How to Interpret a Concentration-Time Graph Approaching Equilibrium

A sealed container starts with colourless dinitrogen tetroxide gas, \(\ce{N2O4}\). Brown nitrogen dioxide, \(\ce{NO2}\), gradually forms. If you plot the concentration of each gas against time, one curve falls, the other rises, and eventually both become horizontal.

Has the reaction stopped?

No. If the system is closed, the temperature is constant, and the reaction is reversible, a sustained horizontal section indicates dynamic equilibrium. The forward and reverse reactions are still occurring, but at equal rates. As a result, neither concentration changes overall.

This is the key to interpreting a concentration-time graph: read the gradient to determine the net reaction rate, then use the system boundaries to decide whether a plateau represents equilibrium.

01Start with the shape of the graph

Consider the reversible reaction:

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

Suppose the container initially contains only \(\ce{N2O4}\).

As the reaction proceeds in the forward direction:

• The concentration of \(\ce{N2O4}\) decreases because it is being consumed.
• The concentration of \(\ce{NO2}\) increases because it is being produced.
• Both curves gradually become less steep as the system approaches equilibrium.

Eventually, both curves become horizontal.

There are three important features to recognise.

Notice that the two concentrations do not have to become equal.

In this example, equilibrium is reached with:

\[ [\ce{N2O4}]=0.060\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]=0.080\ \mathrm{mol\,L^{-1}} \]

They are different, but the system can still be at equilibrium.

What does the gradient actually mean?

A concentration-time graph is useful because its gradient represents the rate of change of concentration.

For example:

• A steep upwards gradient means the concentration is increasing quickly.
• A shallow upwards gradient means it is increasing slowly.
• A horizontal line means there is no net change in concentration.

This is similar to interpreting the gradient of a displacement-time graph in Physics. The important difference is that the vertical axis here represents concentration, so the gradient has units of \(\mathrm{mol\,L^{-1}\,s^{-1}}\), rather than velocity.

If you need to refresh the general graph-reading principle, see Displacement-Time Graphs: How to Read Motion in HSC Physics.

However, there is an important limitation.

A horizontal concentration-time graph tells you that the net rate of concentration change is zero. It does not tell you that the individual forward and reverse reaction rates are zero.

To understand why, we need to consider both reactions separately.

02Connect the concentration curves to forward and reverse rates

For the reaction:

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

There are two simultaneous reactions:

• Forward reaction: \(\ce{N2O4}\) decomposes to form \(\ce{NO2}\).
• Reverse reaction: \(\ce{NO2}\) combines to form \(\ce{N2O4}\).

Let:

• \(r_f\) = forward reaction rate.
• \(r_r\) = reverse reaction rate.
• \(r_{\text{net}}\) = net forward reaction rate.

Using this sign convention:

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

If the forward reaction is faster, the net reaction proceeds forwards. If the reverse reaction is faster, the net reaction proceeds backwards.

What happens to the rates as equilibrium approaches?

Initially, the container contains only \(\ce{N2O4}\).

Therefore:

• The forward reaction can occur.
• The reverse reaction cannot occur initially because there is no \(\ce{NO2}\) available to react.

As time passes, \(\ce{N2O4}\) is consumed and \(\ce{NO2}\) accumulates.

In the usual concentration-dependent kinetic model, the forward rate decreases as the reactant concentration falls, while the reverse rate increases as the product concentration rises.

Eventually, the two rates become equal.

At equilibrium:

\[ \boxed{r_f=r_r} \]

Therefore:

\[ r_{\text{net}}=0 \]

This explains why the concentration-time curves become horizontal.

The forward reaction produces \(\ce{NO2}\) at exactly the same rate that the reverse reaction consumes it. Likewise, the reverse reaction produces \(\ce{N2O4}\) at exactly the same rate that the forward reaction consumes it.

There is no overall change in concentration.

The important distinction: the reaction rates are equal, not zero.

Why do the two concentration gradients have different magnitudes?

Look again at the balanced equation:

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

For every mole of \(\ce{N2O4}\) consumed, two moles of \(\ce{NO2}\) are produced.

This means the concentration changes are related by:

\[ \frac{d[\ce{N2O4}]}{dt}=-r_{\text{net}} \]

\[ \frac{d[\ce{NO2}]}{dt}=2r_{\text{net}} \]

The factor of two comes directly from the balanced equation.

