Why a Catalyst Lowers Both Forward and Reverse Activation Energies

Two sealed vessels start with the same reactant, A. Ten minutes later, the vessel containing a catalyst has produced much more B. It looks as though the catalyst favours the forward reaction. But what would happen if both vessels started with B instead? Would the catalyst still help?

Yes. A catalyst lowers both the forward and reverse activation energies because it provides an alternative reaction pathway that can be followed in either direction. Reactants can become products through this pathway, and products can follow it backwards to become reactants.

The catalyst changes the energy barrier, not the energy difference between the starting and finishing substances. The reaction enthalpy, \(\Delta H\), stays the same. At a fixed temperature, the equilibrium position also stays the same.

That’s the apparent contradiction worth investigating: a catalyst can make more product appear in a given time without increasing the amount of product present at equilibrium.

01How could an experiment show that both directions are catalysed?

Consider a hypothetical reversible gas reaction:

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

Each molecule of A can rearrange into one molecule of B, and the reverse conversion is also possible.

We want to distinguish two competing explanations:

• Model 1: The catalyst makes A turn into B faster but does not help the reverse reaction.
• Model 2: The catalyst makes both conversions faster, leaving the equilibrium composition unchanged.

The second model is correct for an ordinary catalyst acting on this reversible system. But how could measurements establish that?

The apparatus

Imagine two identical sealed reaction vessels held at \(25^\circ\mathrm{C}\) in a temperature-controlled enclosure.

Each vessel contains gas and a small bed of solid pellets. The first contains uncoated, chemically inert pellets. The second contains pellets coated with a catalyst.

A sampling valve connects each vessel to a gas chromatograph. This instrument separates A and B in a small withdrawn sample, allowing their concentrations to be estimated from calibrated signals.

The gas is mixed inside each vessel, and samples are taken at regular intervals.

The investigation controls:

• Temperature, pressure, vessel volume, and total starting concentration.
• The amounts and physical arrangement of support material.
• Sampling conditions and the method used to measure A and B.

We also assume the catalyst does not appreciably absorb or store either gas, and that no side reactions occur. These assumptions become important later.

First attempt: start with reactants

Both vessels initially contain:

\[
[A]_0=0.100\ \mathrm{mol\,L^{-1}}
\]

\[
[B]_0=0
\]

Here, square brackets indicate concentration, and the subscript \(0\) means the initial value.

What should happen if the catalyst lowers the forward activation energy?

B should appear more quickly in the catalysed vessel. More A molecules can follow the accessible reaction pathway per unit time.

Suppose the experiment produces the following results.

Illustrative data: All concentrations and rates in this article are invented for teaching. They are internally consistent with a simple reversible first-order model, not measurements from a published experiment.

Initial rates are in \(\mathrm{mol\,L^{-1}\,min^{-1}}\). Concentrations are in \(\mathrm{mol\,L^{-1}}\).

The catalyst produces a fourfold increase in the initial rate of B formation.

After ten minutes, the catalysed mixture also contains considerably more B.

But the final equilibrium concentrations are identical.

The first result demonstrates faster product formation. It does not, by itself, tell us whether the reverse reaction has also been accelerated.

Why not?

Initially, there is no B available to react backwards. We have measured the system mainly under conditions where the forward reaction dominates.

To test the reverse reaction, the experiment needs a second starting condition.

Revised experiment: start with products

Now prepare both vessels with only B:

\[
[A]_0=0
\]

\[
[B]_0=0.100\ \mathrm{mol\,L^{-1}}
\]

Keep everything else the same.

If the catalyst only helps A become B, it should offer no comparable improvement when B is converting back into A.

But if the catalysed pathway works in both directions, B should disappear faster from the catalysed vessel.

The illustrative results are:

The initial disappearance rates are in \(\mathrm{mol\,L^{-1}\,min^{-1}}\), and concentrations are in \(\mathrm{mol\,L^{-1}}\).

The catalyst increases the initial reverse rate by a factor of four as well.

Notice that the final concentration is again \(0.060\ \mathrm{mol\,L^{-1}}\), regardless of which substance was present initially.

