Why Catalysts Speed Up Equilibrium Without Changing Its Position

Imagine two identical sealed reactors containing nitrogen and hydrogen. Both are kept at the same temperature and volume. One contains an iron catalyst; the other does not.

After 10 minutes, the catalysed reactor contains considerably more ammonia. Has the iron catalyst shifted the equilibrium position?

No. It has made the reaction approach equilibrium faster by increasing the rates of both the forward and reverse reactions. If both reactors remain closed under the same conditions and are given enough time, they reach the same equilibrium composition.

The important distinction is between how quickly equilibrium is reached and where equilibrium ends up.

01Why a catalyst speeds up both directions

Consider the reversible reaction used in ammonia synthesis:

\[ \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \]

The forward reaction produces ammonia. The reverse reaction decomposes ammonia into nitrogen and hydrogen.

A catalyst provides an alternative reaction pathway with a lower activation energy. This allows both forward and reverse reactions to occur more quickly.

At first, this might seem to favour the forward reaction. After all, if you start with nitrogen and hydrogen, the ammonia concentration increases.

But remember that the reverse reaction also becomes faster as ammonia forms. The catalyst does not give one direction a permanent advantage.

Think about what happens at equilibrium

Equilibrium is dynamic. The forward and reverse reactions continue occurring, but their rates are equal.

\[ v_f = v_r \]

Here:

• \(v_f\) is the forward reaction rate.
• \(v_r\) is the reverse reaction rate.

Both rates are measured in units such as \(\mathrm{mol\,L^{-1}\,s^{-1}}\).

Imagine a two-way footbridge. If 20 people cross from left to right every minute, and another 20 cross from right to left, the number of people on each side stays constant.

Now imagine widening both sides of the bridge. Perhaps 100 people can cross in each direction every minute. The number of people on each side still stays constant because the two flows remain equal.

A catalyst has a similar effect on a reversible reaction. It increases the rate of movement in both directions without creating a net change at equilibrium.

The analogy has one limitation: molecules are chemically transformed during reactions, rather than simply travelling between two locations. However, the idea of equal opposing rates is exactly what we need.

What happens to the net reaction rate?

To understand why a catalyst speeds up the approach to equilibrium, we need to consider the difference between the forward and reverse rates.

For ammonia synthesis:

\[ \frac{d[\ce{NH3}]}{dt}=2(v_f-v_r) \]

The factor of 2 comes from the balanced equation, which produces two moles of ammonia for every mole of reaction occurring in the forward direction.

A positive value means ammonia is being produced overall. A negative value means ammonia is being consumed overall.

For a simple HSC rate comparison, we can represent the catalyst’s effect at a fixed composition using a common multiplier, \(\alpha\), where \(\alpha>1\).

The catalysed rates become:

\[ v_f'=\alpha v_f \]

\[ v_r'=\alpha v_r \]

Therefore:

\[ \frac{d[\ce{NH3}]}{dt} =2\alpha(v_f-v_r) \]

Notice what this tells us.

If the system is not at equilibrium: The net reaction continues in the same direction as before, but its rate increases.

If the system is already at equilibrium: The forward and reverse rates are equal. Increasing both leaves their difference at zero.

\[ v_f=v_r \]

\[ 2\alpha(v_f-v_r)=0 \]

The catalyst increases the number of successful forward and reverse reactions, but there is still no net change in concentration.

The actual increase in reaction rate depends on the catalyst and reaction conditions. The common multiplier is a useful simplified model, not a universal numerical value.

02How an experiment could demonstrate this

There is a useful experimental question here:

How can we distinguish between a catalyst changing the equilibrium composition and a catalyst simply changing how quickly that composition is reached?

A controlled comparison between catalysed and uncatalysed reactions can answer this.

Setting up the investigation

Consider a proposed investigation using the reversible ammonia synthesis reaction.

We would use identical, sealed, rigid reactors maintained at the same temperature.

Each reactor contains:

• The same volume of gas.
• The same initial amounts of reactants and products within each matched comparison.
• Either an iron catalyst or an inert material of similar volume.
• A temperature sensor and a non-invasive method of measuring ammonia concentration.

The iron catalyst is held in a basket inside the reactor, where the gas can contact its surface.

