Why Open Systems May Not Reach Chemical Equilibrium

A red solution is flowing through a transparent reaction vessel. Fresh reactants enter through one pipe, the reversible reaction produces a red complex, and solution continuously leaves through another pipe.

After a few minutes, the red colour stops getting darker. The concentration of the red product stays constant.

Has the reaction reached dynamic equilibrium?

Not necessarily. The reaction might still be producing product faster than it converts product back into reactants. If the excess product leaves the vessel at exactly the rate it is produced, its concentration will remain constant even though the forward and reverse reaction rates are unequal.

An open system may fail to establish the expected equilibrium because matter continuously crosses its boundary, changing the amounts of reactants and products. Removing products can maintain a net forward reaction, while supplying fresh reactants can prevent the mixture from settling into its equilibrium composition.

The important distinction is between a reaction reaching equilibrium and a reaction reaching a steady state. They can produce similar-looking concentration readings, but the chemistry behind them is different.

01Start with the system boundary

Consider the reversible reaction:

\[ \ce{Fe^3+(aq) + SCN^-(aq) <=> FeSCN^2+(aq)} \]

The product, \(\mathrm{FeSCN}^{2+}\), is a red complex ion. This makes it useful for illustrating how a chemical reaction behaves in a flow-through system.

Imagine a stirred vessel with two feed pipes entering from the left and an outlet on the right. Solutions containing \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) enter the vessel, where they react. The red complex forms inside the vessel, and some of the solution leaves through the outlet.

The first question is not whether the vessel has a lid. It is:

Which substances can cross the boundary of the system?

For this example, define the system as the liquid inside the stirred vessel.

• Fresh reactants enter.
• Reactants and products leave in the outgoing solution.
• The reaction occurs inside the vessel.
• A temperature-control jacket can transfer energy without transferring matter.

This is an open system because matter crosses its boundary.

Now compare it with a sealed flask containing the same reversible reaction. In that flask, reactants and products remain inside. The forward and reverse reactions can continue until their rates become equal.

In the flow-through vessel, however, the composition is continually affected by material entering and leaving.

02Why product removal prevents the expected equilibrium

Think about what happens to the red complex.

The forward reaction produces \(\mathrm{FeSCN}^{2+}\). The reverse reaction consumes it, reforming \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\).

If product is continuously removed through the outlet, fewer product particles remain available for the reverse reaction.

As a result:

• The forward reaction can continue producing the complex.
• The reverse reaction has fewer product particles available.
• The forward reaction rate can remain greater than the reverse reaction rate.
• The system experiences a net forward reaction.

Fresh reactants entering the vessel can also replenish the substances consumed by the forward reaction.

The system is therefore being driven away from the composition it would establish in a sealed container.

Forward and reverse reaction rates

At dynamic equilibrium, the forward and reverse reaction rates are equal:

\[ r_f=r_r \]

Here:

• \(r_f\) is the forward reaction rate.
• \(r_r\) is the reverse reaction rate.

This does not mean that the two concentrations are equal. It means that the forward reaction produces substances at the same rate that the reverse reaction consumes them.

For a flow-through system, we must also consider material entering and leaving.

Using a positive sign for net formation of the product, the rate of change of the amount of product inside the vessel is:

\[ \frac{dn_C}{dt} = r_f-r_r+ \dot n_{C,\mathrm{in}} - \dot n_{C,\mathrm{out}} \]

Here:

• \(n_C\) is the amount of product inside the vessel, in moles.
• \(r_f\) and \(r_r\) are the forward and reverse reaction rates, in \(\mathrm{mol\,min^{-1}}\).
• \(\dot n_{C,\mathrm{in}}\) is the rate at which product enters.
• \(\dot n_{C,\mathrm{out}}\) is the rate at which product leaves.

For simplicity, assume no product enters with the feed.

The expression becomes:

\[ \frac{dn_C}{dt} = r_f-r_r-\dot n_{C,\mathrm{out}} \]

This equation reveals an important difference.

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

However, a flow-through vessel can maintain a constant product concentration when:

\[ r_f-r_r=\dot n_{C,\mathrm{out}} \]

The net production of product is exactly balanced by its removal.