The revised experiment removes the first design’s limitation.

Starting with A tests forward conversion. Starting with B tests reverse conversion. Both sets of measurements show faster conversion with the catalyst, while the final composition remains unchanged.

This is evidence for faster kinetics in both directions, not for a new equilibrium position.

There is a further limitation. Concentration measurements at one temperature do not directly reveal an activation energy or the detailed molecular mechanism. Estimating an experimental activation energy requires suitable kinetic measurements across temperatures and a model connecting rate to temperature.

02Why does a lower-energy pathway work backwards?

The experiment tells us what happens. The energy model explains why.

A chemical reaction involves atoms changing their arrangement. Existing bonds may weaken or break, while new bonds form.

The molecules generally must pass through an energetically demanding arrangement before the conversion can continue.

A catalyst provides a different way for those rearrangements to happen.

For example, a molecule might temporarily attach to a catalytic surface. Its bonds can then rearrange before the product leaves the surface.

The catalyst is regenerated during the overall process.

The important point is that the sequence can also run backwards. A product molecule may interact with the same catalytic surface and follow the reverse sequence towards the original reactant.

A catalyst isn’t a one-way gate. It cannot provide an easier route from A to B while forbidding B from travelling through that same route in reverse.

This is the physical reason the reverse reaction benefits too.

Three energy levels, two activation barriers

For a simple reaction with one transition state, imagine three enthalpy levels:

• \(H_R\): the enthalpy of the reactants.
• \(H_P\): the enthalpy of the products.
• \(H^\ddagger\): the enthalpy of the transition state, the high-energy arrangement at the top of the pathway.

All enthalpy values are expressed per mole of reaction as written, using the same reference level.

The forward activation energy is the difference between the transition state and the reactants.

The reverse activation energy is the difference between that same transition state and the products.

The reaction enthalpy is the difference between the products and reactants.

Mathematically:

\[
\begin{aligned}
E_{a,f}&=H^\ddagger-H_R\\
E_{a,r}&=H^\ddagger-H_P\\
\Delta H&=H_P-H_R
\end{aligned}
\]

Here, \(E_{a,f}\) and \(E_{a,r}\) are the forward and reverse activation energies.

All quantities use \(\mathrm{kJ\,mol^{-1}}\).

The sign convention for the forward reaction is:

• \(\Delta H<0\): exothermic; products have lower enthalpy.
• \(\Delta H>0\): endothermic; products have higher enthalpy.

Subtracting the reverse activation energy from the forward activation energy gives:

\[
E_{a,f}-E_{a,r}=\Delta H
\]

This relationship is particularly useful for HSC energy-profile questions.

It means that if the reactant and product energy levels remain fixed, lowering a single transition-state peak lowers both activation energies by the same amount.

03The misconception: lowering only the forward barrier

Suppose an exothermic reaction has these uncatalysed activation energies:

\[
E_{a,f}=80\ \mathrm{kJ\,mol^{-1}}
\]

\[
E_{a,r}=105\ \mathrm{kJ\,mol^{-1}}
\]

Its enthalpy change is:

\[
\begin{aligned}
\Delta H&=80-105\\
&=-25\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

A student proposes that a catalyst lowers the forward activation energy to \(50\ \mathrm{kJ\,mol^{-1}}\), but leaves the reverse activation energy at \(105\ \mathrm{kJ\,mol^{-1}}\).

That sounds plausible if the catalyst is being used to produce more products.

But compare the implied enthalpy changes.

All values are in \(\mathrm{kJ\,mol^{-1}}\).

The claimed pathway is impossible for the same reactants and products under the simple single-transition-state model.

It implies that adding a catalyst changes the reaction enthalpy from \(-25\) to \(-55\ \mathrm{kJ\,mol^{-1}}\).

But the catalyst hasn’t changed the substances at either end.

The correct reverse activation energy must therefore be:

\[
\begin{aligned}
E_{a,r}&=E_{a,f}-\Delta H\\
&=50-(-25)\\
&=75\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The catalyst lowers both barriers by \(30\ \mathrm{kJ\,mol^{-1}}\).

