Why Forward and Reverse Reaction Rates Become Equal

Imagine sealing a colourless sample of dinitrogen tetroxide, \(\ce{N2O4}\), inside a transparent flask. Over time, the gas turns brown as nitrogen dioxide, \(\ce{NO2}\), forms. Eventually, the colour stops changing.

Has the reaction stopped?

No. At dynamic equilibrium, the forward and reverse reactions are still occurring, but they happen at exactly the same rate. Every unit of reaction progress in the forward direction is matched by an equal unit in the reverse direction. As a result, the concentrations remain constant even though particles continue reacting.

The key is understanding what the system is doing in both directions, rather than assuming that an unchanging concentration means nothing is happening.

01Why do the forward and reverse rates become equal?

Consider the reversible reaction:

\[ \ce{N2O4(g) <=> 2NO2(g)} \]

The forward reaction breaks one \(\ce{N2O4}\) molecule into two \(\ce{NO2}\) molecules. The reverse reaction combines two \(\ce{NO2}\) molecules to reform \(\ce{N2O4}\).

Suppose we begin with only \(\ce{N2O4}\).

At the beginning, there are plenty of \(\ce{N2O4}\) molecules available to decompose, so the forward reaction occurs rapidly. There is initially almost no \(\ce{NO2}\), so there are very few opportunities for the reverse reaction.

As time passes, two changes occur:

• The concentration of \(\ce{N2O4}\) decreases, making forward reactions less frequent.
• The concentration of \(\ce{NO2}\) increases, making reverse reactions more frequent.

The forward reaction slows down while the reverse reaction speeds up. Eventually, their rates become equal.

This is the physical reason the rates become equal: the changing concentrations alter how frequently the particles can undergo each reaction until neither direction produces a net change in composition.

There is an important distinction here. The rates do not become equal because the concentrations of reactants and products become equal. They become equal because the changing concentrations eventually produce matching forward and reverse reaction rates.

The particle-level reason the rates change

Look more closely at the two directions.

For the forward reaction, a \(\ce{N2O4}\) molecule must decompose. For the reverse reaction, two \(\ce{NO2}\) molecules must encounter one another and combine.

This means the two reactions depend on different particle populations.

At constant temperature and volume, decreasing the concentration of \(\ce{N2O4}\) reduces the number of molecules available to decompose. Increasing the concentration of \(\ce{NO2}\) increases the frequency of encounters that can produce \(\ce{N2O4}\).

Eventually, the two processes balance.

For a simplified model in which both directions follow the elementary steps described above, their rates can be represented by:

\[ r_f=k_f[\ce{N2O4}] \]

\[ r_r=k_r[\ce{NO2}]^2 \]

Here:

• \(r_f\) is the forward reaction rate.
• \(r_r\) is the reverse reaction rate.
• \(k_f\) and \(k_r\) are the forward and reverse rate constants at a particular temperature.
• Square brackets represent molar concentration in \(\mathrm{mol\,L^{-1}}\).

The squared concentration in the reverse expression reflects the requirement for two \(\ce{NO2}\) molecules to encounter one another in this simplified elementary-step model.

At equilibrium:

\[ k_f[\ce{N2O4}] = k_r[\ce{NO2}]^2 \]

Notice that the concentrations do not need to be equal. The rate constants and concentrations combine to produce equal reaction rates.

This is a model for this particular reaction mechanism, not a rule that you can automatically derive every reaction’s rate law from its balanced equation.

The important feature is that the rates meet above zero. They do not gradually decrease to zero.

02How can an experiment show that the rates have become equal?

There is a measurement problem here.

We can measure concentrations, colour, and pressure. We cannot simply look inside a flask and count how many forward and reverse reactions happen every second.

So how could we investigate whether the two rates have become equal?

Consider an experiment using the \(\ce{N2O4}/\ce{NO2}\) equilibrium.

Setting up the experiment

Imagine a sealed optical cell containing \(\ce{N2O4}\) gas.

The cell is placed inside a temperature-controlled enclosure. A beam of light passes through the cell and reaches a detector on the opposite side.

This arrangement allows us to monitor the amount of \(\ce{NO2}\) present.

Why \(\ce{NO2}\)? Because it is brown and absorbs visible light, whereas \(\ce{N2O4}\) is colourless. As more \(\ce{NO2}\) forms, the absorbance increases.

