Instantaneous Concentration Changes vs Equilibrium Shifts

A rigid 1.00 L vessel contains a mixture at equilibrium for the reaction:

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

You inject more C into the vessel. Its concentration jumps immediately. A moment later, however, its concentration starts falling as the reaction proceeds.

Did the equilibrium shift at the instant C was added, or did the shift happen afterwards?

Predict what happens to the concentrations of A and B at the instant of injection. If the vessel has a fixed volume and only C enters, neither concentration changes immediately. Their concentrations change later, as the reverse reaction converts C back into A and B.

This is the distinction you need to recognise in HSC Chemistry:

• An instantaneous concentration change happens because a substance enters or leaves the reaction system, or because its volume changes.
• A subsequent equilibrium shift happens because the forward and reverse reaction rates become unequal, causing a net chemical reaction.

The first is caused by the disturbance. The second is the chemical response to that disturbance.

01Separate the disturbance from the reaction

At equilibrium, the forward and reverse reactions are still occurring. Their rates are equal, so there is no net change in concentration.

\[
r_{\text{forward}}=r_{\text{reverse}}
\]

Now imagine adding a product to the reaction mixture.

Two things happen, in chronological order:

1. The product concentration increases immediately because additional product enters the vessel.
2. The reverse reaction becomes faster than the forward reaction, so the system starts consuming product.

These are separate events, even though they happen very close together.

First, define the system boundary

The system boundary tells you what is entering or leaving the reaction mixture.

For our example, imagine a sealed reaction vessel with an injection port. The syringe is outside the system, while the reacting gases inside the vessel form the system.

When C is injected, it crosses the system boundary. Once the injection port is closed, the reaction mixture is closed to further matter transfer.

Concentration is calculated using:

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

where:

• \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\),
• \(n\) is the amount of substance in mol, and
• \(V\) is volume in L.

If the volume stays constant and you add C, only the amount of C increases immediately. Therefore, only its concentration jumps.

If you instead add water to an aqueous equilibrium, the situation is different. The volume increases, so the concentrations of all dissolved species decrease immediately through dilution.

This is why you should always identify the disturbance before predicting the equilibrium shift.

What happens to the concentration graph?

A concentration-time graph gives you a particularly useful way to distinguish the two events.

An instantaneous concentration change appears as a sudden jump in the graph. The subsequent reaction appears as a continuous curve with a changing concentration over time.

For example, after adding C:

• \([C]\) jumps upwards immediately.
• \([A]\) and \([B]\) do not jump.
• The reverse reaction then consumes C, so \([C]\) gradually decreases.
• A and B gradually increase.
• Eventually, all three concentrations become constant again.

The vertical jump represents the external addition, not an instantaneous chemical reaction.

In a real experiment, mixing takes a finite amount of time. The vertical jump is an idealisation used when mixing is much faster than the subsequent reaction.

02Use forward and reverse rates to identify the shift

Now that you have separated the disturbance from the reaction, you need to determine which direction the reaction proceeds.

For a reversible reaction:

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

there are two competing rates:

• \(r_{\text{forward}}\): the rate at which A and B form C.
• \(r_{\text{reverse}}\): the rate at which C forms A and B.

We will use a positive net rate to represent the forward direction:

\[
r_{\text{net}}=r_{\text{forward}}-r_{\text{reverse}}
\]

Therefore:

Notice that the rates, not the relative concentrations, determine whether there is a net reaction.

You could have a mixture containing far more reactants than products and still be at equilibrium. The important condition is equal forward and reverse rates.

Use the reaction quotient to check the direction

You can determine the net direction mathematically using the reaction quotient, \(Q_c\).

For:

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

the reaction quotient is:

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

The equilibrium constant, \(K_c\), has the same expression, but uses equilibrium concentrations at a particular temperature.

The difference is that \(Q_c\) can be calculated using concentrations at any point, including immediately after a disturbance.

