Why Reaction Rate Cannot Be Predicted from ΔH Alone

A reaction can release 80 kJ mol⁻¹ and still happen slowly, while another reaction with a much smaller energy change can happen quickly. That seems backwards until you distinguish two different energy measurements.

The enthalpy change, ΔH, tells you how much the energy of the system changes between reactants and products. The activation energy, \(E_a\), tells you how much energy particles need to reach the activated complex before the reaction can proceed.

**You cannot determine a reaction's rate from ΔH alone because the overall energy change does not tell you the height of the activation barrier.** A reaction can be strongly exothermic but have a large activation energy. It can also be only slightly exothermic but have a much smaller activation energy.

Let's work through the difference, including what happens when the reaction runs in reverse.

## 1. The energy released is not the energy barrier

Imagine pushing a skateboard up a hill and then letting it roll down the other side.

The difference in height between the starting point and the finishing point tells you how much gravitational potential energy is lost. But it doesn't tell you how high the hill rises in between.

A reaction has a similar distinction:

- ΔH describes the energy difference between the reactants and products.
- Activation energy describes the energy difference between the reactants and the transition state.

The analogy is useful for separating the two measurements, but chemical energy profiles are not literal hills. The vertical axis represents enthalpy, and the horizontal axis represents progress along the reaction pathway.

[IMAGE: An inviting scientific illustration of a skateboard travelling over a hill, with the starting and finishing heights marked and the higher intermediate peak clearly visible. Show that the height difference between the endpoints is different from the height needed to reach the peak.]

## 2. Reading an energy profile correctly

An energy profile makes the distinction precise.

The vertical axis represents the relative enthalpy of the reacting system, usually in kJ mol⁻¹. The horizontal axis represents reaction progress. It does not represent time.

There are three important energy levels:

| Symbol | Meaning |
|---|---|
| \(H_R\) | Enthalpy of the reactants |
| \(H^\ddagger\) | Enthalpy of the transition state |
| \(H_P\) | Enthalpy of the products |

The transition state is the short-lived, high-energy arrangement of atoms reached as bonds break and form.

[IMAGE: An accurately annotated single-step exothermic energy profile. Show reactants at 0 kJ mol−1, products at −40 kJ mol−1, and the transition-state peak at +70 kJ mol−1. Include a vertical arrow from reactants to the peak labelled forward activation energy, another from products to the peak labelled reverse activation energy, and a separate downward arrow between reactants and products labelled ΔH. Label both axes. Make the three energy differences visually distinct.]

### Three measurements, three different questions

Reaction enthalpy is the difference between the products and reactants:

\[
\Delta H=H_P-H_R
\]

Forward activation energy is the difference between the transition state and the reactants:

\[
E_{a,\mathrm{forward}}=H^\ddagger-H_R
\]

Reverse activation energy is the difference between the transition state and the products:

\[
E_{a,\mathrm{reverse}}=H^\ddagger-H_P
\]

All three quantities are measured in kJ mol⁻¹.

Notice that ΔH does not contain the transition-state energy. This is exactly why it cannot tell you the activation barrier.

### Two reactions, the same enthalpy change

Consider two hypothetical reactions with the following illustrative energy levels:

| Energy level | Reaction M | Reaction N |
|---|---:|---:|
| Reactants | 0 kJ mol⁻¹ | 0 kJ mol⁻¹ |
| Transition state | +78 kJ mol⁻¹ | +28 kJ mol⁻¹ |
| Products | −42 kJ mol⁻¹ | −42 kJ mol⁻¹ |

Both reactions have the same enthalpy change:

\[
\Delta H=-42-0=-42\ \mathrm{kJ\,mol^{-1}}
\]

However, their forward activation energies are different.

For reaction M:

\[
E_{a,\mathrm{forward}}=78-0=78
\]

For reaction N:

\[
E_{a,\mathrm{forward}}=28-0=28
\]

Both reactions release the same amount of energy per mole, but reaction N has a much smaller forward activation barrier.

At the same temperature, assuming comparable collision factors and reaction conditions, reaction N would generally proceed faster.

The important point is not that a 28 kJ mol⁻¹ barrier produces a particular numerical rate. You cannot calculate that rate from this table. You can only recognise that the smaller barrier makes successful collisions more accessible.