The apparatus measures absorbance, which can be related to concentration using the Beer-Lambert relationship:

\[ A=\varepsilon lc \]

Here:

• \(A\) is absorbance, which has no units.
• \(\varepsilon\) is the molar absorption coefficient.
• \(l\) is the light path length.
• \(c\) is the concentration of the absorbing species.

For this investigation, assume the wavelength, path length, and temperature remain constant. Absorbance is therefore proportional to the concentration of \(\ce{NO2}\).

The cell is sealed, so no reacting matter can enter or leave. This is essential: we want to investigate the reaction itself, not the effects of gases escaping.

What should we observe?

Initially, the cell contains \(\ce{N2O4}\) and almost no \(\ce{NO2}\).

We predict that absorbance will increase as the forward reaction produces more brown \(\ce{NO2}\).

As the reverse reaction becomes more frequent, the increase in absorbance should slow down. Eventually, absorbance should become constant.

Consider the following illustrative measurements.

ILLUSTRATIVE DATA – NOT EXPERIMENTAL MEASUREMENTS

The increase becomes progressively smaller, and eventually the absorbance stops changing within the measurement precision.

This is consistent with the system approaching equilibrium.

However, we need to be careful about what the experiment actually establishes.

A constant absorbance tells us that the concentration of \(\ce{NO2}\) is no longer changing measurably. It does not independently tell us the forward and reverse rates.

That distinction leads to a useful calculation.

03Why the system boundary matters

There is another condition we need to check before concluding that equal forward and reverse rates explain constant concentrations.

The reaction must be considered within a suitable system boundary.

A closed system can exchange energy with its surroundings, but it cannot exchange matter. In our experiment, the gas cell is sealed, so neither \(\ce{N2O4}\) nor \(\ce{NO2}\) can escape.

This matters because an open system can maintain constant concentrations without reaching chemical equilibrium.

Consider what would happen if our gas cell developed a small leak.

As the forward reaction produces \(\ce{NO2}\), some of that gas escapes. The reverse reaction now has fewer \(\ce{NO2}\) molecules available to reform \(\ce{N2O4}\).

It is possible for the amount of \(\ce{NO2}\) inside the cell to remain constant because its rate of production matches its rate of escape.

But that does not mean the forward and reverse reaction rates are equal.

For a simple reaction:

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

the rate of change of B in a system with an outlet and no B inlet can be written as:

\[ \frac{d[B]}{dt} = r_f-r_r-r_{\mathrm{out}} \]

Here, \(r_{\mathrm{out}}\) is the rate at which B leaves the system, expressed as a concentration change per second.

In a sealed vessel:

\[ r_{\mathrm{out}}=0 \]

Therefore, if the concentration of B is constant:

\[ r_f-r_r=0 \]

So:

\[ r_f=r_r \]

In an open vessel, however, a constant concentration only tells us:

\[ r_f-r_r=r_{\mathrm{out}} \]

The forward and reverse rates can be different.

This is the difference between chemical equilibrium and a steady state. A steady state means measurable properties remain constant. It does not necessarily mean the forward and reverse chemical reactions are occurring at equal rates.

04Why a flat concentration graph does not show stopped reactions

Let’s bring the two types of measurement together.

A concentration-time graph shows the overall change in concentration. Its gradient represents the net rate of formation or consumption of a species.

A reaction-rate graph can show the forward and reverse rates separately.

These are not the same thing.

For the reaction:

\[ \ce{N2O4(g) <=> 2NO2(g)} \]

the rate of change of \(\ce{NO2}\) is:

\[ \frac{d[\ce{NO2}]}{dt} = 2(r_f-r_r) \]

At equilibrium:

\[ r_f=r_r \]

Therefore:

\[ \frac{d[\ce{NO2}]}{dt}=0 \]

The concentration graph becomes horizontal because the net rate is zero.

The individual reaction rates, however, remain positive.

This is a common source of confusion in HSC responses.

If you write that the reaction rate is zero at equilibrium, you are mixing up the net reaction rate with the individual forward and reverse rates.

A better explanation is:

At dynamic equilibrium, the forward and reverse reaction rates are equal and non-zero. Therefore, the net rate of reaction is zero, and the concentrations of reactants and products remain constant.