An optical probe measures the ammonia concentration without opening the vessel or removing gas samples.

The system boundary encloses the gas mixture and catalyst. No gas enters or leaves during a trial.

This matters because withdrawing gas would change the composition and potentially disturb the equilibrium being investigated.

What should we predict?

Start with the forward reaction.

The two matched reactors contain identical amounts of nitrogen and hydrogen. Neither initially contains ammonia.

One reactor contains iron; the other does not.

We predict that:

• Ammonia will form more quickly in the reactor containing iron.
• Both reactors will eventually reach the same equilibrium ammonia concentration.

If the catalyst changes the equilibrium position, the final concentrations should differ.

If it only changes the reaction rates, the concentrations should approach the same final values at different speeds.

What would the results look like?

The following is an illustrative dataset, not the result of an actual experiment.

Both reactors begin with:

• \([\ce{N2}]=0.40\ \mathrm{mol\,L^{-1}}\)
• \([\ce{H2}]=1.20\ \mathrm{mol\,L^{-1}}\)
• \([\ce{NH3}]=0.00\ \mathrm{mol\,L^{-1}}\)

Both reactors approach an equilibrium ammonia concentration of approximately \(0.40\ \mathrm{mol\,L^{-1}}\).

However, the catalysed reactor gets there much sooner.

The important experimental flaw: testing only the forward reaction

There is one problem with the investigation so far.

We have demonstrated that the catalyst speeds up ammonia formation. But have we directly demonstrated that it speeds up the reverse reaction?

Not yet.

Initially, our reactors contain no ammonia. The reverse reaction cannot occur at the start because there are no ammonia molecules available to decompose.

As ammonia accumulates, the reverse reaction becomes possible and its rate increases.

To properly investigate the reverse reaction, we need another pair of reactors.

This time, start with ammonia rather than nitrogen and hydrogen.

For a fair comparison, use the same temperature, volume, and total quantities of nitrogen and hydrogen atoms as in the forward-start trials.

The catalysed reactor should now show ammonia disappearing more quickly. However, both reactors should still approach the same equilibrium composition.

This revised investigation tests the catalyst’s effect from both directions.

It demonstrates that a catalyst does not simply accelerate product formation. It accelerates the reversible reaction system.

03The important contrast: an industrial reactor is not always at equilibrium

There is a situation where a catalyst can produce a higher measured product concentration without changing the equilibrium position.

Consider a continuous-flow ammonia reactor.

Nitrogen and hydrogen enter one end. They pass through a bed containing an iron catalyst, and an ammonia-containing gas mixture exits the other end.

The gas does not remain in the reactor indefinitely. It has a limited residence time, meaning a limited amount of time to react before leaving.

Imagine comparing two reactors:

• One contains an iron catalyst.
• The other contains inert packing.

Both receive the same feed at the same temperature, pressure, and flow rate.

The iron-containing reactor may produce a higher ammonia concentration at its outlet.

Does this mean the catalyst has changed the equilibrium position?

No. The catalysed gas mixture has simply progressed further towards equilibrium during its available residence time.

The uncatalysed mixture may leave the reactor before it has had enough time to reach equilibrium.

There is another important distinction here.

A constant outlet concentration does not necessarily mean the gas is at equilibrium. It may simply mean the reactor has reached a steady state, where the inlet and outlet flows maintain a stable composition.

A continuous-flow reactor is an open system. Material continuously crosses its boundary.

In a closed system, by contrast, no reactants or products enter or leave. Given enough time, the reaction approaches its equilibrium composition.

This is why you must consider the system boundary before deciding whether a measured concentration represents equilibrium.

For a more detailed explanation of the final equilibrium composition, read Why a Catalyst Does Not Change Equilibrium Yield.

To understand the alternative reaction pathway and its activation energy, see Catalysts and Energy Profiles: HSC Chemistry Explained.

04Questions and solutions

Question 1

A reversible reaction takes place in two identical sealed vessels at constant temperature and volume.

The initial ammonia concentration in both vessels is \(0.10\ \mathrm{mol\,L^{-1}}\).

