Why Reaction-Progress Curves Become Less Steep Over Time

A gas syringe connected to a reaction flask fills quickly at first, then more slowly. During the first 20 seconds, it collects 24 mL of oxygen. During the next 20 seconds, it collects only another 18 mL.

The reaction is still producing oxygen. So why is the reaction-progress curve becoming less steep?

The key is to distinguish how much product has formed from how quickly it is forming. The height of the curve tells you the amount of product collected. Its gradient tells you the reaction rate.

In a simple reaction at constant temperature, the curve becomes less steep because reactants are gradually consumed. Their concentration falls, successful collisions become less frequent, and less product forms each second.

Let’s work through the experimental evidence and the graph calculations so you can explain this properly in an HSC Chemistry response.

01What does a less steep reaction-progress curve actually mean?

A reaction-progress curve shows how a measurable quantity changes with time.

Depending on the experiment, the vertical axis might show:

• The concentration of a reactant.
• The concentration of a product.
• The volume of gas collected.
• The mass of a reaction mixture.

The important idea is that the gradient represents how quickly the measured quantity changes.

Consider two different experiments.

Experiment A: A reaction produces a gas. The volume collected increases with time, but the curve gradually flattens.

Experiment B: A reactant is being consumed. Its concentration decreases with time, but the curve gradually becomes less steep.

Both experiments show the same underlying behaviour: the reaction rate is decreasing.

The sign of the gradient depends on what you measure.

For a reactant concentration graph, the gradient is negative because the concentration is decreasing. However, reaction rates are conventionally reported as positive values when describing the rate of disappearance.

For example:

\[ r_{\text{disappearance}} = -\frac{\Delta[\text{reactant}]}{\Delta t} \]

The negative sign converts the negative gradient into a positive rate.

One distinction is worth remembering:

A reaction-progress curve is not a reaction-rate-versus-time graph.

The progress curve shows a quantity changing over time. You calculate the reaction rate from its gradient.

02How an experiment produces a flattening curve

Let’s investigate the decomposition of hydrogen peroxide.

The reaction is:

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

Hydrogen peroxide decomposes into water and oxygen gas. Potassium iodide can be added as a catalyst to increase the reaction rate.

We can measure the oxygen produced using a gas syringe.

Setting up the experiment

Imagine the apparatus from left to right.

1. A conical flask contains a measured volume of hydrogen peroxide solution.
2. The flask sits in a thermostatically controlled water bath to maintain a constant temperature.
3. A measured amount of potassium iodide catalyst is introduced through an injection port.
4. A bung seals the flask.
5. A delivery tube connects the flask to a gas syringe.
6. As oxygen forms, it travels through the tube and pushes the syringe plunger outwards.

The timer starts when the catalyst is introduced. The volume of oxygen is recorded at regular time intervals.

What should we expect to observe?

If the reaction proceeds at a constant rate, the syringe should collect the same additional volume of oxygen during every equal time interval.

For example, it might collect 20 mL every 20 seconds.

That would produce a straight-line volume-time graph.

If the reaction slows down, however, each successive 20-second interval will produce less oxygen than the previous one. The graph will gradually flatten.

Illustrative experimental data

Suppose the following measurements are obtained.

The values below are illustrative, not measurements from an actual experiment.

Notice the pattern.

During the first 20 seconds, 24 mL of oxygen forms.

During the next 20 seconds, only 18 mL forms.

During the final 20 seconds in the table, only 3 mL forms.

The reaction is producing progressively less oxygen per second.

There is an important experimental distinction here.

The gas syringe measures the amount of oxygen collected, not the oxygen production rate directly. At constant temperature and pressure, gas volume is proportional to the amount of gas in moles, so the volume-time gradient provides a useful measure of the production rate.

The units in this experiment are mL/s, not mol/L/s. Converting to a molar reaction rate requires additional information, including the gas conditions and reaction stoichiometry.

What the experimental design needs to control

A flattening curve demonstrates that the measured rate is decreasing. It does not automatically prove that falling reactant concentration is the only cause.