Compare the two values:

This gives you a reliable method:

1. Calculate the concentrations immediately after the disturbance.
2. Calculate \(Q_c\) using those concentrations.
3. Compare \(Q_c\) with \(K_c\).
4. Identify which reaction direction is faster.
5. Explain how the concentrations change until equilibrium is restored.

At constant temperature, \(K_c\) does not change when you add or remove a reacting substance. The disturbance changes \(Q_c\), and the reaction proceeds until \(Q_c=K_c\) again.

One important limitation: the coefficients in a balanced equation determine the powers in the equilibrium expression, but they do not automatically determine the reaction rate law. You need a supplied rate law or an explicitly stated elementary-step assumption to calculate numerical rates.

For more on how collision frequency affects forward and reverse rates, see Collision Theory and Equilibrium Shifts: HSC Chemistry Explained.

04A contrasting case: Concentrations change, but equilibrium does not shift

Here is a situation that exposes another common misconception.

Consider:

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

Suppose the mixture is initially at equilibrium. Its volume is suddenly reduced to half its original value, while temperature remains constant.

What happens?

All three gas concentrations double immediately because the same number of moles now occupies half the volume.

A student might predict that the equilibrium must shift because the concentrations have changed.

However, look at the reaction quotient:

\[
Q_c=\frac{[HI]^2}{[H_2][I_2]}
\]

After compression:

\[
Q_c’=
\frac{(2[HI])^2}
{(2[H_2])(2[I_2])}
\]

Simplifying:

\[
Q_c’=
\frac{4[HI]^2}
{4[H_2][I_2]}
\]

Therefore:

\[
Q_c’=Q_c=K_c
\]

The equilibrium condition is preserved.

There is an instantaneous increase in all three concentrations, but no subsequent equilibrium shift.

This is an important distinction: an instantaneous concentration change does not automatically mean an equilibrium shift will follow.

The effect depends on how the disturbance changes the reaction quotient.

For a more detailed explanation of this pressure example, see When Compression Shifts Equilibrium-and When It Doesn’t.

06A quick diagnostic table for common disturbances

The following table assumes that temperature stays constant, volume is fixed, and only the named reacting species is added or removed.

The important point is that the instantaneous change and the subsequent reaction can have opposite effects on the same substance.

For example, adding a product makes its concentration rise immediately, but the subsequent reverse reaction consumes some of that product.

One more system-boundary trap: Removing a sample

Suppose you withdraw 10% of a well-mixed aqueous solution from a vessel.

You have removed 10% of the dissolved substances, but you have also removed 10% of the solution volume.

The concentrations of the remaining solution are therefore unchanged.

If the system was initially at equilibrium, its reaction quotient remains equal to its equilibrium constant. There is no shift simply because you removed a representative sample.

This is different from selectively removing one particular reactant or product, where its concentration changes relative to the other species.

Always ask what actually crossed the system boundary.

07A reliable method for HSC equilibrium questions

When you encounter a question involving an instantaneous concentration change, use this sequence.

1. Identify the disturbance.

Was a substance added, removed, or diluted? Was the vessel compressed or expanded? Did temperature change?

2. Establish the system boundary.

Identify which substances enter or leave the reacting system and whether the volume changes.

3. Calculate the immediate concentrations.

Do this before considering any chemical reaction.

For example, if a product is added at constant volume, increase its concentration without changing the other concentrations.

4. Calculate the reaction quotient.

Use the concentrations immediately after the disturbance.

5. Compare \(Q_c\) with \(K_c\).

Identify which directional reaction rate is greater.

6. Explain the subsequent concentration changes.

State which species are consumed and which are produced.

7. Identify the new equilibrium condition.

The reaction continues until the forward and reverse rates are equal again, or equivalently, \(Q_c=K_c\) at the unchanged temperature.

A strong HSC explanation follows this sequence rather than simply stating that the system shifts to oppose the disturbance.