[IMAGE: Two contrasting energy profiles for hypothetical reactions M and N. Both begin at zero and finish at −42 kJ mol−1. Reaction M has a peak at +78, while reaction N has a peak at +28. Clearly show that the endpoint difference is identical while the peak heights differ.]

This also explains why the amount of energy released is not a reliable measure of how quickly a reaction happens.

## 3. What would an experiment actually measure?

Imagine being asked to investigate an exothermic reaction between magnesium and hydrochloric acid:

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

You want to find out two things:

1. How much energy does the reaction release per mole of magnesium?
2. How quickly does the reaction occur?

Would one thermometer answer both questions?

No. You need two different measurements.

### First measurement: reaction enthalpy

Start with an insulated cup calorimeter.

Place a known quantity of hydrochloric acid inside a polystyrene cup. Fit a lid containing a temperature probe, and add the measured magnesium sample.

The thermometer records the temperature change as the reaction proceeds.

[IMAGE: A clearly recognisable insulated polystyrene cup calorimeter containing aqueous hydrochloric acid, with a lid, thermometer probe and stirring rod. Show a small magnesium ribbon sample being added. Label the solution, insulated cup and temperature probe. Make the apparatus connections and positions scientifically accurate.]

What should you observe if the reaction is exothermic?

The solution should warm up because energy is transferred from the reacting system to its surroundings.

You can use the temperature change to estimate the heat absorbed by the solution:

\[
q=mc\Delta T
\]

Here:

- \(q\) is the heat transferred, in joules.
- \(m\) is the mass of the solution, in grams.
- \(c\) is the specific heat capacity, in J g⁻¹ °C⁻¹.
- \(\Delta T\) is the temperature change, in °C.

At constant pressure, the heat transferred to the reaction is approximately the reaction enthalpy for the amount that reacted:

\[
q_{\mathrm{reaction}}=-q_{\mathrm{solution}}
\]

The negative sign matters. If the solution gains heat, the reaction loses heat.

This method estimates ΔH. It does not directly measure activation energy or tell you how many successful collisions occur each second.

### Worked example: Calculating reaction enthalpy from calorimetry

A student reacts 0.00250 mol of magnesium with excess hydrochloric acid in an insulated cup.

The following illustrative measurements are recorded:

- Mass of solution: 50.0 g
- Specific heat capacity: 4.18 J g⁻¹ °C⁻¹
- Initial temperature: 21.0 °C
- Final temperature: 26.6 °C

Assume negligible heat loss and that the cup absorbs negligible energy. Estimate the reaction enthalpy per mole of magnesium.

**Step 1**: Calculate the temperature change.

\[
\Delta T=26.6-21.0=5.6^\circ\mathrm C
\]

**Step 2**: Calculate the heat absorbed by the solution.

\[
\begin{aligned}
q_{\mathrm{solution}}&=mc\Delta T\\
&=50.0\times4.18\times5.6\\
&=1170.4\ \mathrm J\\
&=1.17\ \mathrm{kJ}
\end{aligned}
\]

**Step 3**: Determine the heat transferred by the reaction.

The solution gained 1.17 kJ, so the reaction released 1.17 kJ.

\[
q_{\mathrm{reaction}}=-1.17\ \mathrm{kJ}
\]

**Step 4**: Calculate the molar enthalpy change.

\[
\begin{aligned}
\Delta H&=\frac{q_{\mathrm{reaction}}}{n(\mathrm{Mg})}\\
&=\frac{-1.17}{0.00250}\\
&=-468\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The estimated reaction enthalpy is:

\[
\boxed{\Delta H=-468\ \mathrm{kJ\,mol^{-1}}}
\]

The negative sign confirms that the reaction is exothermic.

But notice what we haven't calculated: the reaction rate.

The temperature change tells us about the heat transferred. It does not tell us whether the reaction took 10 seconds or 10 minutes.

### Revised measurement: reaction rate

Now use a separate, matched experiment.

Place the same reactants in a conical flask. Fit a rubber bung connected to a delivery tube, with the other end connected to a gas syringe.

As magnesium reacts, hydrogen gas travels from the flask through the tube and pushes the gas syringe plunger outwards.

Record the gas volume at regular time intervals.

[IMAGE: A scientifically accurate gas-collection apparatus showing a conical flask containing magnesium and hydrochloric acid, sealed with a bung, connected by a delivery tube to a horizontal gas syringe. Show hydrogen travelling from the flask into the syringe and pushing the plunger outwards. Label the flask, delivery tube, hydrogen gas and gas syringe.]