For more practice interpreting the shape and gradient of these curves, see How to Interpret a Concentration-Time Graph Approaching Equilibrium.

05How could we tell that reactions are still happening?

Our colourimeter experiment has established that the macroscopic composition becomes constant.

But there is a limitation: the instrument measures the total amount of \(\ce{NO2}\), not the identity of individual molecules.

Imagine two possibilities:

• The reaction has genuinely stopped.
• The forward and reverse reactions are continuing at equal rates.

Both could produce an unchanged absorbance reading.

How could we distinguish them?

One conceptual approach is to use an isotopic tracer.

Following labelled molecules

Imagine preparing a sealed vessel that has already reached equilibrium.

We replace a very small proportion of the ordinary \(\ce{NO2}\) with isotopically labelled \(\ce{NO2}\), using nitrogen-15, written as \({}^{15}\ce{NO2}\). The total concentration of \(\ce{NO2}\) is kept effectively unchanged.

The labelled molecules behave chemically almost identically to the ordinary molecules. We assume the small isotope effect does not measurably disturb the equilibrium.

We then use an analytical method capable of distinguishing labelled and unlabelled molecules.

What should happen if the equilibrium is genuinely dynamic?

The labelled \(\ce{NO2}\) can combine with another \(\ce{NO2}\) molecule to form an isotopically labelled \(\ce{N2O4}\) molecule.

That labelled \(\ce{N2O4}\) can subsequently decompose, releasing labelled \(\ce{NO2}\) again.

The isotope label moves between chemical species even though the total concentrations of \(\ce{N2O4}\) and \(\ce{NO2}\) remain constant.

This would provide evidence of continuing molecular conversion.

The first experiment measured the overall composition. The tracer approach investigates molecular exchange itself.

This is a conceptual extension of the measurement problem, rather than a required school-laboratory procedure. Its value is understanding why constant macroscopic measurements do not mean that microscopic chemical changes have stopped.

06A reliable way to decide whether rates are equal

When you encounter a question about dynamic equilibrium, work through these checks.

The final two points are particularly important.

A reaction that has stopped has zero forward and reverse rates. A dynamic equilibrium has equal, non-zero forward and reverse rates.

For a broader comparison of these two situations, see Static vs Dynamic Equilibrium in HSC Chemistry.

07Questions and solutions

Question 1

A sealed vessel contains the reversible reaction:

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

At constant temperature, the forward and reverse reaction rates are both:

\[ 4.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The concentrations are:

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

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

Calculate the net reaction rate and explain whether the system can be at dynamic equilibrium despite the unequal concentrations.

Solution 1

The net reaction rate is zero, and the system is at dynamic equilibrium under the stated conditions.

The net reaction rate is the difference between the forward and reverse rates:

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

Substituting:

\[ r_{\mathrm{net}} = 4.0\times10^{-3} - 4.0\times10^{-3} \]

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

The forward and reverse reactions are continuing at equal, non-zero rates. Therefore, A is converted into B at the same rate that B is converted back into A.

The concentrations remain constant.

There is no requirement for the concentrations of A and B to be equal. The equilibrium condition concerns reaction rates, not the relative amounts of reactants and products.

Question 2

A sealed vessel contains the reversible reaction:

\[ \ce{2A(g) <=> B(g)} \]

Over a period of 20.0 seconds, the following concentration changes are measured:

(a) Calculate the average net reaction rate over this interval.

(b) Verify that the concentration changes are consistent with the balanced equation.

(c) Determine whether the system is at dynamic equilibrium during this interval. Explain your reasoning.

Solution 2

The average net reaction rate is \(2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}\) in the forward direction, so the system is not at equilibrium.

Step 1: Calculate the change in concentration of A.

\[ \Delta[A]=0.72-0.80 \]

\[ =-0.08\ \mathrm{mol\,L^{-1}} \]

Step 2: Calculate the net reaction rate using A.

The balanced equation shows that two moles of A are consumed for every mole of B produced.