For example, if the net forward reaction rate is \(0.004\ \mathrm{mol\,L^{-1}\,s^{-1}}\), then:

• \(\ce{N2O4}\) decreases at \(0.004\ \mathrm{mol\,L^{-1}\,s^{-1}}\).
• \(\ce{NO2}\) increases at \(0.008\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

Both gradients become zero at equilibrium.

Be careful: the gradient of a concentration-time curve represents the net rate of concentration change, not the forward reaction rate on its own.

A concentration-time graph cannot independently tell you the individual forward and reverse rates when both reactions are occurring.

03How an experiment produces a concentration-time graph

Understanding how the graph is measured makes its limitations much clearer.

Consider the question an experimenter wants to answer:

How can we determine whether a reversible reaction has reached equilibrium without disturbing the reaction mixture?

One approach is to monitor the concentration of a coloured gas using light absorption.

Setting up the experiment

For the \(\ce{N2O4}/\ce{NO2}\) system:

• \(\ce{N2O4}\) is colourless.
• \(\ce{NO2}\) is brown and absorbs visible light.

A sealed gas cell is placed inside a temperature-controlled jacket. A light source and wavelength selector sit on one side of the cell. Light passes through the gas mixture and reaches a detector on the opposite side.

The detector records how much light is absorbed over time.

The measurement uses the Beer-Lambert relationship:

\[ A=\varepsilon \ell c \]

Where:

• \(A\) = absorbance.
• \(\varepsilon\) = molar absorptivity.
• \(\ell\) = light path length through the gas.
• \(c\) = concentration of the absorbing species.

At a fixed wavelength and path length, absorbance is proportional to the concentration of \(\ce{NO2}\).

A calibration allows the absorbance measurements to be converted into concentrations.

Because the cell remains sealed, the reaction mixture can be monitored without removing gas samples.

Why does the system need to be closed?

Imagine using a different method.

You could remove a small gas sample every ten seconds and analyse its composition. However, removing gas changes the amount of material and pressure inside the container.

That changes the conditions of the reaction you are trying to observe.

The measurement itself could disturb the equilibrium.

Monitoring absorbance through the sealed cell avoids this problem.

The system boundary is the gas contained inside the cell:

• Matter cannot cross the boundary.
• Light can enter and leave.
• Heat can be exchanged with the temperature-controlled jacket.

This is a closed system, not an isolated system.

Keeping the temperature constant is particularly important because the equilibrium composition of \(\ce{N2O4}\) and \(\ce{NO2}\) depends on temperature.

Converting the measurement into both concentrations

The detector directly measures the concentration of \(\ce{NO2}\), but we can calculate the concentration of \(\ce{N2O4}\) using conservation of matter.

Suppose the cell initially contains:

\[ [\ce{N2O4}]_0=0.100\ \mathrm{mol\,L^{-1}} \]

and no \(\ce{NO2}\).

From the balanced equation, one mole of \(\ce{N2O4}\) produces two moles of \(\ce{NO2}\).

Therefore:

\[ [\ce{N2O4}]+\frac{[\ce{NO2}]}{2}=0.100 \]

This relationship holds throughout the reaction, provided the system remains closed, its volume stays constant, and no other reactions occur.

Rearranging:

\[ [\ce{N2O4}]=0.100-\frac{[\ce{NO2}]}{2} \]

We can now calculate both concentrations from the measured \(\ce{NO2}\) concentration.

The following values are illustrative experimental data.

Notice that the concentrations change rapidly at first, but their changes become progressively smaller.

Between 160 and 240 seconds, neither concentration changes at the displayed precision.

This provides experimental evidence that the system has reached equilibrium.

In a real experiment, measurements have finite precision. You should therefore look for a sustained plateau across several readings rather than treating one repeated measurement as definitive proof.

04A reliable method for interpreting any concentration-time graph

When you encounter an unfamiliar graph in an HSC Chemistry question, use the following sequence.

Step 1: Identify the system boundary.

Is the reaction occurring in a closed system?

Check whether matter can enter or leave. Also check whether the temperature or volume changes during the experiment.

A flat concentration-time graph in an open system does not necessarily indicate equilibrium.

Step 2: Identify which concentrations are increasing and decreasing.

An increasing concentration indicates net formation of that species.

A decreasing concentration indicates net consumption.

Use the balanced equation to identify the direction of the net reaction.

Step 3: Examine the gradients.

Look at how steep each curve is.

• Steep gradients indicate rapid net concentration changes.
• Shallower gradients indicate slower net concentration changes.
• Horizontal sections indicate zero net concentration change.