For example, temperature might change during the experiment. Since temperature also affects reaction rate, an uncontrolled experiment cannot separate these effects.

A better design uses:

• A thermostatically controlled water bath.
• The same initial hydrogen peroxide concentration and volume.
• The same catalyst concentration.
• A leak-free gas collection system.
• Consistent temperature and pressure conditions for gas-volume readings.

A thermometer can confirm that the reaction temperature remains stable.

The revised design makes the concentration explanation much more convincing because an important competing variable has been controlled.

03Why does the reaction rate decrease?

Now we can explain what is happening at the particle level.

Hydrogen peroxide molecules are being consumed as the reaction proceeds.

At the beginning, the solution contains a relatively high concentration of hydrogen peroxide. There are many reactant molecules in a given volume, so productive encounters occur frequently.

As the reaction continues, more hydrogen peroxide is converted into water and oxygen.

The concentration of hydrogen peroxide decreases.

There are now fewer hydrogen peroxide molecules in the same volume of solution. This reduces the frequency of productive encounters, so fewer successful reactions occur each second.

The reaction rate therefore decreases.

Understanding this through collision theory

Collision theory explains reaction rates by considering how particles interact.

For a reaction to occur through a collision, the reacting particles generally need:

• Sufficient energy to overcome the activation energy.
• A suitable orientation when their structures require it.

The activation energy, \(E_a\), is the minimum energy required for a successful reaction pathway.

Think of a crowded school canteen. When lots of students are moving around in the same space, encounters are frequent. When most students leave, encounters become less frequent.

Reactant particles behave somewhat similarly. Fewer particles in a given volume means fewer opportunities for collisions.

The analogy has a limitation, though. Molecules move randomly, and not every collision produces a reaction. The energy and orientation of the collision also matter.

Does the temperature decrease as the reaction proceeds?

Not necessarily.

This is a common misconception.

You might predict that the particles gradually lose kinetic energy as the reaction uses up its reactants. However, if the reaction is maintained at a constant temperature, the average kinetic energy of the particles remains constant.

The fraction of particles with energy greater than or equal to the activation energy also remains approximately constant.

What changes is the number of reactant particles available to collide.

Consequently, even if the proportion of collisions with sufficient energy stays the same, the total number of successful collisions per second decreases.

This is why concentration and temperature must be treated as separate factors.

What about the catalyst?

A catalyst provides an alternative reaction pathway with a lower activation energy.

This increases the proportion of collisions that can successfully produce a reaction.

However, a catalyst does not replace reactant molecules that have already been consumed.

The reaction can therefore be faster with a catalyst than without one, while still becoming progressively slower as the reactant concentration decreases.

A catalyst changes how quickly a reaction proceeds. It does not guarantee a constant reaction rate.

Is the reaction rate always guaranteed to decrease?

No.

The explanation above applies to a simple batch reaction under controlled conditions, where reactants are consumed and other important factors remain constant.

Other factors can change the shape of a progress curve.

For example:

• Increasing temperature can increase the reaction rate.
• Increasing reactant concentration can increase the frequency of productive collisions.
• Increasing the surface area of a solid reactant can increase the reaction rate.
• A catalyst can increase the reaction rate without being consumed.
• Some reactions involve autocatalysis, where a product helps accelerate the reaction.

In a heterogeneous reaction, the exposed surface area of a solid may also decrease as it reacts.

The flattening curve is therefore a common pattern, not a universal law.

04Average rate and instantaneous rate are different

Now let’s connect the physical explanation to the calculations you need to perform.

There are two different ways to describe the rate of a reaction from a progress curve.

Average reaction rate

The average rate measures how much a quantity changes over a specified time interval.

For a gas-volume-time graph:

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

where:

• \(\Delta V\) is the change in gas volume.
• \(\Delta t\) is the time interval.

For a reactant concentration-time graph:

\[ r_{\text{avg}} = -\frac{\Delta[\text{reactant}]}{\Delta t} \]

The negative sign accounts for the decreasing reactant concentration.