The following is illustrative experimental data, not a record of an actual experiment.

| Time (s) | Hydrogen volume (cm³) |
|---:|---:|
| 0 | 0 |
| 10 | 18 |
| 20 | 31 |
| 30 | 40 |
| 60 | 55 |

We can calculate the average rate of hydrogen production over the first 20 seconds:

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

\[
\begin{aligned}
r_{\mathrm{avg}}&=\frac{31-0}{20-0}\\
&=1.55\ \mathrm{cm^3\,s^{-1}}
\end{aligned}
\]

The average hydrogen production rate is 1.55 cm³ s⁻¹ over this interval.

This measurement tells us how quickly the reaction produces gas under the experimental conditions.

It still does not directly measure activation energy.

### What did the two experiments establish?

| Experiment | Measurement | What it tells us |
|---|---|---|
| Cup calorimetry | Temperature change | Heat transferred and estimated ΔH |
| Gas syringe | Gas volume over time | Reaction rate |
| Energy profile | Relative energy levels | Activation barriers and reaction enthalpy |

The original calorimeter design could establish whether the reaction was exothermic and estimate how much energy was released. It could not establish how quickly the reaction occurred.

The revised rate experiment addresses that limitation by measuring product formation over time.

Both measurements are useful, but they answer different questions.

There are also experimental limitations. Calorimetry can underestimate heat release if energy escapes to the surroundings. The calculation above assumes the solution has the same specific heat capacity as water and that the cup absorbs negligible heat. Gas collection can be affected by leaks, and the experiment must control reactant quantities, temperature, and surface area.

The central conclusion remains: a measurement of ΔH alone cannot establish a reaction's rate.

## 4. What changes when the reaction runs in reverse?

So far, we've considered the forward reaction. But many reactions are reversible.

Consider:

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

The forward reaction converts A into B. The reverse reaction converts B into A.

For a simple single-step pathway, both directions pass through the same transition state.

This gives us an important relationship between forward activation energy, reverse activation energy, and ΔH.

Starting with:

\[
E_{a,\mathrm{forward}}=H^\ddagger-H_R
\]

and:

\[
E_{a,\mathrm{reverse}}=H^\ddagger-H_P
\]

We also know:

\[
\Delta H=H_P-H_R
\]

Rearranging gives:

\[
\boxed{E_{a,\mathrm{reverse}}=
E_{a,\mathrm{forward}}-\Delta H}
\]

This equation is useful, but you must keep the sign of ΔH.

### Exothermic forward reaction

For an exothermic reaction, ΔH is negative.

The products have lower enthalpy than the reactants. Therefore, the reverse reaction must climb further to reach the same transition state.

This means:

\[
E_{a,\mathrm{reverse}}>E_{a,\mathrm{forward}}
\]

[IMAGE: A single-step exothermic reversible energy profile. Reactants are at 0 kJ mol−1, products at −35 kJ mol−1, and the transition state at +55 kJ mol−1. Show the forward and reverse activation-energy arrows measured from their respective starting levels, plus the separate negative ΔH arrow. Label the forward and reverse directions.]

### Endothermic forward reaction

For an endothermic reaction, ΔH is positive.

The products have higher enthalpy than the reactants. The reverse reaction therefore starts from a higher energy level and has a smaller barrier to overcome.

This means:

\[
E_{a,\mathrm{reverse}}<E_{a,\mathrm{forward}}
\]

The two cases can be summarised:

| Forward reaction | Sign of ΔH | Relationship |
|---|---|---|
| Exothermic | Negative | Reverse activation energy is greater |
| Endothermic | Positive | Reverse activation energy is smaller |

### Worked example: Finding the reverse barrier for an endothermic reaction

A reversible, single-step reaction has a forward activation energy of 58 kJ mol⁻¹. The forward reaction absorbs 21 kJ mol⁻¹.

Calculate the reverse activation energy.

**Step 1**: Identify the sign of the enthalpy change.

The reaction absorbs energy, so it is endothermic:

\[
\Delta H=+21\ \mathrm{kJ\,mol^{-1}}
\]

**Step 2**: Select the relationship.

\[
E_{a,\mathrm{reverse}}=
E_{a,\mathrm{forward}}-\Delta H
\]

**Step 3**: Substitute the values.

\[
\begin{aligned}
E_{a,\mathrm{reverse}}&=58-(+21)\\
&=37\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

**Step 4**: Interpret the result.

The reverse activation energy is 37 kJ mol⁻¹, which is smaller than the forward activation energy of 58 kJ mol⁻¹.