After 20 minutes:

• The vessel without a catalyst has an ammonia concentration of \(0.22\ \mathrm{mol\,L^{-1}}\).
• The vessel with a catalyst has an ammonia concentration of \(0.34\ \mathrm{mol\,L^{-1}}\).

Both vessels eventually reach an ammonia concentration of \(0.40\ \mathrm{mol\,L^{-1}}\).

Calculate the average rate of ammonia formation in each vessel over the first 20 minutes. Explain what the results demonstrate about the catalyst.

Solution 1

The catalyst produces a higher average rate of ammonia formation, but both vessels reach the same equilibrium concentration.

Step 1: Calculate the average rate without the catalyst.

\[ \text{Rate}=\frac{\Delta[\ce{NH3}]}{\Delta t} \]

\[ =\frac{0.22-0.10}{20} \]

\[ =0.0060\ \mathrm{mol\,L^{-1}\,min^{-1}} \]

Step 2: Calculate the average rate with the catalyst.

\[ =\frac{0.34-0.10}{20} \]

\[ =0.012\ \mathrm{mol\,L^{-1}\,min^{-1}} \]

Step 3: Compare the rates.

\[ \frac{0.012}{0.0060}=2 \]

The catalysed reaction has twice the average ammonia formation rate during the first 20 minutes.

Both vessels eventually reach \(0.40\ \mathrm{mol\,L^{-1}}\), demonstrating that the catalyst changes the rate of approach to equilibrium, not the equilibrium position.

The tempting conclusion is that more ammonia in the catalysed vessel at 20 minutes means a higher equilibrium yield. However, the two vessels have not reached equilibrium at that point. Their concentrations are still changing.

Question 2

A reversible reaction has reached dynamic equilibrium in a sealed vessel.

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

\[ 0.015\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

A catalyst is added, increasing both rates by a factor of 5.

Calculate the net rate of change in ammonia concentration immediately before and after adding the catalyst, assuming the reaction is:

\[ \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \]

Explain why the ammonia concentration does not change immediately after adding the catalyst.

Solution 2

The net rate of ammonia concentration change is zero both before and after adding the catalyst.

Step 1: Calculate the net rate before adding the catalyst.

\[ \frac{d[\ce{NH3}]}{dt}=2(v_f-v_r) \]

\[ =2(0.015-0.015) \]

\[ =0\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Step 2: Calculate the new forward and reverse rates.

The catalyst increases both by a factor of 5:

\[ v_f'=5(0.015)=0.075 \]

\[ v_r'=5(0.015)=0.075 \]

Step 3: Calculate the new net rate.

\[ \frac{d[\ce{NH3}]}{dt} =2(0.075-0.075) \]

\[ =0\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The forward and reverse reactions are now occurring more rapidly, but their rates remain equal.

Therefore, there is no net production or consumption of ammonia.

The common misconception is that a faster forward reaction must increase the product concentration. That would only be true if the forward rate exceeded the reverse rate. At equilibrium, both increase together, so the concentration remains constant.

Question 3

A student investigates the reverse reaction of ammonia synthesis using two identical sealed vessels at constant temperature and volume.

Both vessels initially contain \(0.70\ \mathrm{mol\,L^{-1}}\) ammonia, with no nitrogen or hydrogen present.

One vessel contains an iron catalyst, while the other contains inert packing.

After 6 minutes, the ammonia concentrations are:

Both vessels eventually reach an equilibrium ammonia concentration of \(0.35\ \mathrm{mol\,L^{-1}}\).

(a) Calculate the average rate of ammonia disappearance in each vessel over the first 6 minutes.

(b) Determine how many times greater the average rate is in the catalysed vessel.

(c) Explain why this investigation provides evidence that a catalyst speeds up the reverse reaction.

Solution 3

The catalysed vessel has three times the average ammonia disappearance rate, providing evidence that the catalyst accelerates the reverse reaction.

Step 1: Calculate the average rate of ammonia disappearance with the catalyst.

For disappearance, use the magnitude of the concentration decrease:

\[ \text{Rate} =\frac{0.70-0.52}{6} \]

\[ =0.030\ \mathrm{mol\,L^{-1}\,min^{-1}} \]

Step 2: Calculate the average rate without the catalyst.

\[ \text{Rate} =\frac{0.70-0.64}{6} \]

\[ =0.010\ \mathrm{mol\,L^{-1}\,min^{-1}} \]

Step 3: Compare the rates.

\[ \frac{0.030}{0.010}=3 \]

The catalysed vessel has three times the average rate of ammonia disappearance.