This is a useful consistency test: in a simple energy profile, a proposed catalysed route that changes \(E_{a,f}-E_{a,r}\) without changing the overall reaction is incorrect.

Where the one-peak model becomes incomplete

Real catalytic mechanisms can involve several elementary reactions and temporary intermediates. Their energy profiles may have multiple peaks.

Each elementary step has its own activation barrier. The experimentally observed activation energy of a complex mechanism can also depend on which steps control the overall rate.

So the statement that the forward and reverse barriers fall by exactly the same numerical amount applies to the idealised single-transition-state comparison. Don’t assume every experimentally measured apparent activation energy in a multistep mechanism behaves that simply.

What remains true for a catalyst acting on an unchanged reversible reaction at fixed temperature is that it facilitates the reaction pathway without changing the reaction’s intrinsic thermodynamic equilibrium.

For more practice interpreting energy profiles and multistep pathways, see Catalysts and Energy Profiles: HSC Chemistry Explained.

04Why faster forward and reverse reactions don’t shift equilibrium

Return to the experimental vessels.

At equilibrium, the forward and reverse reactions are still occurring.

Their rates are equal:

\[
r_f=r_r
\]

Here, \(r_f\) is the forward reaction rate and \(r_r\) is the reverse reaction rate.

As a result, the net reaction rate is zero:

\[
r_{\mathrm{net}}=r_f-r_r=0
\]

This doesn’t mean molecules have stopped converting.

It means B is being produced and consumed at equal rates.

The catalyst can increase both opposing rates while maintaining their equality at equilibrium.

For the illustrative experiment, the catalyst increased the forward and reverse rate constants by the same factor of four.

At a fixed temperature, the equilibrium ratio remained unchanged.

This gives us a precise way to separate three ideas:

There is one distinction worth keeping clear.

An unchanged \(\Delta H\) is not, by itself, a complete explanation for an unchanged equilibrium constant. Equilibrium depends on the thermodynamic stability of reactants and products, including entropy effects.

A catalyst leaves that thermodynamic relationship unchanged. It changes the route and rate of conversion, rather than the relative equilibrium stability of the chemical states.

Temperature changes and the physical removal of substances are different. They can alter the composition of an equilibrium system.

That distinction is central to the final problems.

05Questions and solutions

Question 1

An exothermic reaction has an uncatalysed forward activation energy of \(70\ \mathrm{kJ\,mol^{-1}}\) and an enthalpy change of \(-20\ \mathrm{kJ\,mol^{-1}}\).

A catalyst lowers the forward activation energy to \(45\ \mathrm{kJ\,mol^{-1}}\).

Assume both pathways have one transition-state peak.

Calculate:

a. The uncatalysed reverse activation energy.

b. The catalysed reverse activation energy.

c. The decrease in activation energy for each direction.

Solution 1

The uncatalysed reverse activation energy is \(90\ \mathrm{kJ\,mol^{-1}}\), the catalysed reverse activation energy is \(65\ \mathrm{kJ\,mol^{-1}}\), and both directions have their activation energies reduced by \(25\ \mathrm{kJ\,mol^{-1}}\).

Part a: Uncatalysed reverse activation energy

Use:

\[
E_{a,r}=E_{a,f}-\Delta H
\]

Substitute:

\[
\begin{aligned}
E_{a,r}&=70-(-20)\\
&=90\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The reverse barrier is larger because this exothermic reaction produces substances at a lower enthalpy.

Part b: Catalysed reverse activation energy

\[
\begin{aligned}
E_{a,r}&=45-(-20)\\
&=65\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Part c: Compare the reductions

Forward reduction:

\[
70-45=25\ \mathrm{kJ\,mol^{-1}}
\]

Reverse reduction:

\[
90-65=25\ \mathrm{kJ\,mol^{-1}}
\]

Both barriers decrease equally in this simple model.

The common mistake is treating the negative \(\Delta H\) as though it should be subtracted as a positive number. Doing that would incorrectly make the exothermic reaction’s reverse barrier smaller than its forward barrier.