This makes sense because the products start at a higher enthalpy than the reactants.

The most common mistake here is to add 21 instead of subtracting it. Always use the signed value of ΔH.

For more practice with this calculation, see [How to Calculate Reverse Activation Energy Using Delta H](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-find-reverse-activation-energy-from-forward-activation-energy-and-delta-h/).

### A larger reverse barrier does not automatically mean a slower net reaction

There's one more distinction worth understanding.

A larger activation energy generally makes an individual reaction direction less accessible at the same temperature, assuming comparable kinetic factors.

However, the actual forward and reverse rates also depend on the concentrations of the reactants in each direction.

At equilibrium, the forward and reverse reaction rates are equal.

That does not mean their activation energies are equal. It means the concentrations and kinetic behaviour balance so that both directions occur at the same rate.

We'll return to this idea in the final question.

## 5. The contrasting case: A very exothermic reaction that needs a spark

Consider the reaction between hydrogen and oxygen:

\[
\ce{2H2(g) + O2(g) -> 2H2O(l)}
\]

Under standard conditions, the reaction enthalpy is approximately:

\[
\Delta H=-572\ \mathrm{kJ\,mol^{-1}}
\]

for the reaction as written.

That's a substantial release of energy.

You might predict that hydrogen and oxygen should react extremely quickly whenever they are mixed. After all, the products have considerably lower enthalpy than the reactants.

But that prediction ignores the activation barrier.

At ordinary temperatures, a hydrogen-oxygen mixture can react extremely slowly without an ignition source. An initial input of energy is needed to initiate the reaction. Once combustion begins, the energy released helps sustain the rapid reaction.

[IMAGE: An educational illustration of hydrogen and oxygen molecules approaching a reaction pathway, with a small ignition spark initiating the process and water molecules as products. Include a separate simple energy-profile inset showing a strongly exothermic reaction with a substantial activation barrier. Do not depict an actual experiment or suggest that the illustration is a photograph.]

The reaction is strongly exothermic, but that does not mean it has a small activation energy.

The spark helps overcome the initial kinetic barrier. It does not make the reaction enthalpy more negative.

This is the distinction you need to recognise in exam questions:

- Thermodynamics describes the energy difference between reactants and products.
- Kinetics describes how quickly the reaction proceeds.
- Activation energy helps explain the kinetic barrier.

A reaction can be energetically favourable but kinetically slow.

## 6. A useful decision process for exam questions

When you're given ΔH and asked about reaction rate, don't jump straight to a conclusion.

Work through this sequence:

| Information supplied | What you can conclude | What you cannot conclude |
|---|---|---|
| ΔH only | Whether the reaction is exothermic or endothermic | The reaction rate |
| Forward activation energy | The forward energy barrier | The exact numerical rate |
| ΔH and forward activation energy | Reverse activation energy for the same single-step pathway | The exact forward or reverse rate |
| Rate measurements over time | How quickly the reaction proceeds under those conditions | Activation energy from one rate measurement alone |
| Multiple rate measurements at different temperatures | How the rate changes with temperature | ΔH from rate data alone |

Keep the assumptions in mind. The reverse activation-energy relationship uses the same single-step pathway and the same physical states in both directions. A multistep mechanism has several transition states and intermediates, so each elementary activation barrier must be measured from its preceding energy minimum.

For broader explanations of how temperature, concentration, surface area, and catalysts affect successful collisions, see [How to Explain Reaction-Rate Changes in HSC Chemistry](https://tutorgum.com/learn/resources/how-to-explain-reaction-rate-trends-without-just-listing-the-factors/).

## Questions and solutions

### Question 1

Two reactions, P and Q, are carried out at the same temperature. Each begins with the same initial amount of its limiting reactant.

Their illustrative data are:

| Measurement | Reaction P | Reaction Q |
|---|---:|---:|
| Enthalpy change | −75 kJ mol⁻¹ | −18 kJ mol⁻¹ |
| Reactant converted in first 10 s | 20% | 65% |

Calculate the average percentage conversion per second for each reaction.

Identify which reaction is faster and which releases more energy per mole. Explain why the data contradict the prediction that the more exothermic reaction must be faster.