Step 4: Explain the chemical reasoning.

Initially, the vessels contain ammonia but no nitrogen or hydrogen.

Therefore, the reverse reaction can occur immediately, while the forward reaction initially has no reactants available.

The decrease in ammonia concentration is therefore evidence of the reverse reaction occurring.

The catalyst allows ammonia to decompose more rapidly, producing nitrogen and hydrogen.

Both vessels eventually reach the same equilibrium ammonia concentration because the catalyst changes the reaction rates, not the equilibrium position.

This experiment addresses a limitation of a forward-start investigation: observing faster ammonia formation alone does not directly demonstrate faster ammonia decomposition.

Question 4

Two continuous-flow reactors are used for ammonia synthesis.

Both receive identical nitrogen and hydrogen feeds at the same temperature, pressure, and flow rate.

One reactor contains an iron catalyst, while the other contains inert packing.

After operating for sufficient time, their outlet ammonia concentrations become stable:

• Iron catalyst reactor: 22% ammonia.
• Inert reactor: 9% ammonia.

A student concludes:

“The iron catalyst has shifted the equilibrium position towards ammonia because the outlet contains a higher percentage of ammonia.”

Evaluate this conclusion. Explain the difference between a steady-state outlet composition and an equilibrium composition. Predict what would happen if equal feed-equivalent samples from both outlets were placed in separate sealed vessels with the same temperature, volume, and total quantities of nitrogen and hydrogen atoms, and allowed to reach equilibrium.

Solution 4

The student’s conclusion is incorrect. The higher ammonia concentration demonstrates that the catalyst has increased the extent of reaction during the available residence time, not that it has changed the equilibrium position.

Step 1: Consider the system boundary.

Both reactors are continuous-flow systems.

Nitrogen and hydrogen continuously enter, while the ammonia-containing mixture continuously leaves.

Therefore, neither reactor is a closed system.

Step 2: Explain the higher ammonia concentration.

The iron catalyst increases the rates of the forward and reverse reactions.

During the limited residence time, the catalysed mixture progresses further towards equilibrium than the uncatalysed mixture.

Consequently, more ammonia can be present at the outlet.

Step 3: Distinguish steady state from equilibrium.

A steady state occurs when the measured composition at a particular location remains constant over time.

This can happen because the inlet and outlet flows maintain stable conditions.

Dynamic equilibrium, however, requires equal forward and reverse reaction rates, with no net chemical change in the closed reaction system.

A stable outlet concentration does not prove that the gas has reached equilibrium.

Step 4: Predict what happens in the sealed vessels.

Assuming both samples contain the same total quantities of nitrogen and hydrogen atoms, are held at the same temperature and volume, and undergo no material loss, they will eventually reach the same equilibrium composition.

Their starting compositions may differ, but the equilibrium composition is determined by the equilibrium conditions and the total amounts of each element.

The catalyst changes how quickly that composition is reached, not the final equilibrium position.

The important lesson is that a higher measured product concentration does not automatically indicate a shift in equilibrium. You must first determine whether the system has actually reached equilibrium.

Question 5

A sealed, rigid vessel contains nitrogen, hydrogen, and ammonia at dynamic equilibrium:

\[ \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \]

The vessel is maintained at constant temperature and contains an iron catalyst.

Initially, before adding the catalyst, the forward and reverse reaction rates are both:

\[ 0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

At \(t=0\), the catalyst is added. It doubles both reaction rates.

At \(t=10\ \mathrm{s}\), an ammonia-selective membrane pump begins removing ammonia from the vessel while the catalyst remains present.

At the first instant after a small amount of ammonia has been removed, the corresponding uncatalysed reaction rates at this new composition would be:

\[ v_f=0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

\[ v_r=0.015\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The catalyst continues to double both rates.