Step 4: Look for a sustained plateau.

For a reversible reaction in a closed system at constant temperature, a sustained plateau in all relevant species indicates equilibrium.

Step 5: State what happens to the reaction rates.

At dynamic equilibrium:

\[ \boxed{r_f=r_r} \]

The individual rates are equal and non-zero.

Do not write that the reaction has stopped.

Common mistakes to avoid

05Worked examples

06The important exception: A flat graph does not always mean equilibrium

Here is a situation that exposes a common misconception.

Imagine a reaction occurring in a continuously stirred tank. Reactant A is continuously pumped into the tank, and a mixture containing A and product B continuously flows out.

The reaction is reversible:

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

Suppose the concentration of A and B inside the tank remains constant.

Does this mean the system has reached chemical equilibrium?

Not necessarily.

A constant concentration could result from two competing processes:

1. The chemical reaction converts A into B.
2. The continuous flow supplies and removes material at rates that balance the chemical changes.

The concentrations can remain constant even when the forward and reverse reaction rates are unequal.

This situation is called a steady state.

The distinction is:

This is why you must check the system boundary before interpreting a plateau.

A concentration-time graph shows what happens to concentrations. The boundary conditions help explain why.

07Questions and solutions

Question 1

A reversible reaction occurs in a sealed container at constant temperature:

\[ \ce{P <=> Q} \]

The following data are collected.

a) By what time has equilibrium been reached?

b) Explain what happens to the forward and reverse reaction rates.

c) Must the concentrations of P and Q be equal at equilibrium?

Solution 1

Equilibrium has been reached by 60 seconds, and the concentrations of P and Q do not need to be equal.

a) Between 60 and 80 seconds, neither concentration changes.

This shows that the graph has reached a sustained plateau. The measurements cannot establish the exact instant equilibrium was first reached, but it was reached by 60 seconds.

b) At equilibrium:

\[ r_f=r_r \]

The forward and reverse reactions continue at equal, non-zero rates.

The rate at which P is consumed equals the rate at which P is produced. The same applies to Q.

c) No. The equilibrium concentrations are:

\[ [P]=0.52\ \mathrm{mol\,L^{-1}} \]

\[ [Q]=0.48\ \mathrm{mol\,L^{-1}} \]

They are different, but this is completely consistent with dynamic equilibrium.

The defining condition is equal forward and reverse rates, not equal concentrations.

Question 2

Consider the reversible reaction:

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

In a closed container at constant temperature, the concentration of A decreases from \(0.120\ \mathrm{mol\,L^{-1}}\) to \(0.084\ \mathrm{mol\,L^{-1}}\) over 10.0 seconds.

Over the same interval, the concentration of B increases from \(0.040\ \mathrm{mol\,L^{-1}}\) to \(0.058\ \mathrm{mol\,L^{-1}}\).

a) Calculate the average net forward reaction rate.

b) Explain why the graph cannot be used to determine the average forward reaction rate on its own.

Solution 2

The average net forward reaction rate is \(1.8\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}\), but the forward reaction rate alone cannot be determined.

a) Start with the balanced equation:

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

Two moles of A are consumed for every mole of B produced.

The net reaction rate can therefore be calculated using either species:

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

Substitute the values for A:

\[ r_{\text{net}} = -\frac{1}{2} \left( \frac{0.084-0.120}{10.0} \right) \]

\[ = -\frac{1}{2}(-0.0036) \]

\[ \boxed{r_{\text{net}}=0.0018\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

We can check this using B:

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

\[ = \frac{0.058-0.040}{10.0} \]

\[ \boxed{r_{\text{net}}=0.0018\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

Both methods give the same result.

b) The calculated value represents:

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

The concentration-time graph measures the overall change in concentration.

It cannot separate the contribution of the forward reaction from the contribution of the reverse reaction.

Since the net reaction proceeds forwards, the forward rate must be greater than the reverse rate during this interval.

However, their individual values cannot be calculated from the supplied data.

Question 3

A sealed container at constant temperature initially contains only nitrogen dioxide gas at a concentration of:

\[ [\ce{NO2}]_0=0.200\ \mathrm{mol\,L^{-1}} \]

The reversible reaction is:

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

After the concentration-time graph reaches a plateau, the measured concentration of nitrogen dioxide is:

\[ [\ce{NO2}]_{\text{eq}}=0.080\ \mathrm{mol\,L^{-1}} \]

Assume that only \(\ce{N2O4}\) and \(\ce{NO2}\) are present.

a) Calculate the equilibrium concentration of \(\ce{N2O4}\).

b) Identify the net reaction direction during the approach to equilibrium.

c) Explain how the system can reach the same equilibrium composition as a container initially containing \(0.100\ \mathrm{mol\,L^{-1}}\) of \(\ce{N2O4}\), at the same temperature and volume.