### Solution 1

Reaction Q is faster, even though reaction P is more exothermic.

Calculate the average percentage conversion per second.

For reaction P:

\[
r_P=\frac{20}{10}
=2.0\%\ \mathrm{s^{-1}}
\]

For reaction Q:

\[
r_Q=\frac{65}{10}
=6.5\%\ \mathrm{s^{-1}}
\]

Reaction Q has an average percentage conversion rate 3.25 times that of reaction P over the first 10 seconds.

However, reaction P releases more energy per mole because its enthalpy change is more negative:

\[
-75<-18
\]

The tempting prediction is that P must be faster because it releases more energy. But the data show that energy released and reaction speed are different properties.

The enthalpy change tells us the overall energy difference. It does not establish the activation energy or the number of successful collisions per second.

We also cannot calculate either reaction's activation energy from these conversion percentages alone.

### Question 2

A hypothetical reversible reaction has the following relative enthalpy levels:

| Energy level | Enthalpy (kJ mol⁻¹) |
|---|---:|
| Reactants | 0 |
| Transition state | +63 |
| Products | +22 |

[IMAGE: A single-peak endothermic energy profile with reactants at 0 kJ mol−1, transition state at +63 kJ mol−1 and products at +22 kJ mol−1. Label the three horizontal energy levels and both axes. Do not include activation-energy arrows or the ΔH arrow, as students must determine these.]

Calculate:

(a) The forward reaction enthalpy.

(b) The forward activation energy.

(c) The reverse activation energy.

(d) Explain which direction has the smaller activation barrier.

### Solution 2

The forward reaction is endothermic, with a forward activation energy of 63 kJ mol⁻¹ and a reverse activation energy of 41 kJ mol⁻¹.

**(a) Calculate ΔH.**

Use:

\[
\Delta H=H_P-H_R
\]

Substitute:

\[
\Delta H=22-0
\]

\[
\boxed{\Delta H=+22\ \mathrm{kJ\,mol^{-1}}}
\]

The positive value indicates that the forward reaction absorbs energy.

**(b) Calculate forward activation energy.**

\[
E_{a,\mathrm{forward}}=H^\ddagger-H_R
\]

\[
=63-0
\]

\[
\boxed{E_{a,\mathrm{forward}}=63\ \mathrm{kJ\,mol^{-1}}}
\]

**(c) Calculate reverse activation energy.**

\[
E_{a,\mathrm{reverse}}=H^\ddagger-H_P
\]

\[
=63-22
\]

\[
\boxed{E_{a,\mathrm{reverse}}=41\ \mathrm{kJ\,mol^{-1}}}
\]

**(d) Compare the barriers.**

The reverse reaction has the smaller activation energy.

The tempting mistake is to assume that reversing a reaction must increase its activation energy. That is not always true.

Here, the products have a higher enthalpy than the reactants. The reverse reaction therefore starts closer to the transition-state energy level.

### Question 3

A student investigates the reaction between zinc and hydrochloric acid:

\[
\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}
\]

Two separate, matched experiments are conducted.

In the calorimetry experiment:

- Mass of solution = 100.0 g
- Specific heat capacity = 4.18 J g⁻¹ °C⁻¹
- Temperature increase = 1.85 °C
- Amount of zinc reacted = 0.00500 mol

In a separate gas-syringe experiment, the following illustrative measurements are obtained:

| Time (s) | Hydrogen volume (cm³) |
|---:|---:|
| 0 | 0 |
| 10 | 30 |
| 20 | 52 |
| 30 | 68 |
| 40 | 80 |

Assume negligible heat loss in the calorimetry experiment, constant pressure, and that the solution has the specific heat capacity of water.

[IMAGE: Two separate experimental setups shown side by side. On the left, an insulated cup calorimeter containing zinc and hydrochloric acid, with a temperature probe. On the right, a conical flask containing the same reactants connected through a bung and delivery tube to a gas syringe. Clearly distinguish the two apparatus arrangements and show the hydrogen travelling into the syringe.]

(a) Calculate the approximate molar enthalpy change of the reaction.

(b) Calculate the average rate of hydrogen production over the first 20 seconds.

(c) Explain why the calorimetry result cannot be used to calculate the reaction rate.

(d) If the same amount of zinc reacts completely in both experiments, explain why the final amount of hydrogen can be the same even if the reactions proceed at different rates.