An average rate does not tell you the exact rate at every moment within the interval.

Instantaneous reaction rate

The instantaneous rate measures how quickly the reaction is proceeding at one particular instant.

Graphically, it is determined by calculating the gradient of a tangent to the progress curve at the chosen time.

For a gas-volume-time graph:

\[ r_{\text{inst}}= \text{gradient of tangent} \]

For a reactant concentration-time graph:

\[ r_{\text{inst}} = -\text{gradient of tangent} \]

A tangent is a straight line that touches the curve at the point of interest and follows its local direction.

This distinction matters because the reaction rate is changing throughout the experiment.

If you calculate an average rate between 0 and 40 seconds, you are combining the reaction’s behaviour across that entire interval. You are not finding its exact rate at 20 seconds.

For a detailed method of calculating gas-volume gradients, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

For more practice with tangent gradients, see How to Find Instantaneous Rate Using a Tangent in HSC Chemistry.

05When a flattening curve has a different explanation

So far, we have considered a reaction that proceeds towards completion.

As reactants are consumed, their concentrations decrease, and the reaction rate approaches zero.

However, there is another important situation in which a reaction-progress curve becomes horizontal.

That situation is chemical equilibrium.

Consider a reversible reaction:

\[ \ce{A <=> B} \]

Initially, the forward reaction produces B.

As the concentration of B increases, the reverse reaction becomes increasingly significant.

Eventually, the forward and reverse reaction rates become equal.

At this point, the concentrations stop changing.

The reaction has reached dynamic equilibrium.

This creates an important distinction:

At completion, the reaction rate may approach zero because the limiting reactant has been consumed. At equilibrium, the net reaction rate is zero because the forward and reverse rates are equal.

These are not the same thing.

At equilibrium:

\[ r_{\text{forward}}=r_{\text{reverse}} \]

The forward and reverse reactions are still occurring. They simply produce and consume substances at equal rates.

This also explains why a catalyst does not shift the equilibrium position.

A catalyst increases the rates of both the forward and reverse reactions. The system reaches equilibrium faster, but the equilibrium concentrations remain unchanged at a fixed temperature.

A flat curve therefore does not automatically mean that no chemical reactions are occurring.

You need to consider the reaction system before deciding what the plateau represents.

06Questions and solutions

The following questions build from straightforward graph interpretation to more demanding combinations of reaction rates, experimental variables, and equilibrium.

Question 1

A student investigates the production of oxygen gas during a chemical reaction.

The following measurements are recorded:

Calculate the average rate of oxygen production between 20 and 60 seconds.

Explain why this value is an average rate rather than an instantaneous rate.

Solution 1

The average oxygen-production rate is \(0.90\text{ mL/s}\). It represents the rate across the entire 40-second interval, not the rate at one particular instant.

Use:

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

Substitute the values:

\[ \begin{aligned} r_{\text{avg}} &=\frac{63-27}{60-20}\\ &=\frac{36}{40}\\ &=0.90\text{ mL/s} \end{aligned} \]

The answer is:

\[ \boxed{0.90\text{ mL/s}} \]

This is an average because the calculation considers the total volume change over 40 seconds.

To determine the instantaneous rate at 40 seconds, you would need the gradient of a tangent to the volume-time curve at that point.

Question 2

A reaction occurs according to:

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

The concentration of A is recorded at regular intervals.

(a) Calculate the average rate of disappearance of A between 0 and 60 seconds.

(b) Calculate the average rate of disappearance of A between 60 and 120 seconds.

(c) Calculate the average rate of formation of B between 60 and 120 seconds.

(d) Explain why the rate of formation of B is not numerically equal to the rate of disappearance of A.

Solution 2

The average rate of disappearance of A is \(7.00\times10^{-3}\text{ mol L}^{-1}\text{s}^{-1}\) during the first interval and \(2.50\times10^{-3}\text{ mol L}^{-1}\text{s}^{-1}\) during the second interval. The average rate of formation of B during the second interval is \(1.25\times10^{-3}\text{ mol L}^{-1}\text{s}^{-1}\).