Question 2

A hypothetical reversible reaction follows:

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

Assume the elementary forward and reverse rates obey:

\[
r_f=k_f[A]
\]

\[
r_r=k_r[B]
\]

Here, \(k_f\) and \(k_r\) are the forward and reverse rate constants in \(\mathrm{min^{-1}}\). Rates are in \(\mathrm{mol\,L^{-1}\,min^{-1}}\).

At a fixed temperature, the uncatalysed rate constants are:

\[
k_f=0.030\ \mathrm{min^{-1}}
\]

\[
k_r=0.020\ \mathrm{min^{-1}}
\]

A catalyst increases the forward rate constant to:

\[
k_{f,\mathrm{cat}}=0.120\ \mathrm{min^{-1}}
\]

The catalyst does not change the equilibrium constant.

A new reaction mixture initially contains:

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

\[
[B]=0.120\ \mathrm{mol\,L^{-1}}
\]

a. Calculate the catalysed reverse rate constant.

b. Calculate the initial forward and reverse rates in the catalysed mixture.

c. Determine the net direction of reaction immediately after adding the catalyst.

d. Compare the magnitude of the catalysed net rate with the uncatalysed net rate for this starting mixture.

A student argues that because the catalysed forward rate constant is larger than the reverse rate constant, B must initially increase. Explain why that argument fails.

Solution 2

The catalysed reverse rate constant is \(0.080\ \mathrm{min^{-1}}\), and the mixture initially converts B into A. Its net reaction rate is four times as large with the catalyst, even though the catalyst makes the forward conversion faster too.

Part a: Determine the reverse rate constant

At equilibrium, the two rates are equal:

\[
k_f[A]_{\mathrm{eq}}=k_r[B]_{\mathrm{eq}}
\]

Rearranging gives:

\[
K_c=\frac{k_f}{k_r}
\]

For the uncatalysed system:

\[
\begin{aligned}
K_c&=\frac{0.030}{0.020}\\
&=1.5
\end{aligned}
\]

Because the catalyst does not change \(K_c\):

\[
\begin{aligned}
k_{r,\mathrm{cat}}&=\frac{0.120}{1.5}\\
&=0.080\ \mathrm{min^{-1}}
\end{aligned}
\]

Part b: Calculate the catalysed rates

Forward:

\[
\begin{aligned}
r_f&=0.120\times0.030\\
&=0.0036\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

Reverse:

\[
\begin{aligned}
r_r&=0.080\times0.120\\
&=0.0096\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

Part c: Determine the net direction

\[
\begin{aligned}
r_{\mathrm{net}}&=r_f-r_r\\
&=0.0036-0.0096\\
&=-0.0060\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

The negative sign means net conversion is occurring from B to A.

The mixture contains enough B that reverse conversion exceeds forward conversion, despite the larger forward rate constant.

Part d: Compare with the uncatalysed reaction

Without the catalyst:

\[
\begin{aligned}
r_f&=0.030\times0.030\\
&=0.0009\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

\[
\begin{aligned}
r_r&=0.020\times0.120\\
&=0.0024\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
r_{\mathrm{net}}&=0.0009-0.0024\\
&=-0.0015\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

The ratio of the magnitudes is:

\[
\frac{0.0060}{0.0015}=4
\]

The catalyst makes the net reverse conversion four times faster.

The student’s argument fails because reaction rates depend on both rate constants and concentrations. Comparing \(k_f\) and \(k_r\) alone ignores the large amount of B initially present.

A catalyst can therefore make B disappear faster, even when it substantially increases the forward rate constant.

Question 3

A sealed vessel contains the hypothetical reaction:

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

At a fixed temperature, the mixture is at equilibrium with:

\[
[A]=0.040\ \mathrm{mol\,L^{-1}}
\]

\[
[B]=0.060\ \mathrm{mol\,L^{-1}}
\]

The uncatalysed rate constants are:

\[
k_f=0.030\ \mathrm{min^{-1}}
\]

\[
k_r=0.020\ \mathrm{min^{-1}}
\]

Both rate constants increase by a factor of four when a catalyst is added.