Therefore:

\[ r_{\mathrm{net}} = -\frac{1}{2} \frac{\Delta[A]}{\Delta t} \]

Substituting:

\[ r_{\mathrm{net}} = -\frac{1}{2} \left( \frac{-0.08}{20.0} \right) \]

\[ \boxed{ r_{\mathrm{net}} =2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

Step 3: Verify using the change in B.

\[ \Delta[B]=0.14-0.10 \]

\[ =0.04\ \mathrm{mol\,L^{-1}} \]

The rate calculated using B is:

\[ r_{\mathrm{net}} = \frac{\Delta[B]}{\Delta t} \]

\[ = \frac{0.04}{20.0} \]

\[ =2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Both calculations agree, confirming that the measured changes are consistent with the reaction stoichiometry.

Step 4: Determine whether equilibrium has been reached.

The net reaction rate is positive in the forward direction.

A is decreasing while B is increasing. Therefore, the forward reaction is occurring faster than the reverse reaction.

The system has not reached dynamic equilibrium during this interval.

Question 3

A sealed optical cell contains a mixture of \(\ce{N2O4}\) and \(\ce{NO2}\) at constant temperature.

A colourimeter records an unchanged absorbance over a period of 15 minutes.

A student concludes:

“The reaction has stopped because the colour is no longer changing. Both the forward and reverse reaction rates must be zero.”

Evaluate this conclusion.

Describe how an isotopic tracer could provide evidence that the forward and reverse reactions are continuing, even though the total concentrations remain constant.

State one assumption required for the tracer experiment.

Solution 3

The student’s conclusion is incorrect. Constant absorbance does not mean that both reaction rates are zero.

The colourimeter measures the absorbance associated with \(\ce{NO2}\). If the absorbance remains constant, the concentration of \(\ce{NO2}\) is not changing measurably.

However, this could occur because the forward and reverse reactions are proceeding at equal rates.

To investigate whether molecular conversion is continuing, a small proportion of ordinary \(\ce{NO2}\) could be replaced with nitrogen-15-labelled \(\ce{NO2}\).

The labelled molecules can be distinguished from ordinary molecules using an appropriate analytical technique.

If the forward and reverse reactions continue, the labelled nitrogen can move between \(\ce{NO2}\) and \(\ce{N2O4}\):

• Labelled \(\ce{NO2}\) combines with another \(\ce{NO2}\) molecule to form labelled \(\ce{N2O4}\).
• Labelled \(\ce{N2O4}\) subsequently decomposes, releasing labelled \(\ce{NO2}\).

This demonstrates ongoing chemical conversion.

The total concentrations can remain constant because the forward and reverse reactions occur at equal rates. The identities of individual molecules can change without producing a net change in the overall composition.

An important assumption is that introducing the tracer does not significantly alter the equilibrium composition or reaction rates. This can be approximated by using a sufficiently small tracer quantity and assuming negligible isotope effects.

Question 4

A sealed vessel containing the following reaction has reached dynamic equilibrium:

\[ \ce{H2(g) + I2(g) <=> 2HI(g)} \]

A student adds a catalyst while keeping the temperature and volume constant.

The student predicts that:

• The forward reaction rate increases.
• The reverse reaction rate remains unchanged.
• The equilibrium shifts towards HI.

Explain why this prediction is incorrect. Describe what happens to the forward and reverse rates and the equilibrium concentrations.

Solution 4

The catalyst increases the rates of both the forward and reverse reactions, but it does not shift the equilibrium or change the equilibrium concentrations.

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

This allows a greater proportion of reacting particles to undergo successful reactions per unit time.

Importantly, the catalyst affects both directions of the reversible reaction.

Before the catalyst is added:

\[ r_f=r_r \]

After the catalyst is added, both reaction rates increase. The forward and reverse rates remain equal because the system was already at equilibrium and the catalyst does not change the equilibrium composition.

Therefore:

\[ r_f=r_r>0 \]

The concentrations of \(\ce{H2}\), \(\ce{I2}\), and \(\ce{HI}\) remain unchanged.

The student’s mistake is assuming that a catalyst selectively increases the forward reaction rate. It speeds up the process of reaching equilibrium, not the formation of products at the expense of reactants.

The system simply undergoes more rapid molecular exchange while maintaining the same overall composition.

Question 5

A constant-volume reactor contains the reversible gaseous reaction:

\[ \ce{A(g) <=> 2B(g)} \]

The reactor operates at constant temperature. A selective inlet supplies A, while a selective outlet removes B. The external flows are adjusted to keep the reactor volume constant.