(a) Calculate the average rate between 0 and 60 seconds.

Use:

\[ r_{\text{disappearance}} = -\frac{\Delta[\ce{A}]}{\Delta t} \]

Substitute:

\[ \begin{aligned} r_{\text{disappearance}} &=-\frac{0.48-0.90}{60-0}\\ &=\frac{0.42}{60}\\ &=0.00700 \end{aligned} \]

Therefore:

\[ \boxed{ 7.00\times10^{-3} \text{ mol L}^{-1}\text{s}^{-1} } \]

(b) Calculate the average rate between 60 and 120 seconds.

\[ \begin{aligned} r_{\text{disappearance}} &=-\frac{0.33-0.48}{120-60}\\ &=\frac{0.15}{60}\\ &=0.00250 \end{aligned} \]

Therefore:

\[ \boxed{ 2.50\times10^{-3} \text{ mol L}^{-1}\text{s}^{-1} } \]

The rate of disappearance has decreased because the concentration of A is lower as the reaction proceeds.

(c) Calculate the average rate of formation of B.

The balanced equation is:

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

Two moles of A disappear for every one mole of B produced.

Therefore:

\[ r_{\text{formation of B}} = \frac{1}{2}r_{\text{disappearance of A}} \]

\[ \begin{aligned} r_{\text{formation of B}} &=\frac{1}{2}(0.00250)\\ &=0.00125 \end{aligned} \]

Therefore:

\[ \boxed{ 1.25\times10^{-3} \text{ mol L}^{-1}\text{s}^{-1} } \]

(d) Explain the difference.

The two rates are different because the stoichiometric coefficients are different.

For every two moles of A consumed, only one mole of B forms.

A common mistake is to assume that the concentration changes of all substances must have the same numerical rate. They do not.

The balanced equation determines the relationship between their rates.

Question 3

Calcium carbonate chips react with excess hydrochloric acid in an open conical flask placed on a digital balance.

The reaction is:

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

The carbon dioxide escapes from the flask, causing the measured mass to decrease.

The following illustrative measurements are recorded:

(a) Calculate the average rate of mass loss between 0 and 60 seconds.

(b) Calculate the average rate of mass loss between 60 and 120 seconds.

(c) Explain why the mass-time curve becomes less steep.

(d) A student concludes that the decreasing hydrochloric acid concentration must be the only reason for the decreasing reaction rate. Evaluate this conclusion.

Solution 3

The average rate of mass loss decreases from \(0.00750\text{ g/s}\) to \(0.00400\text{ g/s}\). The curve becomes less steep because the reaction produces carbon dioxide more slowly. However, decreasing acid concentration cannot automatically be identified as the only cause.

(a) Calculate the average rate between 0 and 60 seconds.

The mass decreases from 125.60 g to 125.15 g.

The positive rate of mass loss is:

\[ r_{\text{mass loss}} = -\frac{\Delta m}{\Delta t} \]

Substitute:

\[ \begin{aligned} r_{\text{mass loss}} &=-\frac{125.15-125.60}{60-0}\\ &=\frac{0.45}{60}\\ &=0.00750\text{ g/s} \end{aligned} \]

Therefore:

\[ \boxed{0.00750\text{ g/s}} \]

(b) Calculate the average rate between 60 and 120 seconds.

The mass decreases from 125.15 g to 124.91 g.

\[ \begin{aligned} r_{\text{mass loss}} &=-\frac{124.91-125.15}{120-60}\\ &=\frac{0.24}{60}\\ &=0.00400\text{ g/s} \end{aligned} \]

Therefore:

\[ \boxed{0.00400\text{ g/s}} \]

(c) Explain the decreasing gradient.

Carbon dioxide is produced more quickly at the beginning because the initial reactant concentrations are relatively high.