### Solution 3

The reaction has an estimated enthalpy change of −154.7 kJ mol⁻¹, and the average hydrogen production rate over the first 20 seconds is 2.6 cm³ s⁻¹.

**(a) Calculate the molar enthalpy change.**

First, calculate the heat absorbed by the solution:

\[
q=mc\Delta T
\]

\[
\begin{aligned}
q&=100.0\times4.18\times1.85\\
&=773.3\ \mathrm J\\
&=0.7733\ \mathrm{kJ}
\end{aligned}
\]

The solution gains heat, so the reaction releases heat:

\[
q_{\mathrm{reaction}}=-0.7733\ \mathrm{kJ}
\]

Calculate the enthalpy change per mole of zinc:

\[
\begin{aligned}
\Delta H&=\frac{-0.7733}{0.00500}\\
&=-154.66\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta H=-154.7\ \mathrm{kJ\,mol^{-1}}}
\]

The reaction is exothermic.

**(b) Calculate the average rate of hydrogen production.**

Use the gas volumes at 0 and 20 seconds:

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

\[
\begin{aligned}
r_{\mathrm{avg}}&=\frac{52-0}{20-0}\\
&=2.6\ \mathrm{cm^3\,s^{-1}}
\end{aligned}
\]

The average hydrogen production rate is:

\[
\boxed{2.6\ \mathrm{cm^3\,s^{-1}}}
\]

**(c) Explain why ΔH cannot determine the rate.**

The calorimeter measures the heat transferred during the reaction. It gives information about the overall energy change per mole.

It does not measure how much product forms in a given time.

Reaction rate requires a measurement of change over time, such as gas volume produced per second. The activation barrier also cannot be calculated from ΔH alone.

**(d) Explain why the final hydrogen volumes can be equal.**

The final amount of hydrogen depends on the amount of limiting reactant that reacts, not on how quickly it reacts.

From the balanced equation:

\[
\ce{1Zn -> 1H2}
\]

One mole of zinc produces one mole of hydrogen when the reaction goes to completion.

If both experiments begin with the same amount of zinc and hydrochloric acid is in excess, they can produce the same final amount of hydrogen.

One reaction may simply take longer to reach completion.

This is another important distinction: reaction rate describes how quickly products form, while the final product amount depends on the reaction's extent and the available reactants.

### Question 4

A reversible reaction proceeds through two elementary steps:

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

Its illustrative energy profile has the following relative enthalpy levels:

| Position | Enthalpy (kJ mol⁻¹) |
|---|---:|
| Reactant A | 0 |
| First transition state | +48 |
| Intermediate B | +38 |
| Second transition state | +66 |
| Product C | −12 |

Assume each step has one transition state, and that the elementary steps have comparable kinetic factors and intermediate concentrations for the purpose of identifying the slower step from their activation barriers.

[IMAGE: A two-peak reaction energy profile showing A at 0, first transition state at +48, intermediate B at +38, second transition state at +66, and product C at −12 kJ mol−1. Label A, B, C, TS1 and TS2, along with the reaction-progress and relative-enthalpy axes. Do not draw activation-energy arrows or identify the rate-determining step.]

(a) Calculate the overall enthalpy change.

(b) Calculate the forward activation energy for each elementary step.

(c) Calculate the activation energy for each elementary step in the reverse direction.

(d) Identify the slower elementary step in each direction under the stated assumptions.

(e) Explain why choosing the highest transition-state energy does not necessarily identify the slowest forward step.

### Solution 4

The overall reaction is exothermic, with ΔH = −12 kJ mol⁻¹. The first forward step has the larger activation barrier, while the first reverse step has the larger reverse barrier.

**(a) Calculate the overall enthalpy change.**

The overall reaction begins at A and finishes at C:

\[
\begin{aligned}
\Delta H&=H_C-H_A\\
&=-12-0\\
&=\boxed{-12\ \mathrm{kJ\,mol^{-1}}}
\end{aligned}
\]

**(b) Calculate the forward activation energies.**

For the first step, A to B:

\[
E_{a,1}=48-0
\]

\[
\boxed{E_{a,1}=48\ \mathrm{kJ\,mol^{-1}}}
\]

For the second step, B to C:

\[
E_{a,2}=66-38
\]

\[
\boxed{E_{a,2}=28\ \mathrm{kJ\,mol^{-1}}}
\]

The first forward step has the larger activation barrier.