Solution 3

The equilibrium concentration of \(\ce{N2O4}\) is \(0.060\ \mathrm{mol\,L^{-1}}\), and the net reaction initially proceeds in the reverse direction.

a) Use conservation of matter.

The balanced equation shows that one mole of \(\ce{N2O4}\) contains the same amount of nitrogen dioxide units as two moles of \(\ce{NO2}\).

Therefore:

\[ [\ce{N2O4}]+\frac{[\ce{NO2}]}{2}=C_0 \]

Initially, only \(\ce{NO2}\) is present.

Thus:

\[ C_0=\frac{0.200}{2} \]

\[ C_0=0.100\ \mathrm{mol\,L^{-1}} \]

At equilibrium:

\[ [\ce{N2O4}]+\frac{0.080}{2}=0.100 \]

\[ [\ce{N2O4}]+0.040=0.100 \]

Therefore:

\[ \boxed{[\ce{N2O4}]_{\text{eq}}=0.060\ \mathrm{mol\,L^{-1}}} \]

b) The net reaction initially proceeds backwards.

The concentration of \(\ce{NO2}\) decreases while the concentration of \(\ce{N2O4}\) increases.

Initially, there is no \(\ce{N2O4}\), so the forward reaction cannot occur. The reverse reaction consumes \(\ce{NO2}\) to form \(\ce{N2O4}\).

As \(\ce{N2O4}\) forms, the forward reaction becomes possible.

Eventually, both reactions occur at equal rates.

c) Both experiments have the same temperature, volume, and total concentration of nitrogen oxide units.

The initial conditions are different, but the equilibrium composition is determined by the equilibrium conditions and the total amount of material available.

Consequently, both systems can reach:

\[ [\ce{N2O4}]_{\text{eq}}=0.060\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]_{\text{eq}}=0.080\ \mathrm{mol\,L^{-1}} \]

This is an important observation: a reversible reaction can approach the same equilibrium state from either direction when the temperature and total composition constraints are identical.

Question 4

A sealed container with a movable piston initially contains an equilibrium mixture of nitrogen dioxide and dinitrogen tetroxide at constant temperature.

The concentrations are:

\[ [\ce{N2O4}]=0.060\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]=0.080\ \mathrm{mol\,L^{-1}} \]

The container is suddenly compressed to half its original volume. The temperature remains constant.

For the reaction:

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

Describe the concentration-time graph for each gas immediately after compression and as the system approaches its new equilibrium.

Explain why the initial change in concentration should not be confused with a chemical reaction.

Solution 4

Both concentrations initially double, after which the net reaction proceeds in the reverse direction.

There are two distinct stages.

Stage 1: Immediate compression

The volume is halved, but the number of moles of each gas has not yet changed.

Using:

\[ c=\frac{n}{V} \]

where \(c\) is concentration, \(n\) is the amount of substance, and \(V\) is volume, halving the volume doubles both concentrations.

Immediately after compression:

\[ [\ce{N2O4}]=0.120\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]=0.160\ \mathrm{mol\,L^{-1}} \]

Both concentration-time curves therefore show an instantaneous upward jump.

This jump is caused by the decrease in volume, not by a chemical reaction.

Stage 2: The system approaches its new equilibrium

The pressure has increased.

Consider the number of gas particles on each side of the balanced equation:

• Left side: 1 mole of gaseous \(\ce{N2O4}\).
• Right side: 2 moles of gaseous \(\ce{NO2}\).

The system responds to the compression by favouring the side with fewer moles of gas.

The net reaction therefore proceeds backwards.

As a result:

• \(\ce{NO2}\) is consumed, so its concentration decreases from \(0.160\ \mathrm{mol\,L^{-1}}\).
• \(\ce{N2O4}\) is produced, so its concentration increases from \(0.120\ \mathrm{mol\,L^{-1}}\).

Eventually, both curves become horizontal again.

At the new equilibrium:

\[ r_f=r_r \]

The important graph-reading distinction is that an instantaneous vertical jump can represent an external change in volume, whereas the subsequent sloping sections represent changes caused by the chemical reaction.