As the reaction proceeds, hydrochloric acid is consumed. Its concentration decreases, reducing the frequency of productive collisions at the calcium carbonate surface.

Consequently, carbon dioxide is released more slowly, and the measured mass decreases at a slower rate.

The mass-time curve becomes progressively less steep.

(d) Evaluate the student’s conclusion.

The conclusion is incomplete because the surface area of the calcium carbonate chips can also change.

As the chips dissolve, their exposed surface area may decrease. This reduces the area available for collisions between the acid and the solid reactant.

Therefore, both decreasing acid concentration and changing solid surface area may contribute to the declining reaction rate.

The data alone cannot establish that acid concentration is the only cause.

To investigate concentration independently, the experiment would need to control other relevant variables, particularly temperature and exposed solid surface area. A homogeneous reaction would also avoid the changing solid-surface-area problem.

Question 4

Two experiments investigate the decomposition of hydrogen peroxide using the same initial hydrogen peroxide concentration, volume, and catalyst concentration.

The reaction is:

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

In Trial A, the reaction is maintained at a constant temperature of \(25^\circ\text{C}\).

In Trial B, the reaction mixture warms from \(25^\circ\text{C}\) to \(31^\circ\text{C}\) during the first 40 seconds. It then remains at \(31^\circ\text{C}\).

The gas syringe is maintained at the same collection temperature and pressure in both trials, so thermal expansion of the collected gas does not affect the comparison.

The following illustrative measurements are obtained:

(a) Calculate the average oxygen-production rate between 20 and 40 seconds for both trials.

(b) Explain why Trial B has a greater average reaction rate during this interval, even though its reactant concentration is also decreasing.

(c) Explain why the reaction rate in Trial B subsequently decreases.

(d) Identify the most important experimental control needed to investigate the effect of reactant concentration independently of temperature.

Solution 4

Trial A has an average oxygen-production rate of \(0.80\text{ mL/s}\), while Trial B has an average rate of \(1.25\text{ mL/s}\) during the 20-40 second interval. The difference is explained by the temperature increase in Trial B, which temporarily increases the frequency of successful collisions.

(a) Calculate the average rates.

For Trial A:

\[ \begin{aligned} r_{\text{avg}} &=\frac{38-22}{40-20}\\ &=\frac{16}{20}\\ &=0.80\text{ mL/s} \end{aligned} \]

For Trial B:

\[ \begin{aligned} r_{\text{avg}} &=\frac{47-22}{40-20}\\ &=\frac{25}{20}\\ &=1.25\text{ mL/s} \end{aligned} \]

Therefore:

\[ \boxed{r_A=0.80\text{ mL/s}} \]

\[ \boxed{r_B=1.25\text{ mL/s}} \]

(b) Explain the higher rate in Trial B.

The temperature of Trial B increases during the reaction.

As temperature increases, the average kinetic energy of the particles increases. A greater proportion of particles now have energy equal to or greater than the activation energy.

This increases the number of successful collisions per second.

Although the hydrogen peroxide concentration is decreasing, the temperature increase temporarily offsets the effect of reactant consumption.

The reaction therefore becomes faster during this interval.

(c) Explain the subsequent decrease in rate.

Between 40 and 80 seconds, the temperature remains constant at \(31^\circ\text{C}\).

However, hydrogen peroxide continues to be consumed.

Its concentration decreases, reducing the frequency of productive collisions.

The average oxygen-production rates in Trial B are:

\[ r_{40-60} = \frac{65-47}{20} = 0.90\text{ mL/s} \]

\[ r_{60-80} = \frac{75-65}{20} = 0.50\text{ mL/s} \]

The decreasing rates show that reactant depletion is now the dominant effect.

(d) Identify the experimental control.

The reaction temperature must be maintained at a constant value using a thermostatically controlled water bath.

This prevents temperature changes from influencing the reaction rate.

With temperature controlled, the effect of changing reactant concentration can be investigated more reliably.