The concentration stays constant, but the reaction is still proceeding in the forward direction overall.

This is a steady state, not dynamic equilibrium.

03A useful experiment: Does a constant colour prove equilibrium?

How could we distinguish the two situations experimentally?

A useful investigation would compare two vessels containing the same reversible reaction at the same temperature.

The apparatus

The first vessel is a sealed, stirred flask. A temperature-controlled water bath surrounds it, allowing energy to enter or leave while preventing matter from escaping.

The second vessel is a continuously stirred flow cell. Separate feed lines introduce the reactants, and an outlet removes solution at the same volumetric rate as the incoming solution. An optical probe measures the concentration of the red complex.

Both vessels begin with the same initial mixture.

What should happen?

In the sealed flask, the forward reaction initially produces more complex than the reverse reaction consumes.

As the complex concentration increases, the reverse reaction becomes faster. Eventually, the two reaction rates become equal.

The concentration then reaches its equilibrium value.

In the flow cell, fresh reactants continue entering and product continues leaving. Its product concentration may also settle at a constant value.

However, this does not necessarily mean the forward and reverse rates have become equal.

Illustrative data

Suppose the following conditions are established in the flow cell.

All numerical values are illustrative model data, not measurements from an actual experiment.

What do the reaction rates tell us?

The net rate of product formation is:

\[ \begin{aligned} r_{\mathrm{net}} &=r_f-r_r\\ &=8.0-3.0\\ &=5.0\ \mathrm{\mu mol\,min^{-1}} \end{aligned} \]

The reaction is therefore producing 5.0 µmol of additional complex each minute.

Now calculate how quickly the complex leaves the vessel.

\[ \begin{aligned} \dot n_{C,\mathrm{out}} &=0.0100\ \mathrm{L\,min^{-1}}\\ &\quad\times 0.50\ \mathrm{mmol\,L^{-1}}\\ &=0.00500\ \mathrm{mmol\,min^{-1}}\\ &=5.00\ \mathrm{\mu mol\,min^{-1}} \end{aligned} \]

The amount of complex leaving is exactly equal to its net production.

Therefore:

\[ \frac{dn_C}{dt}=5.0-5.0=0 \]

The concentration of the red complex remains constant.

But notice that:

\[ r_f=8.0\ne 3.0=r_r \]

The vessel has reached a steady state, not dynamic equilibrium.

We can also check the reactants. Each feed contains 1.50 mmol/L of its respective reactant, while the vessel contains 1.00 mmol/L.

The net supply of each reactant is:

\[ \begin{aligned} \dot n_{\mathrm{in}}-\dot n_{\mathrm{out}} &=0.0100(1.50-1.00)\\ &=0.00500\ \mathrm{mmol\,min^{-1}}\\ &=5.00\ \mathrm{\mu mol\,min^{-1}} \end{aligned} \]

This matches the net consumption of each reactant by the reaction.

The material balance works for all three species. Their concentrations can remain constant even though the reaction continues with unequal forward and reverse rates.

What was missing from the original experiment?

Simply watching the red colour was not enough.

The optical probe measured the concentration of the complex, but it did not establish whether the forward and reverse reaction rates were equal.

The first investigation could not distinguish two different explanations for the constant reading:

1. The reaction had reached dynamic equilibrium.
2. The reaction was continuing, but product removal balanced its net production.

A revised experiment would close both inlet and outlet valves simultaneously while maintaining the same temperature.

This stops matter from crossing the boundary. The reaction mixture is now a closed system.

If the complex concentration subsequently increases, the original flow-cell composition was not at equilibrium. The reaction was still proceeding in the forward direction.

The distinction becomes clear when we compare the concentration-time behaviour.

The important experimental lesson is that a flat concentration-time graph does not, by itself, establish chemical equilibrium.

04Use the reaction quotient to check the direction of change

Reaction rates tell us whether the forward and reverse reactions are balanced. The reaction quotient provides another way to determine whether a mixture has reached its expected equilibrium composition.