**(c) Calculate the reverse activation energies.**

The reverse reaction begins at C.

For the first reverse step, C to B:

\[
E_{a,\mathrm{reverse},1}=66-(-12)
\]

\[
\boxed{E_{a,\mathrm{reverse},1}=78\ \mathrm{kJ\,mol^{-1}}}
\]

For the second reverse step, B to A:

\[
E_{a,\mathrm{reverse},2}=48-38
\]

\[
\boxed{E_{a,\mathrm{reverse},2}=10\ \mathrm{kJ\,mol^{-1}}}
\]

**(d) Identify the slower elementary step.**

In the forward direction, the first step has the larger barrier:

\[
48>28
\]

Therefore, the first forward step is the slower step under the stated assumptions.

In the reverse direction, the first step, C to B, has the larger barrier:

\[
78>10
\]

Therefore, C to B is the slower reverse step under the stated assumptions.

**(e) Explain why the highest peak can be misleading.**

The second transition state has the highest absolute enthalpy at +66 kJ mol⁻¹.

A tempting approach is to identify it as the slowest forward step simply because it is the highest point on the profile.

But activation energy is measured from the starting energy level of each elementary step.

The second step begins at intermediate B, which already has an enthalpy of +38 kJ mol⁻¹.

Its activation barrier is therefore only:

\[
66-38=28\ \mathrm{kJ\,mol^{-1}}
\]

The first step starts at zero and has a barrier of 48 kJ mol⁻¹.

The correct comparison uses local activation barriers, not just the absolute heights of the peaks.

In a real multistep mechanism, intermediate concentrations and the reaction mechanism can also affect the overall rate. The comparison here uses the simplifying assumptions supplied in the question.

### Question 5

A reversible, single-step reaction is represented by:

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

Its forward reaction has:

\[
\Delta H=-24\ \mathrm{kJ\,mol^{-1}}
\]

\[
E_{a,\mathrm{forward}}=36\ \mathrm{kJ\,mol^{-1}}
\]

At 25 °C, an equilibrium mixture contains:

\[
[A]=0.20\ \mathrm{mol\,L^{-1}}
\]

\[
[B]=0.80\ \mathrm{mol\,L^{-1}}
\]

A separate kinetic experiment at the same temperature measures the initial rates in solutions containing equal concentrations of A and B, each at 1.0 mol L⁻¹:

| Direction | Initial rate |
|---|---:|
| Forward | 6.0 arbitrary rate units s⁻¹ |
| Reverse | 1.5 arbitrary rate units s⁻¹ |

For this question, assume that each reaction direction's rate is directly proportional to the concentration of its own reactant. The rate measurements use the same temperature and reaction mechanism as the equilibrium mixture.

A student makes the following claims:

1. The reverse activation energy must be smaller because the forward reaction is exothermic.
2. The forward reaction must always be faster because its activation energy is smaller.
3. Adding more A makes the reaction more exothermic.
4. Adding a catalyst changes the equilibrium composition because it changes the activation energies.

Evaluate all four claims.

Then consider what happens when 0.60 mol L⁻¹ of A is added to the equilibrium mixture at constant volume.

Calculate the ratio of the initial forward and reverse rates immediately after the addition. Explain what happens next, and distinguish the effects of adding reactant from the effects of adding a catalyst.

[IMAGE: An annotated problem setup showing a reversible single-step energy profile with A at 0 kJ mol−1, B at −24 kJ mol−1 and the transition state at +36 kJ mol−1. Beside it, show two simple concentration comparisons: the original equilibrium mixture with A at 0.20 M and B at 0.80 M, and the mixture immediately after adding A with both at 0.80 M. Label concentrations only; do not show calculated activation energies, rates or the direction of subsequent change.]

### Solution 5

The reverse activation energy is 60 kJ mol⁻¹. Initially, the equilibrium mixture has equal forward and reverse rates despite the different activation energies. After adding A, the forward rate becomes four times the reverse rate. Neither the addition of A nor the addition of a catalyst changes the reaction enthalpy.

**Step 1: Calculate the reverse activation energy.**

Use:

\[
E_{a,\mathrm{reverse}}
=E_{a,\mathrm{forward}}-\Delta H
\]

Substitute:

\[
\begin{aligned}
E_{a,\mathrm{reverse}}&=36-(-24)\\
&=36+24\\
&=\boxed{60\ \mathrm{kJ\,mol^{-1}}}
\end{aligned}
\]

The reverse activation energy is larger, not smaller.