Assume there are no other reactions, and all rates below are expressed per litre of reactor volume per second.

At a particular moment, the measured internal reaction rates are:

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

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

The external streams contribute:

• A enters at \(0.015\ \mathrm{mol\,L^{-1}\,s^{-1}}\).
• B leaves at \(0.030\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

The concentrations of both A and B are observed to remain constant.

(a) Use the reaction stoichiometry to verify whether the external flows can maintain constant concentrations of both A and B.

(b) Is the reactor at dynamic equilibrium? Explain why or why not.

(c) Both external streams are suddenly shut off. Assuming the reaction rates do not change instantaneously when the streams stop, calculate the initial rates of concentration change of A and B. Identify the initial direction of the net reaction.

Solution 5

The reactor is at a steady state, not dynamic equilibrium. When the external flows stop, the reaction initially proceeds in the forward direction, consuming A and producing B.

This question requires us to account for both the chemical reaction and the movement of matter across the system boundary.

Step 1: Calculate the net reaction rate inside the reactor.

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

\[ =0.040-0.025 \]

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

The reaction is proceeding in the forward direction.

Step 2: Calculate the net change in A due to the chemical reaction.

The forward reaction consumes one mole of A per reaction event, while the reverse reaction produces one mole of A.

Therefore:

\[ \left(\frac{d[A]}{dt}\right)_{\mathrm{reaction}} =-0.015\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The external supply of A contributes:

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

Combining the two:

\[ \frac{d[A]}{dt} =-0.015+0.015 \]

\[ \boxed{\frac{d[A]}{dt}=0} \]

The concentration of A remains constant.

Step 3: Calculate the net change in B due to the chemical reaction.

This is where the coefficient of 2 matters.

The forward reaction produces two moles of B per reaction event, while the reverse reaction consumes two moles.

Therefore:

\[ \left(\frac{d[B]}{dt}\right)_{\mathrm{reaction}} =2(r_f-r_r) \]

\[ =2(0.015) \]

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

The outlet removes B at exactly this rate:

\[ \frac{d[B]}{dt} =0.030-0.030 \]

\[ \boxed{\frac{d[B]}{dt}=0} \]

The concentration of B also remains constant.

The external flows are therefore consistent with the observed constant concentrations.

Step 4: Determine whether the reactor is at dynamic equilibrium.

It is not.

Although both concentrations remain constant:

\[ r_f=0.040 \]

\[ r_r=0.025 \]

The rates are unequal.

The forward reaction is producing B faster than the reverse reaction consumes it. The excess B is simply being removed through the outlet.

The system is at a steady state because external flows balance the net chemical changes.

Step 5: Consider what happens when both external streams stop.

The concentrations cannot change instantaneously simply because the valves close. Therefore, the reaction rates initially retain their previous values.

However, the external contributions are now zero.

For A:

\[ \frac{d[A]}{dt} =-r_f+r_r \]

\[ =-0.040+0.025 \]

\[ \boxed{ \frac{d[A]}{dt} =-0.015\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

For B:

\[ \frac{d[B]}{dt} =2r_f-2r_r \]

\[ =2(0.040)-2(0.025) \]

\[ \boxed{ \frac{d[B]}{dt} =+0.030\ \mathrm{mol\,L^{-1}\,s^{-1}} } \]

A begins decreasing, while B begins increasing.

The net reaction is initially forward.

As the concentrations change, the forward and reverse rates will also change. If the sealed system is maintained at constant temperature and no other processes occur, it can subsequently approach dynamic equilibrium when the two rates become equal.

The key mistake would be to assume that constant concentrations always mean equal reaction rates. They only mean that the total production and consumption of each species balance. You must also account for anything entering or leaving the system.

08Where to go next

You now have the distinction needed to understand equilibrium properly: equal forward and reverse rates produce zero net chemical change, not zero chemical activity.

The next useful question is what happens when the system is disturbed.

If extra reactant is added, the volume changes, or the temperature increases, the forward and reverse rates may no longer be equal. The system then changes composition until a new equilibrium is established.

That is the reasoning behind Le Chatelier’s principle. In particular, understanding the difference between an immediate concentration change and the subsequent equilibrium shift makes it much easier to predict what happens after a disturbance.

See Instantaneous Concentration Changes vs Equilibrium Shifts.