The important lesson is that a reaction-progress curve reflects the combined effects of all changing experimental conditions. A curve that becomes steeper does not automatically mean that reactant concentration has increased.

Question 5

Consider the reversible reaction:

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

A rigid, sealed vessel is maintained at a constant temperature.

The reaction has reached equilibrium, and the measured concentrations are:

\[ [\ce{A}]=0.40\text{ mol/L} \]

\[ [\ce{B}]=0.60\text{ mol/L} \]

At this equilibrium, the forward and reverse reaction rates are both:

\[ 0.012\text{ mol L}^{-1}\text{s}^{-1} \]

The equilibrium concentration ratio is:

\[ K_c=\frac{[\ce{B}]}{[\ce{A}]}=1.50 \]

For this question, assume the forward and reverse steps are elementary, so their respective rates are proportional to the concentrations of A and B.

At \(t=100\text{ s}\), two changes are made simultaneously:

• A selective trap removes exactly half of the gaseous B, without changing the vessel volume or temperature. The trapped B no longer participates in the reaction.
• A catalyst is introduced. It doubles both the forward and reverse reaction rates at any given concentrations.

Assume no other substances enter or leave the vessel, and the remaining gaseous A and B continue to react according to the original reversible equation.

(a) Determine the concentrations of A and B immediately after the changes.

Calculate the new forward rate, reverse rate, and net rate of formation of B immediately after the changes.

(b) Calculate the new equilibrium concentrations of A and B.

(c) Describe how the concentration-time curve for B changes immediately after the disturbance and as the system approaches its new equilibrium.

(d) Explain why the original horizontal section of the concentration-time curve did not mean that the reaction had stopped.

(e) Explain why adding the catalyst does not change the new equilibrium concentrations.

Solution 5

Immediately after the disturbance, the concentrations are \([\ce{A}]=0.40\text{ mol/L}\) and \([\ce{B}]=0.30\text{ mol/L}\). The forward rate becomes \(0.024\text{ mol L}^{-1}\text{s}^{-1}\), while the reverse rate becomes \(0.012\text{ mol L}^{-1}\text{s}^{-1}\).

The net reaction therefore proceeds forwards, producing B.

The new equilibrium concentrations are:

\[ \boxed{[\ce{A}]=0.28\text{ mol/L}} \]

\[ \boxed{[\ce{B}]=0.42\text{ mol/L}} \]

The catalyst changes how quickly this equilibrium is reached, not its final composition.

(a) Calculate the concentrations and rates immediately after the disturbance.

The original concentration of A is unchanged because the trap selectively removes B.

Therefore:

\[ [\ce{A}]=0.40\text{ mol/L} \]

The original concentration of B is halved:

\[ [\ce{B}]=\frac{0.60}{2} \]

\[ [\ce{B}]=0.30\text{ mol/L} \]

Now consider the reaction rates.

Before the disturbance, the forward and reverse rates were both:

\[ 0.012\text{ mol L}^{-1}\text{s}^{-1} \]

Because A is unchanged, the forward rate initially remains unchanged before the catalyst is considered.

However, B has been halved.

Since the reverse reaction rate is proportional to the concentration of B, the reverse rate is also halved.

Therefore, before adding the catalyst:

\[ r_{\text{forward}}=0.012 \]

\[ r_{\text{reverse}}=0.006 \]

Both rates are in \(\text{mol L}^{-1}\text{s}^{-1}\).

The catalyst doubles both rates.

Thus:

\[ r_{\text{forward}}=2(0.012) \]

\[ r_{\text{forward}}=0.024 \]

And:

\[ r_{\text{reverse}}=2(0.006) \]

\[ r_{\text{reverse}}=0.012 \]

The net rate of formation of B is:

\[ \begin{aligned} r_{\text{net}} &=r_{\text{forward}}-r_{\text{reverse}}\\ &=0.024-0.012\\ &=0.012 \end{aligned} \]

Therefore:

\[ \boxed{ r_{\text{net}} = 0.012\text{ mol L}^{-1}\text{s}^{-1} } \]

The forward reaction is now faster than the reverse reaction, so B begins to form overall.