An investigator replaces \(0.010\ \mathrm{mol\,L^{-1}}\) of ordinary B with the same concentration of isotopically labelled B, written \(B^*\).

The replacement leaves the total concentration of B unchanged.

Assume the label is carried into \(A^*\) when \(B^*\) converts to A, and that isotopic labelling does not affect reaction rates.

a. Calculate the initial rate of formation of \(A^*\) in each vessel.

b. Explain whether \(A^*\) can appear while the total concentrations of A and B remain unchanged.

c. Explain why this experiment provides evidence about the reverse reaction that an ordinary equilibrium concentration measurement cannot provide.

Solution 3

Labelled \(A^*\) appears four times faster in the catalysed vessel, even though the total concentrations of A and B remain constant. The isotope acts as a tracer of molecular conversion that ordinary concentration measurements cannot reveal.

Part a: Calculate the initial formation rates

Initially, all the label is in B, so:

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

There is initially no labelled A to convert back into labelled B.

Without the catalyst:

\[
\begin{aligned}
r_{B^*\to A^*}&=k_r[B^*]\\
&=0.020\times0.010\\
&=0.00020\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

With the catalyst, the reverse rate constant is:

\[
k_{r,\mathrm{cat}}=4(0.020)=0.080\ \mathrm{min^{-1}}
\]

Therefore:

\[
\begin{aligned}
r_{B^*\to A^*}&=0.080\times0.010\\
&=0.00080\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

The rate of initial labelled-A formation increases fourfold.

Part b: Explain the unchanged concentrations

For ordinary A and B, calculate the total forward and reverse rates without the catalyst.

Forward:

\[
\begin{aligned}
r_f&=0.030\times0.040\\
&=0.0012\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

Reverse:

\[
\begin{aligned}
r_r&=0.020\times0.060\\
&=0.0012\ \mathrm{mol\,L^{-1}\,min^{-1}}
\end{aligned}
\]

They are equal.

With the catalyst, both become:

\[
4(0.0012)=0.0048\ \mathrm{mol\,L^{-1}\,min^{-1}}
\]

The net rate remains zero.

Molecules are still converting in both directions. The label simply allows us to follow some of those conversions.

As labelled B turns into labelled A, other A molecules turn into B. The total concentrations remain constant because the overall forward and reverse rates are equal.

Part c: Explain the evidence

An ordinary concentration measurement sees no net change at equilibrium. It cannot distinguish a static system from one with ongoing molecular conversion.

The appearance of \(A^*\) demonstrates that reverse conversion is occurring despite the constant total concentrations.

Its faster appearance in the catalysed vessel supports increased reverse reaction activity.

The tempting prediction is that nothing should change once equilibrium has been reached. That confuses zero net conversion with zero molecular conversion.

Dynamic equilibrium involves continuing reactions whose opposing rates balance.

Question 4

The reversible gas reaction:

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

is exothermic in the forward direction.

In this illustrative system, its equilibrium constants are:

A sealed \(1.00\ \mathrm{L}\) vessel initially contains \(0.100\ \mathrm{mol}\) of A and B combined, at equilibrium at \(20^\circ\mathrm{C}\).

The following changes are made:

1. The mixture is heated to \(40^\circ\mathrm{C}\), and a catalyst is added simultaneously.
2. The system is allowed to reach equilibrium at \(40^\circ\mathrm{C}\).
3. The vessel is cooled back to \(20^\circ\mathrm{C}\), with the catalyst still present.
4. The system is allowed to reach equilibrium again.

Assume A and B are the only chemical species in the reaction, neither is lost or adsorbed, and the vessel volume stays constant.

a. Calculate the equilibrium concentrations of A and B at all three equilibrium states.

b. Determine the net direction of reaction immediately after heating and immediately after cooling.

c. A student observes that B decreases after the catalyst is added and concludes that the catalyst favours the reverse reaction. Explain why the observation does not support that conclusion.

d. Describe a control experiment that separates the effect of temperature from the effect of catalysis.

Solution 4

The equilibrium concentration of B changes from \(0.080\) to \(0.050\) and back to \(0.080\ \mathrm{mol\,L^{-1}}\). Heating causes the initial net reverse reaction, while cooling causes the subsequent net forward reaction. The catalyst speeds up the approach to each temperature’s equilibrium without changing either equilibrium constant.