Consider the general reversible reaction:

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

Its equilibrium constant expression is:

\[ K_c=\frac{[C]}{[A][B]} \]

The reaction quotient has the same mathematical expression:

\[ Q_c=\frac{[C]}{[A][B]} \]

The difference is that \(K_c\) describes the equilibrium condition at a particular temperature, while \(Q_c\) describes the current mixture.

For this reaction, both quantities have units of \(\mathrm{L\,mol^{-1}}\), when concentrations are expressed in \(\mathrm{mol\,L^{-1}}\).

The comparison gives three possibilities:

At a constant temperature, removing a product does not change \(K_c\).

Instead, it decreases the product concentration, changing \(Q_c\).

If the product is continuously removed, the reaction quotient can remain below the equilibrium constant. The reaction continues in the forward direction rather than settling at the expected equilibrium composition.

06When an open container can still behave like a closed system

There is an important qualification to the idea that an open system cannot establish equilibrium.

The word open does not automatically mean that the reacting substances are continuously escaping.

Consider a beaker containing the reversible reaction:

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

Assume that:

• A, B, and C are non-volatile.
• No solution is added or removed.
• The surrounding air does not participate in the reaction.
• Evaporation and other material transfers are negligible.
• Temperature remains constant.

The beaker is physically uncovered, but none of the reacting species is crossing the boundary in an appreciable amount.

For this particular reaction, the mixture behaves effectively like a closed system and can establish chemical equilibrium.

This is why identifying the actual matter transfer is more useful than simply looking at whether a container has a lid.

For HSC Chemistry, the standard distinction remains:

• A closed system prevents matter from entering or leaving and allows a reversible reaction to establish dynamic equilibrium.
• An open system allows matter transfer, which can prevent the reaction mixture from maintaining the expected equilibrium.
• A steady-state system can maintain constant concentrations through balanced material flows even when the reaction rates are unequal.

The key is to explain what is happening to the reacting substances, not just label the container.

07A practical decision process for HSC questions

When a question describes an open reaction system, use this sequence.

1. Identify the system boundary.

State exactly what is inside the system and what can cross its boundary.

For example, does the vessel lose a gaseous product? Does fresh reactant continuously enter?

2. Identify the effect on the reaction.

If product is removed, its concentration decreases and the reverse reaction becomes less favourable relative to the forward reaction.

If reactants are continuously supplied, the forward reaction can continue using the incoming material.

3. Compare the forward and reverse rates.

Do not assume they are equal just because concentrations remain constant.

If the forward rate is greater than the reverse rate, there is net forward reaction.

4. Account for material entering and leaving.

Check whether the net production of a substance is balanced by its removal.

If it is, the concentration may remain constant even without equilibrium.

5. State the correct condition.

If the forward and reverse reaction rates are equal, the reaction has reached dynamic equilibrium.

If unequal reaction rates are balanced by material entering or leaving, the system is at steady state.

A strong HSC explanation might read:

The system does not establish the expected dynamic equilibrium because product is continuously removed across the system boundary. This reduces the amount of product available for the reverse reaction, allowing the forward reaction rate to remain greater than the reverse reaction rate. The concentration may remain constant if the net rate of product formation equals its rate of removal, but this represents a steady state rather than chemical equilibrium.

08Questions and solutions

Question 1

A reversible reaction occurs in a continuously stirred vessel:

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

The forward reaction rate is \(14.0\ \mathrm{\mu mol\,min^{-1}}\), and the reverse reaction rate is \(9.0\ \mathrm{\mu mol\,min^{-1}}\).

No C enters the vessel, but C leaves at a rate of \(5.0\ \mathrm{\mu mol\,min^{-1}}\). Fresh A and B enter at rates that maintain their concentrations inside the vessel.

The concentration of C remains constant.

Determine whether the system has reached dynamic equilibrium or steady state. Justify your answer using the reaction rates.

Solution 1

The system has reached a steady state, not dynamic equilibrium.

The net forward reaction rate is:

\[ \begin{aligned} r_{\mathrm{net}} &=r_f-r_r\\ &=14.0-9.0\\ &=5.0\ \mathrm{\mu mol\,min^{-1}} \end{aligned} \]

The net production of C is therefore \(5.0\ \mathrm{\mu mol\,min^{-1}}\).