(b) Calculate the new equilibrium concentrations.

The equilibrium constant does not change because the temperature remains constant and a catalyst does not change the equilibrium position.

Therefore:

\[ K_c=\frac{[\ce{B}]}{[\ce{A}]}=1.50 \]

Immediately after B is removed:

\[ [\ce{A}]+[\ce{B}] = 0.40+0.30 \]

\[ [\ce{A}]+[\ce{B}]=0.70 \]

Because the reaction converts one mole of A into one mole of B, the total concentration of gaseous A and B remains \(0.70\text{ mol/L}\) as the reaction proceeds.

Let the new equilibrium concentration of A be \(x\).

Then:

\[ [\ce{A}]=x \]

\[ [\ce{B}]=1.50x \]

Using the total concentration:

\[ x+1.50x=0.70 \]

\[ 2.50x=0.70 \]

\[ x=0.28 \]

Therefore:

\[ [\ce{A}]=0.28\text{ mol/L} \]

And:

\[ [\ce{B}]=1.50(0.28) \]

\[ [\ce{B}]=0.42\text{ mol/L} \]

The new equilibrium concentrations are:

\[ \boxed{[\ce{A}]=0.28\text{ mol/L}} \]

\[ \boxed{[\ce{B}]=0.42\text{ mol/L}} \]

Notice that B does not return to its original concentration of 0.60 mol/L.

The trap permanently removed some B from the reacting gas mixture. The new equilibrium must therefore satisfy the original equilibrium ratio using the smaller remaining total concentration.

(c) Describe the concentration-time curve for B.

The curve has three distinct stages.

First, before the disturbance, the concentration of B is constant at 0.60 mol/L because the system is at equilibrium.

Second, at 100 seconds, the selective trap removes half of B. The measured concentration therefore drops immediately from 0.60 mol/L to 0.30 mol/L.

Third, the forward reaction becomes faster than the reverse reaction. B is produced faster than it is consumed, so its concentration rises towards 0.42 mol/L.

The curve eventually becomes horizontal again when the new equilibrium is reached.

An important detail is that the sudden vertical drop is caused by the external removal of B, not by the chemical reaction itself. It should not be interpreted as an instantaneous reaction rate.

At the new equilibrium, the forward and reverse rates are equal again.

In fact, using the given elementary-rate assumptions, both rates are \(0.0168\text{ mol L}^{-1}\text{s}^{-1}\). They are non-zero, even though the concentration-time curve is horizontal.

(d) Explain why the original plateau did not mean the reaction had stopped.

The original plateau represented dynamic equilibrium.

At that point:

\[ r_{\text{forward}}=r_{\text{reverse}} \]

Both reactions continued at:

\[ 0.012\text{ mol L}^{-1}\text{s}^{-1} \]

B was being produced and consumed at equal rates, so its concentration remained constant.

The absence of a concentration change does not mean the absence of molecular reactions.

This is the critical distinction between a reaction reaching completion and a reversible reaction reaching equilibrium.

(e) Explain why the catalyst does not change the new equilibrium concentrations.

A catalyst increases the rates of both the forward and reverse reactions.

It does not change the equilibrium constant at a fixed temperature.

Therefore, it does not change the final ratio:

\[ \frac{[\ce{B}]}{[\ce{A}]}=1.50 \]

The catalyst simply allows the system to reach that ratio more quickly.

The tempting prediction is that increasing the forward reaction rate must produce more B at equilibrium. However, the catalyst also increases the reverse reaction rate.

The system still reaches the same equilibrium composition.

This is why interpreting a reaction-progress curve requires more than looking at its steepness. You must consider the reaction conditions, the substances being measured, and whether the system is approaching completion or dynamic equilibrium.

The next useful step is learning to distinguish a reaction that is slowing because its reactants are being depleted from one that is approaching equilibrium. Both can produce a horizontal curve, but the chemistry happening underneath is fundamentally different.