Part a: Calculate the equilibrium concentrations

Because the reaction has a 1:1 ratio and the vessel volume is fixed:

\[
[A]+[B]=0.100\ \mathrm{mol\,L^{-1}}
\]

At \(20^\circ\mathrm{C}\):

\[
\begin{aligned}
K_c&=\frac{[B]}{[A]}=4.0\\
[B]&=4[A]
\end{aligned}
\]

Using conservation:

\[
\begin{aligned}
[A]+4[A]&=0.100\\
5[A]&=0.100\\
[A]&=0.020\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Therefore:

\[
[B]=0.080\ \mathrm{mol\,L^{-1}}
\]

At \(40^\circ\mathrm{C}\), \(K_c=1.0\), so the equilibrium concentrations are equal:

\[
[A]=[B]=0.050\ \mathrm{mol\,L^{-1}}
\]

After cooling back to \(20^\circ\mathrm{C}\), the equilibrium constant returns to 4.0.

The final concentrations therefore return to:

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

\[
[B]=0.080\ \mathrm{mol\,L^{-1}}
\]

Part b: Determine the net directions

Immediately after heating, the mixture still has the old composition:

\[
[A]=0.020,\qquad [B]=0.080
\]

Its concentration reaction quotient is:

\[
\begin{aligned}
Q_c&=\frac{[B]}{[A]}\\
&=\frac{0.080}{0.020}\\
&=4.0
\end{aligned}
\]

But at \(40^\circ\mathrm{C}\), the equilibrium constant is only 1.0.

Since \(Q_c>K_c\), there is too much B relative to A for equilibrium at the new temperature.

Net reaction therefore proceeds backwards, converting B into A.

After cooling, the mixture initially has:

\[
[A]=[B]=0.050\ \mathrm{mol\,L^{-1}}
\]

Its reaction quotient is now 1.0, while \(K_c=4.0\) at \(20^\circ\mathrm{C}\).

Since \(Q_c<K_c\), net reaction proceeds forwards, producing more B.

This agrees with Le Chatelier’s principle: heating favours the endothermic reverse direction, while cooling favours the exothermic forward direction.

Part c: Explain the incorrect conclusion

Two changes occurred simultaneously: the mixture was heated, and a catalyst was added.

The temperature change altered the equilibrium constant and therefore changed the equilibrium composition.

The catalyst accelerated the reaction towards that new equilibrium.

A decrease in B is not evidence that the catalyst selectively favours the reverse reaction. It is consistent with the mixture responding to the higher temperature.

The tempting mistake is assigning every observed change to the catalyst while overlooking the second changed variable.

Part d: Design the control

Compare two otherwise identical systems held at \(40^\circ\mathrm{C}\), one with the catalyst and one without.

Both should approach the same equilibrium concentrations:

\[
[A]=[B]=0.050\ \mathrm{mol\,L^{-1}}
\]

The catalysed vessel should generally reach that composition faster.

A corresponding comparison at \(20^\circ\mathrm{C}\) would confirm that the equilibrium composition is unchanged by the catalyst at that temperature.

The investigation must compare catalytic and uncatalysed conditions at the same temperature. Otherwise, the effects cannot be separated reliably.

Question 5

A researcher studies the hypothetical reversible aqueous reaction:

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

At a fixed temperature, the equilibrium constant for the free dissolved species is:

\[
K_c=\frac{[B]_{\mathrm{free}}}{[A]_{\mathrm{free}}}=2.0
\]

A \(1.00\ \mathrm{L}\) vessel initially contains \(0.120\ \mathrm{mol}\) of dissolved A and no B.

The researcher compares two systems.

System I: No solid additive

The reaction reaches equilibrium without a catalyst. Neither A nor B is removed from the solution.

System II: Catalytic beads

A solid material provides an alternative, faster reaction pathway. However, the solid also reversibly binds some B to its surface.