This exactly matches the rate at which C leaves the vessel.

Consequently, the amount of C inside the vessel does not change.

However, the forward and reverse reaction rates are not equal:

\[ 14.0\ne9.0 \]

The tempting mistake is to assume that constant concentration automatically means equilibrium. The missing consideration is the product leaving the system.

The constant concentration results from net chemical production balancing material removal, rather than equal forward and reverse reaction rates.

Question 2

An uncovered beaker contains a reversible aqueous reaction:

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

Assume that all three substances are non-volatile, no solution enters or leaves the beaker, the surrounding air does not participate in the reaction, and the temperature remains constant.

A student argues:

“The beaker is open, so the reaction cannot establish equilibrium.”

Evaluate this statement. Explain how the actual movement of matter across the system boundary affects your conclusion.

Solution 2

The student’s statement is too broad. Under the stated assumptions, the reacting mixture can establish chemical equilibrium because none of the reacting species is crossing its boundary.

Although the beaker is physically uncovered, A, B, and C remain in the solution.

The surrounding air does not participate in the reaction, and there is no appreciable evaporation or material transfer involving the reacting substances.

Therefore, for this reaction, the solution behaves effectively as a closed system.

The forward and reverse reactions can continue until their rates become equal.

The important distinction is that the physical appearance of a container does not, by itself, determine whether the reacting mixture can reach equilibrium.

In a real experiment, evaporation, absorption of gases, or the addition of substances could change this conclusion. Those material transfers would need to be considered.

Question 3

A sealed vessel contains the reaction:

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

The reaction has reached dynamic equilibrium.

A selective membrane is then opened. It allows C to diffuse out of the vessel into an external region where its concentration is maintained at a very low value. A and B cannot cross the membrane.

Assume the temperature remains constant.

Explain:

(a) What happens to the reaction quotient immediately after product begins leaving?

(b) How do the forward and reverse reaction rates respond?

(c) Does the equilibrium constant change?

(d) Can the original equilibrium be maintained while product continues to escape?

Solution 3

The reaction moves in the forward direction, and the original equilibrium cannot be maintained while C is continuously removed.

(a) Reaction quotient

The reaction quotient is:

\[ Q_c=\frac{[C]}{[A][B]} \]

As C leaves the vessel, its concentration decreases.

The concentrations of A and B have not yet changed significantly, so the reaction quotient decreases.

Since the system was initially at equilibrium, its initial reaction quotient equalled the equilibrium constant.

After product removal:

\[ Q_c<K_c \]

(b) Reaction rates

The reduced concentration of C makes the reverse reaction less favourable.

The forward reaction rate becomes greater than the reverse reaction rate, producing a net forward reaction.

(c) Equilibrium constant

The equilibrium constant does not change.

At constant temperature, \(K_c\) remains constant even when the concentrations of substances change.

Removing C changes \(Q_c\), not \(K_c\).

(d) Maintaining equilibrium

The original equilibrium cannot be maintained while C is continuously removed and no corresponding material transfer restores the equilibrium composition.

The forward reaction continues producing C, but the membrane allows that product to escape.

If the membrane is closed, the system becomes closed to matter again. The reaction can then proceed towards a new equilibrium.

Question 4

Two identical vessels contain the same reversible reaction at the same temperature.

Vessel P is sealed.

Vessel F has a continuous reactant supply and an outlet that removes the reaction mixture.

A concentration sensor records the concentration of product C in each vessel. After some time, both concentration-time readings become horizontal.

A student concludes:

“Both reactions have reached dynamic equilibrium because the product concentrations are no longer changing.”

Explain why this conclusion is not justified.

Then propose a modification to the experimental design that would distinguish equilibrium from steady state in Vessel F. Describe what you would expect to observe if Vessel F was originally at steady state rather than equilibrium.

Solution 4

The student’s conclusion is not justified because constant concentration does not prove that the forward and reverse reaction rates are equal.

In Vessel P, the system is closed to matter. If the reaction is reversible and sufficient time has passed, the constant concentrations can indicate dynamic equilibrium.

In Vessel F, the concentrations could remain constant because net product formation balances product removal.