The forward reaction is exothermic, so the products have lower enthalpy. The reverse reaction must climb further to reach the same transition state.

This makes claim 1 incorrect.

**Step 2: Explain why the reaction rates are initially equal.**

This is the crucial part of the question.

The reverse activation energy is larger, but the actual reaction rates also depend on the concentrations of A and B.

At equal concentrations of 1.0 mol L⁻¹, the measured rates are:

- Forward: 6.0 units s⁻¹
- Reverse: 1.5 units s⁻¹

The forward rate is four times the reverse rate under those equal-concentration conditions.

However, the equilibrium mixture does not contain equal concentrations.

It contains four times as much B as A.

Because the rates are proportional to their respective reactant concentrations, calculate the rates in the equilibrium mixture.

Forward rate:

\[
r_f=6.0\times\frac{0.20}{1.0}
\]

\[
r_f=1.2\ \mathrm{units\,s^{-1}}
\]

Reverse rate:

\[
r_r=1.5\times\frac{0.80}{1.0}
\]

\[
r_r=1.2\ \mathrm{units\,s^{-1}}
\]

Therefore:

\[
\boxed{r_f=r_r=1.2\ \mathrm{units\,s^{-1}}}
\]

The system is at equilibrium because the forward and reverse rates are equal.

This does not mean the reactions have stopped. Both directions continue to occur, but there is no net change in the concentrations.

Claim 2 is incorrect because a smaller activation energy does not guarantee a faster observed reaction under every set of conditions.

The concentrations also matter.

**Step 3: Calculate the effect of adding A.**

Initially:

\[
[A]=0.20\ \mathrm{mol\,L^{-1}}
\]

\[
[B]=0.80\ \mathrm{mol\,L^{-1}}
\]

Adding 0.60 mol L⁻¹ of A at constant volume gives:

\[
[A]_{\mathrm{new}}=0.20+0.60=0.80\ \mathrm{mol\,L^{-1}}
\]

The concentration of B remains initially unchanged:

\[
[B]_{\mathrm{new}}=0.80\ \mathrm{mol\,L^{-1}}
\]

The system has been disturbed, so it is no longer at equilibrium.

Calculate the new initial forward rate:

\[
r_f=6.0\times\frac{0.80}{1.0}
\]

\[
r_f=4.8\ \mathrm{units\,s^{-1}}
\]

Calculate the new initial reverse rate:

\[
r_r=1.5\times\frac{0.80}{1.0}
\]

\[
r_r=1.2\ \mathrm{units\,s^{-1}}
\]

The ratio is:

\[
\begin{aligned}
\frac{r_f}{r_r}&=\frac{4.8}{1.2}\\
&=\boxed{4:1}
\end{aligned}
\]

The forward reaction initially proceeds four times as fast as the reverse reaction.

There is now a net conversion of A into B.

As B forms and A is consumed, the rates change until a new equilibrium is established.

**Step 4: Evaluate the enthalpy claim.**

Adding A does not change ΔH.

Reaction enthalpy is a property of the chemical reaction between specified reactants and products in specified physical states.

Changing the amount or concentration of a reactant changes the reaction conditions, not the enthalpy change per mole for the same reaction under the same reference conditions.

Claim 3 is incorrect.

**Step 5: Evaluate the catalyst claim.**

A catalyst provides an alternative pathway with a lower activation barrier.

For a reversible reaction, it increases the rates of both forward and reverse reactions.

It does not change:

- The enthalpy change, ΔH.
- The equilibrium constant.
- The final equilibrium composition under the same conditions.

The catalyst allows the system to reach equilibrium more quickly, but it does not move the equilibrium position.

Claim 4 is incorrect.

The important distinction is that adding A changes the mixture's composition and produces a net forward reaction, whereas adding a catalyst increases the rates of both directions without changing the equilibrium position.

This question combines the ideas that ΔH, activation energy, concentration, and reaction rate describe different aspects of a chemical system. None can simply be substituted for another.

## Where to go next

Once you can distinguish the overall energy change from the activation barrier, the next useful step is explaining why reaction rates change when temperature, concentration, surface area, or catalysts change.

The key is to connect each factor to the frequency of successful collisions, rather than assuming that a more exothermic reaction must be faster.