 40 seconds, the acid concentration in flask L is twice that in flask P.

That difference is the first clue that comparing their raw instantaneous rates doesn’t isolate the effect of surface area.

(b) Use the supplied relationship:

\[ r=kAc \]

For the two flasks, \(k\) is the same. Dividing the two rate expressions gives:

\[ \frac{r_{\mathrm{P}}}{r_{\mathrm{L}}} = \frac{A_{\mathrm{P}}c_{\mathrm{P}}} {A_{\mathrm{L}}c_{\mathrm{L}}} \]

Rearrange:

\[ \frac{A_{\mathrm{P}}}{A_{\mathrm{L}}} = \frac{r_{\mathrm{P}}}{r_{\mathrm{L}}} \frac{c_{\mathrm{L}}}{c_{\mathrm{P}}} \]

Substitute:

\[ \begin{aligned} \frac{A_{\mathrm{P}}}{A_{\mathrm{L}}} &=\frac{1.00}{1.20} \times\frac{0.200}{0.100}\\ &\approx1.67 \end{aligned} \]

The powder still has about 1.67 times the accessible surface area of the large pieces under this model.

The large pieces have the higher instantaneous rate because they are reacting in a solution containing twice the HCl concentration.

This is the central trap: a higher observed rate does not prove a larger surface area when the chemical environment is different.

(c) The fresh acid adds:

\[ \begin{aligned} n_{\mathrm{added}}(\ce{HCl}) &=0.400\times0.0200\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

The total liquid volume in each flask becomes 40.0 mL, or 0.0400 L.

For P:

\[ \begin{aligned} c'_{\mathrm{P}} &=\frac{0.00200+0.00800}{0.0400}\\ &=0.250\ \mathrm{mol/L} \end{aligned} \]

For L:

\[ \begin{aligned} c'_{\mathrm{L}} &=\frac{0.00400+0.00800}{0.0400}\\ &=0.300\ \mathrm{mol/L} \end{aligned} \]

Here, the prime indicates the concentration immediately after the addition.

The accessible surface area hasn’t changed during that instant, and the same proportionality constant applies.

Therefore, each instantaneous rate changes in proportion to its acid concentration.

For P:

\[ \begin{aligned} r'_{\mathrm{P}} &=1.00\times\frac{0.250}{0.100}\\ &=2.50\ \mathrm{mL/s} \end{aligned} \]

For L:

\[ \begin{aligned} r'_{\mathrm{L}} &=1.20\times\frac{0.300}{0.200}\\ &=1.80\ \mathrm{mL/s} \end{aligned} \]

The powder now has the higher instantaneous rate again.

Why?

Before the addition, the large-piece flask had twice the acid concentration of the powder flask.

After the addition, that concentration advantage has become much smaller: 0.300 mol/L compared with 0.250 mol/L.

The powder’s larger accessible surface area now outweighs the remaining concentration difference.

These new rates apply immediately after mixing under the given model. They should not be extrapolated unchanged over the rest of either reaction because surface area and concentration will continue changing.

(d) The second student’s claim is incorrect.

Adding identical quantities of acid does not necessarily produce identical final concentrations when the flasks contained different amounts of acid before the addition.

The concentrations remain 0.250 mol/L and 0.300 mol/L, respectively.

To compare accessible surface area while truly controlling acid concentration at that moment, the two systems would need the same acid concentration, not merely the same additional dose.

Finally, both flasks started with 0.400 g of pure calcium carbonate, approximately 0.00400 mol.

After the acid addition, both have enough hydrochloric acid to consume all remaining calcium carbonate.

The total amount of carbon dioxide formed in each will therefore be approximately 0.00400 mol.

Its equivalent gas volume is:

\[ \begin{aligned} V(\ce{CO2}) &=0.00400\times24.5\\ &=0.0980\ \mathrm{L}\\ &=98.0\ \mathrm{mL} \end{aligned} \]

Both eventually reach approximately 98 mL of total collected carbon dioxide, assuming complete collection.

The powder began faster, became slower at a particular moment, and then became faster again when the acid conditions changed.

None of those rate changes altered the final stoichiometric amount of carbon dioxide available from the original calcium carbonate.

That is why a reaction-rate comparison must specify both the time of comparison and the conditions at that time.

07What to use when the next gas graph is less straightforward

For an unfamiliar calcium carbonate experiment, first decide whether you’re being asked about the amount of gas formed or the speed at which it forms.

If you’re comparing rates, use the same time interval for average rates or tangents at the same time for instantaneous rates. Then ask whether differences in exposed surface area, acid concentration, or remaining reactant could explain the result.

If you’re comparing final gas volumes, check the limiting reactant before making a claim about particle size.

For a closer look at reading gradients accurately, continue with How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

The useful next step is to connect the shape of a reaction-rate curve to how the reaction mixture changes over time. Powdered calcium carbonate makes that connection particularly clear: more exposed surface makes the initial reaction faster, but the conditions responsible for the rate never remain completely fixed as the solid reacts.