The two vessels can therefore produce similar concentration-time graphs for different reasons.

A suitable modification is to close both the inlet and outlet of Vessel F simultaneously, while maintaining its temperature.

This prevents matter from crossing the boundary.

If Vessel F was originally at steady state, the concentrations will generally change after the flow stops because the forward and reverse reaction rates were previously unequal.

The reaction will continue towards equilibrium until the forward and reverse rates become equal.

If the original mixture was already at chemical equilibrium, closing the material-flow connections would not produce a net reaction shift, provided the composition, volume, and temperature remain unchanged.

This revised experiment tests the actual chemical state rather than relying only on a constant concentration reading.

Question 5

A well-mixed flow reactor contains the reversible reaction:

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

The reactor has a volume of \(0.100\ \mathrm{L}\). Both the inlet and outlet flow rates are \(0.0100\ \mathrm{L\,min^{-1}}\).

The feed contains:

• \([A]=12.0\ \mathrm{mmol\,L^{-1}}\)
• \([B]=12.0\ \mathrm{mmol\,L^{-1}}\)
• \([C]=0\)

The measured concentrations inside the reactor remain constant at:

• \([A]=10.0\ \mathrm{mmol\,L^{-1}}\)
• \([B]=10.0\ \mathrm{mmol\,L^{-1}}\)
• \([C]=2.00\ \mathrm{mmol\,L^{-1}}\)

An independent kinetic analysis gives:

• Forward reaction rate: \(0.0300\ \mathrm{mmol\,min^{-1}}\)
• Reverse reaction rate: \(0.0100\ \mathrm{mmol\,min^{-1}}\)

At this temperature, the equilibrium constant is:

\[ K_c=62.5\ \mathrm{L\,mol^{-1}} \]

The inlet and outlet valves are then closed simultaneously, isolating the reaction mixture without changing its volume or temperature.

Assume the reaction proceeds without side reactions and eventually reaches equilibrium.

(a) Calculate the reaction quotient of the original flow-reactor mixture and determine its favoured direction.

(b) Verify that the original constant concentrations are consistent with the supplied reaction rates and material flows.

(c) Calculate the equilibrium concentrations of A, B, and C after the reactor is isolated.

(d) Explain why the original flow reactor could maintain constant concentrations even though it was not at equilibrium.

Solution 5

The original reactor is at steady state, not dynamic equilibrium. Once isolated, the reaction proceeds in the forward direction until the concentrations reach a new equilibrium composition:

\[ [A]=[B]=8.00\ \mathrm{mmol\,L^{-1}} \]

\[ [C]=4.00\ \mathrm{mmol\,L^{-1}} \]

The reasoning requires us to account for the reaction quotient, the material flows, and the change in system boundary.

(a) Calculate the reaction quotient

For the reaction:

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

the reaction quotient is:

\[ Q_c=\frac{[C]}{[A][B]} \]

Convert the concentrations into \(\mathrm{mol\,L^{-1}}\):

\[ \begin{aligned} [A]&=0.0100\ \mathrm{mol\,L^{-1}}\\ [B]&=0.0100\ \mathrm{mol\,L^{-1}}\\ [C]&=0.00200\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Substitute:

\[ \begin{aligned} Q_c &=\frac{0.00200}{0.0100\times0.0100}\\ &=20.0\ \mathrm{L\,mol^{-1}} \end{aligned} \]

Compare this with the equilibrium constant:

\[ Q_c=20.0<K_c=62.5 \]

The forward reaction is favoured.

This means that the original composition is not at equilibrium.

(b) Verify the material balance

First, calculate the net reaction rate.

\[ \begin{aligned} r_f-r_r &=0.0300-0.0100\\ &=0.0200\ \mathrm{mmol\,min^{-1}} \end{aligned} \]

The net reaction produces \(0.0200\) mmol of C each minute.

Now calculate the rate at which C leaves the reactor:

\[ \begin{aligned} \dot n_{C,\mathrm{out}} &=0.0100\times2.00\\ &=0.0200\ \mathrm{mmol\,min^{-1}} \end{aligned} \]

No C enters with the feed.