At equilibrium, the amount of surface-bound B is exactly half the amount of free dissolved B:

\[
n(B_{\mathrm{bound}})
=\frac12 n(B_{\mathrm{free}})
\]

The beads do not bind A.

The catalyst does not change the intrinsic equilibrium constant for dissolved A and B. The surface binding is an additional physical process.

Assume the volume of the solution remains \(1.00\ \mathrm{L}\), no chemical material is lost, and the adsorption relationship above holds at equilibrium.

For the simple uncatalysed reaction-energy profile:

\[
E_{a,f}=70\ \mathrm{kJ\,mol^{-1}}
\]

\[
E_{a,r}=95\ \mathrm{kJ\,mol^{-1}}
\]

With the catalytic material, the forward pathway has an activation energy of \(40\ \mathrm{kJ\,mol^{-1}}\).

Assume the simplified catalysed energy profile also has one transition-state peak.

a. Calculate \(\Delta H\) and both catalysed activation energies.

b. Calculate the equilibrium amounts of free A, free B, and bound B in each system.

c. Determine the total amount of B, including bound B, in each system. The researcher claims the catalyst has changed the equilibrium constant because System II contains a different total amount of B. Evaluate this claim.

d. After System II reaches equilibrium, all bound B is released into the same solution without any loss, and the beads are removed. Temperature and volume remain constant. Determine the initial net reaction direction after release and the final equilibrium concentrations.

Explain why the answer to part d is consistent with the conclusion in part c.

Solution 5

The catalysed activation energies are \(40\ \mathrm{kJ\,mol^{-1}}\) forwards and \(65\ \mathrm{kJ\,mol^{-1}}\) backwards. System II contains more total B but less free B because some B is held on the beads. The catalyst has not changed \(K_c\); the additional binding process has changed how the material is distributed.

After the bound B is released and the beads are removed, the solution initially reacts backwards until it returns to the same equilibrium composition as System I.

Part a: Determine the energy quantities

For the uncatalysed pathway:

\[
\begin{aligned}
\Delta H&=E_{a,f}-E_{a,r}\\
&=70-95\\
&=-25\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The reaction is exothermic.

The catalysed forward activation energy is supplied as:

\[
E_{a,f}=40\ \mathrm{kJ\,mol^{-1}}
\]

The catalysed reverse activation energy is:

\[
\begin{aligned}
E_{a,r}&=E_{a,f}-\Delta H\\
&=40-(-25)\\
&=65\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The catalyst lowers both activation energies by \(30\ \mathrm{kJ\,mol^{-1}}\).

It does not change the reaction enthalpy.

Part b: Determine the equilibrium amounts

For System I, no B is bound.

The equilibrium condition is:

\[
[B]=2[A]
\]

The total amount of A and B is \(0.120\ \mathrm{mol}\).

Since the solution volume is \(1.00\ \mathrm{L}\):

\[
[A]+[B]=0.120\ \mathrm{mol\,L^{-1}}
\]

Therefore:

\[
\begin{aligned}
[A]+2[A]&=0.120\\
3[A]&=0.120\\
[A]&=0.040\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

And:

\[
[B]=0.080\ \mathrm{mol\,L^{-1}}
\]

So System I contains:

• Free A: \(0.040\ \mathrm{mol}\).
• Free B: \(0.080\ \mathrm{mol}\).
• Bound B: \(0\ \mathrm{mol}\).

Now consider System II.

Let \(x\) be the amount of free A in moles.

The vessel has a volume of \(1.00\ \mathrm{L}\), so its numerical concentration in \(\mathrm{mol\,L^{-1}}\) is also \(x\).

Because \(K_c=2.0\):

\[
n(B_{\mathrm{free}})=2x
\]

The adsorption condition gives:

\[
\begin{aligned}
n(B_{\mathrm{bound}})
&=\frac12(2x)\\
&=x
\end{aligned}
\]

Conservation of the total amount of A-derived material requires:

\[
\begin{aligned}
x+2x+x&=0.120\\
4x&=0.120\\
x&=0.030\ \mathrm{mol}
\end{aligned}
\]

Therefore, System II contains:

• Free A: \(0.030\ \mathrm{mol}\).
• Free B: \(0.060\ \mathrm{mol}\).
• Bound B: \(0.030\ \mathrm{mol}\).