Question 5

This question tests whether you can distinguish chemical equilibrium from a steady state when the concentration-time graphs look identical.

Two reactors contain the reversible reaction:

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

Both reactors show horizontal concentration-time curves after 60 seconds, with:

\[ [A]=0.60\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.40\ \mathrm{mol\,L^{-1}} \]

Reactor I is a sealed batch container maintained at constant temperature. No matter enters or leaves.

Reactor II is a continuously stirred tank with a constant volume of \(2.0\ \mathrm{L}\). A feed containing \(1.00\ \mathrm{mol\,L^{-1}}\) of A and no B enters at \(0.20\ \mathrm{L\,s^{-1}}\). An equal volumetric flow rate leaves the tank, carrying the well-mixed reactor contents.

At the steady concentrations in Reactor II, the reaction rates are:

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

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

Assume both reactors are well mixed, the reaction is the only chemical process, and there are no side reactions.

a) Which reactor is at dynamic chemical equilibrium?

b) Calculate the net forward reaction rate in Reactor II.

c) Demonstrate how the concentrations in Reactor II can remain constant despite the unequal forward and reverse reaction rates.

d) Explain why the two concentration-time graphs cannot be interpreted in the same way.

Solution 5

Reactor I is at dynamic chemical equilibrium, while Reactor II is at a steady state but not at chemical equilibrium.

This is because Reactor II has a net forward reaction occurring, even though its concentrations remain constant.

a) Reactor I is at dynamic equilibrium.

It is a closed system maintained at constant temperature.

Since both concentrations remain constant and the reaction is reversible, the forward and reverse rates are equal:

\[ r_f=r_r \]

Reactor II is different because matter continuously enters and leaves the vessel.

A constant concentration does not necessarily mean the chemical reaction has reached equilibrium.

b) Calculate the net forward reaction rate in Reactor II.

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

Substitute:

\[ r_{\text{net}}=0.060-0.020 \]

Therefore:

\[ \boxed{r_{\text{net}}=0.040\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]

The reaction is proceeding in the forward direction.

c) Show how the flow balances the reaction.

First, calculate the volumetric flow rate per unit reactor volume.

Let \(F\) represent volumetric flow rate and \(V\) represent reactor volume.

\[ D=\frac{F}{V} \]

\[ D=\frac{0.20}{2.0} \]

\[ D=0.10\ \mathrm{s^{-1}} \]

Now consider A.

The feed supplies A at a rate per unit reactor volume of:

\[ 0.10\times1.00 = 0.100\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The outlet removes A at:

\[ 0.10\times0.60 = 0.060\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The net supply of A from the flow is therefore:

\[ 0.100-0.060=0.040 \]

The chemical reaction consumes A at:

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

The incoming excess A exactly balances the amount consumed by the reaction.

Therefore, the concentration of A remains constant.

Now consider B.

There is no B entering the tank.

The outlet removes B at:

\[ 0.10\times0.40 = 0.040\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The net chemical reaction produces B at:

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

Again, the rates balance exactly.

The concentration of B also remains constant.

We can express both balances together:

\[ \frac{d[A]}{dt} = D([A]_{\text{in}}-[A])-r_{\text{net}} \]

\[ =0.10(1.00-0.60)-0.040 \]

\[ =0 \]

And:

\[ \frac{d[B]}{dt} = D([B]_{\text{in}}-[B])+r_{\text{net}} \]

\[ =0.10(0-0.40)+0.040 \]

\[ =0 \]

Both concentration gradients are zero, even though:

\[ r_f\neq r_r \]

d) The two graphs look identical, but the underlying processes are different.

In Reactor I, the concentrations remain constant because the forward and reverse reaction rates are equal.

In Reactor II, the concentrations remain constant because the chemical reaction is balanced by the continuous supply and removal of material.

Reactor II is at a steady state, not chemical equilibrium.

The tempting interpretation is that horizontal concentration-time curves always indicate equilibrium. The missing information is the system boundary.

This is why a good HSC Chemistry response should explain both the concentration changes and the conditions under which they occur.

08Where to go next

Once you can identify equilibrium from concentration-time graphs, the next step is predicting what happens when the system is disturbed.

Changes in concentration, pressure, volume, or temperature can alter the net reaction direction and produce a new equilibrium position.

This leads directly into Le Chatelier’s principle and equilibrium constants, \(K_c\). The concentration-time graph shows how the system approaches its new equilibrium, while \(K_c\) helps describe the equilibrium composition at a given temperature.