The pump removes ammonia at an equivalent concentration rate of:

\[ 0.025\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Assume:

• The pump selectively removes ammonia, not nitrogen or hydrogen.
• The vessel temperature and volume remain constant.
• No other reactions occur.
• The pump is eventually switched off, leaving the vessel sealed.
• Pumping is stopped while the remaining mixture has \(Q_c<K_c\).

(a) Calculate the net rate of ammonia concentration change immediately after adding the catalyst but before the pump begins operating.

(b) Calculate the net chemical rate of ammonia production immediately after pumping begins.

(c) Calculate the actual rate of change of ammonia concentration in the vessel while the pump is operating.

(d) Explain whether the vessel is at dynamic equilibrium while the pump is running.

(e) Predict the direction of the reaction after the pump is switched off. Explain whether the catalyst changes the new equilibrium position.

Solution 5

The catalyst initially increases both reaction rates without changing the equilibrium. Once ammonia is removed, the forward reaction becomes favoured. Interestingly, the measured ammonia concentration can still decrease because the pump removes ammonia faster than the chemical reaction produces it.

Step 1: Calculate the net rate immediately after adding the catalyst.

Before adding the catalyst, both rates are:

\[ v_f=v_r=0.020 \]

The catalyst doubles both:

\[ v_f'=0.040 \]

\[ v_r'=0.040 \]

Using the net ammonia production equation:

\[ \frac{d[\ce{NH3}]}{dt} =2(v_f'-v_r') \]

\[ =2(0.040-0.040) \]

\[ =0\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The catalyst increases both rates, but the system remains at equilibrium.

There is no net change in ammonia concentration.

Step 2: Calculate the net chemical rate after ammonia removal begins.

At the new composition, the uncatalysed rates would be:

\[ v_f=0.020 \]

\[ v_r=0.015 \]

The catalyst doubles both:

\[ v_f'=0.040 \]

\[ v_r'=0.030 \]

Therefore:

\[ \frac{d[\ce{NH3}]}{dt}_{\text{chemical}} =2(0.040-0.030) \]

\[ =0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The net chemical reaction is now producing ammonia.

This is because the forward rate is greater than the reverse rate.

Step 3: Account for ammonia being removed by the pump.

The pump removes ammonia at:

\[ 0.025\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The actual concentration change must account for both chemical production and external removal:

\[ \frac{d[\ce{NH3}]}{dt}_{\text{observed}} = \text{chemical production rate} - \text{removal rate} \]

Substituting:

\[ =0.020-0.025 \]

\[ =-0.005\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The ammonia concentration decreases at \(0.005\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

This is the surprising part.

The chemical reaction is proceeding in the forward direction, producing ammonia overall. Yet the measured ammonia concentration is falling because the pump removes ammonia faster than the reaction produces it.

A falling product concentration does not necessarily mean the reverse reaction is favoured. Material can also cross the system boundary.

Step 4: Determine whether the system is at equilibrium during pumping.

The vessel is not at dynamic equilibrium while the pump operates.

There are two reasons:

• The forward and reverse chemical reaction rates are unequal.
• Ammonia is continuously crossing the system boundary.

The system is open while pumping occurs.

Even if the measured ammonia concentration happened to remain constant because chemical production balanced external removal, that would represent a non-equilibrium steady state, not necessarily dynamic equilibrium.

Step 5: Predict what happens after pumping stops.

The pump has removed ammonia, reducing the reaction quotient:

\[ Q_c= \frac{[\ce{NH3}]^2} {[\ce{N2}][\ce{H2}]^3} \]

The question states that the remaining mixture has \(Q_c<K_c\).

Therefore, the forward reaction is favoured.

The reaction proceeds towards producing more ammonia until a new equilibrium is established.

The iron catalyst increases both forward and reverse rates, allowing this new equilibrium to be reached sooner.

However, the catalyst does not change the equilibrium constant or the new equilibrium position.

The final equilibrium concentrations may differ from the original ones because ammonia has physically been removed from the vessel.

This question brings together three separate ideas:

• A catalyst changes reaction rates, not the equilibrium position.
• Removing a product changes the composition and can favour the forward reaction.
• A concentration change in an open system must account for material entering or leaving the system boundary.

The distinction between reaction rate, measured concentration change, and equilibrium position is essential when analysing more complicated equilibrium problems.