Therefore, its net production exactly balances its removal.

Now check A:

\[ \begin{aligned} \dot n_{A,\mathrm{in}} &=0.0100\times12.0\\ &=0.120\ \mathrm{mmol\,min^{-1}} \end{aligned} \]

Its rate of removal is:

\[ \begin{aligned} \dot n_{A,\mathrm{out}} &=0.0100\times10.0\\ &=0.100\ \mathrm{mmol\,min^{-1}} \end{aligned} \]

The net supply of A is:

\[ 0.120-0.100=0.0200\ \mathrm{mmol\,min^{-1}} \]

This matches its net consumption by the reaction.

The same calculation applies to B because its feed and internal concentrations are identical to those of A.

The material balance is therefore consistent for all three substances.

The reactor maintains constant concentrations because the incoming reactants replace those consumed, while the outgoing solution removes the net amount of product formed.

(c) Calculate the new equilibrium concentrations

Once both valves are closed, matter can no longer enter or leave.

The system is now closed, and the reaction can move towards equilibrium.

The initial concentrations are:

\[ \begin{aligned} [A]&=0.0100\ \mathrm{mol\,L^{-1}}\\ [B]&=0.0100\ \mathrm{mol\,L^{-1}}\\ [C]&=0.00200\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Since \(Q_c<K_c\), the forward reaction proceeds.

Let \(x\) be the increase in the concentration of C, measured in \(\mathrm{mol\,L^{-1}}\).

Because the reaction consumes A and B in a 1:1 ratio, the equilibrium concentrations are:

Substitute these expressions into the equilibrium constant equation:

\[ K_c=\frac{[C]}{[A][B]} \]

\[ 62.5=\frac{0.00200+x}{(0.0100-x)^2} \]

Rearranging:

\[ 62.5(0.0100-x)^2=0.00200+x \]

Expanding and simplifying:

\[ 62.5x^2-2.25x+0.00425=0 \]

Solving gives two mathematical roots:

\[ x=0.00200 \]

or

\[ x=0.0340 \]

The second root is impossible because it would consume more A and B than are initially available.

Therefore:

\[ x=0.00200\ \mathrm{mol\,L^{-1}} \]

The final equilibrium concentrations are:

\[ \begin{aligned} [A]_{\mathrm{eq}} &=0.0100-0.00200\\ &=0.00800\ \mathrm{mol\,L^{-1}} \end{aligned} \]

\[ \begin{aligned} [B]_{\mathrm{eq}} &=0.0100-0.00200\\ &=0.00800\ \mathrm{mol\,L^{-1}} \end{aligned} \]

\[ \begin{aligned} [C]_{\mathrm{eq}} &=0.00200+0.00200\\ &=0.00400\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Converting to mmol/L:

• A: 8.00 mmol/L
• B: 8.00 mmol/L
• C: 4.00 mmol/L

(d) Explain the original steady state

The original flow reactor was not at equilibrium because its forward reaction rate was greater than its reverse reaction rate.

The net production of C was \(0.0200\ \mathrm{mmol\,min^{-1}}\), exactly matching its removal through the outlet.

The incoming reactants also replaced the amounts consumed by the reaction.

This maintained constant concentrations even though the reaction was proceeding in the forward direction overall.

Once the valves were closed, the material flows stopped. The difference between the forward and reverse reaction rates could no longer be balanced by material crossing the boundary.

The mixture therefore moved towards its actual equilibrium composition, where \(Q_c=K_c\) and the forward and reverse reaction rates are equal.

The crucial insight is that constant concentrations describe what you observe, while equal forward and reverse rates explain why a reaction is at dynamic equilibrium.

09Where this idea leads next

Understanding the difference between equilibrium and steady state makes it easier to interpret concentration-time graphs and predict how a system responds when its conditions change.

The next useful distinction is whether a concentration change comes from a reaction shifting towards a new equilibrium or from a substance physically entering or leaving the system.

For more on the foundations, see Open and closed equilibrium systems: HSC Chemistry guide.

For help interpreting the shape of concentration-time graphs, see How to Interpret a Concentration-Time Graph Approaching Equilibrium.