Part c: Compare total B and evaluate the claim

In System I:

\[
n(B_{\mathrm{total}})=0.080\ \mathrm{mol}
\]

In System II:

\[
\begin{aligned}
n(B_{\mathrm{total}})
&=0.060+0.030\\
&=0.090\ \mathrm{mol}
\end{aligned}
\]

System II contains more total B.

However, it contains less free dissolved B.

This is the surprising result:

All amounts are in moles.

Check the equilibrium constant using free dissolved concentrations.

System I:

\[
K_c=\frac{0.080}{0.040}=2.0
\]

System II:

\[
K_c=\frac{0.060}{0.030}=2.0
\]

Both give the same equilibrium constant.

The additional B in System II arises because the beads bind some product. The binding process reduces the amount of free B, allowing further conversion of A into B before equilibrium is reached.

This is not a catalyst changing the equilibrium constant. It is a material performing two roles: catalysing the reversible reaction and storing some product through adsorption.

The tempting approach is to count only the total B produced and conclude that the catalyst favours products. That overlooks the extra equilibrium process involving the solid surface.

Part d: Release the bound B and remove the beads

Immediately before release, System II contains:

\[
n(A)=0.030\ \mathrm{mol}
\]

\[
n(B_{\mathrm{free}})=0.060\ \mathrm{mol}
\]

\[
n(B_{\mathrm{bound}})=0.030\ \mathrm{mol}
\]

After releasing all bound B into solution and removing the beads without losing material:

\[
n(A)=0.030\ \mathrm{mol}
\]

\[
n(B)=0.090\ \mathrm{mol}
\]

The volume remains \(1.00\ \mathrm{L}\), so the immediate reaction quotient is:

\[
\begin{aligned}
Q_c&=\frac{[B]}{[A]}\\
&=\frac{0.090}{0.030}\\
&=3.0
\end{aligned}
\]

But the equilibrium constant is still:

\[
K_c=2.0
\]

Since:

\[
Q_c>K_c
\]

the solution initially undergoes net reverse reaction:

\[
\ce{B -> A}
\]

Some B converts back into A until the equilibrium ratio is restored.

With no B remaining bound, the final concentrations satisfy:

\[
[A]+[B]=0.120
\]

and:

\[
[B]=2[A]
\]

Therefore:

\[
[A]=0.040\ \mathrm{mol\,L^{-1}}
\]

\[
[B]=0.080\ \mathrm{mol\,L^{-1}}
\]

These are exactly the equilibrium concentrations in System I.

The crucial constraint is conservation of matter combined with the equilibrium condition for the free substances.

While B is adsorbed, the solution and the solid form a larger coupled system. Once the adsorption process is removed, the remaining A and B must satisfy the original solution equilibrium.

Removing the catalyst may make equilibration slower, but it does not change the direction predicted by \(Q_c\) and \(K_c\).

06The next decision: is it the pathway or the equilibrium conditions?

When an exam question combines catalysts, reaction rates, and equilibrium, separate the questions being asked.

Activation energy tells you how difficult a reaction pathway is to follow. Reaction enthalpy tells you the overall enthalpy difference between reactants and products. The equilibrium constant tells you the equilibrium composition relationship at a given temperature.

A catalyst changes the first of these without independently changing the other two.

If a catalysed sample contains more product after five minutes, establish whether it has actually reached equilibrium. If two samples have different final compositions, check temperature, loss of reactants or products, adsorption, and side reactions before attributing the difference to catalysis.

For a simple reversible energy profile, you can always test your barrier calculations against:

\[
\Delta H=E_{a,f}-E_{a,r}
\]

For more practice using that relationship, see How to Calculate Reverse Activation Energy Using Delta H.

The next useful skill is deciding which way a mixture will react after a disturbance. Compare its current reaction quotient, \(Q_c\), with the equilibrium constant, \(K_c\). That tells you the direction of net reaction. The catalyst determines how quickly the system can respond, not which equilibrium it is trying to reach.