08Questions and solutions

The following questions progress from straightforward concentration changes to situations involving competing equilibria and multiple reaction pathways.

Question 1

An equilibrium mixture follows:

\[
\ce{P(aq) + Q(aq) <=> R(aq)}
\]

Initially, the system is at equilibrium with:

\[
[P]=0.40,\quad [Q]=0.50,\quad [R]=0.20
\]

All concentrations are in \(\mathrm{mol\,L^{-1}}\).

At constant temperature and volume, additional R is introduced, causing its concentration to increase immediately to \(0.50\ \mathrm{mol\,L^{-1}}\).

Calculate the new reaction quotient and determine the direction of the subsequent equilibrium shift.

Solution 1

The reaction shifts to the left because the immediate increase in R causes \(Q_c\) to become greater than \(K_c\).

First, calculate the original equilibrium constant:

\[
K_c=\frac{[R]}{[P][Q]}
\]

\[
K_c=\frac{0.20}{0.40\times0.50}
\]

\[
K_c=1.00
\]

Immediately after adding R:

\[
[P]=0.40,\quad [Q]=0.50,\quad [R]=0.50
\]

The new reaction quotient is:

\[
Q_c=\frac{0.50}{0.40\times0.50}
\]

\[
Q_c=2.50
\]

Since:

\[
Q_c>K_c
\]

the reverse reaction is faster than the forward reaction.

Therefore, R is consumed, while P and Q are produced, until equilibrium is restored.

The immediate increase in R is caused by its addition. The subsequent decrease is caused by the reverse chemical reaction.

Question 2

An equilibrium mixture follows:

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

Initially:

\[
[A]=[B]=[C]=1.00\ \mathrm{mol\,L^{-1}}
\]

The system is at equilibrium, with \(K_c=1.00\).

A selective membrane removes \(0.40\ \mathrm{mol\,L^{-1}}\) of A without changing the vessel volume or removing any B or C.

Calculate:

a) The immediate reaction quotient.

b) The direction of the subsequent reaction.

c) The new equilibrium concentrations of A, B, and C.

Solution 2

The reaction shifts to the left, producing more A and B and consuming C.

Part a: Calculate the immediate reaction quotient.

Immediately after removal:

\[
[A]=0.60
\]

\[
[B]=1.00
\]

\[
[C]=1.00
\]

Therefore:

\[
Q_c=\frac{1.00}{0.60\times1.00}
\]

\[
Q_c=1.67
\]

Since \(Q_c>K_c\), the reverse reaction is favoured.

Part b: Identify the direction.

The reverse reaction converts C into A and B.

Therefore, A and B increase through the subsequent chemical reaction, while C decreases.

Part c: Calculate the new equilibrium concentrations.

Let \(x\) represent the concentration of C consumed.

At equilibrium:

\[
K_c=\frac{1.00-x}{(0.60+x)(1.00+x)}
\]

Substituting \(K_c=1.00\):

\[
1.00-x=(0.60+x)(1.00+x)
\]

Expanding:

\[
1.00-x=0.60+1.60x+x^2
\]

Rearranging:

\[
x^2+2.60x-0.40=0
\]

Using the quadratic formula:

\[
x=\frac{-2.60\pm\sqrt{2.60^2+4(0.40)}}{2}
\]

Taking the positive root:

\[
x=0.146
\]

Therefore:

\[
[A]=0.60+0.146=0.746
\]

\[
[B]=1.00+0.146=1.146
\]

\[
[C]=1.00-0.146=0.854
\]

All concentrations are in \(\mathrm{mol\,L^{-1}}\).

Notice that A initially dropped from 1.00 to 0.60, then recovered to 0.746. It did not return to its original concentration.

Question 3

An equilibrium mixture follows the reaction:

\[
\ce{2M(aq) <=> N(aq)}
\]

Initially:

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

\[
[N]=0.32\ \mathrm{mol\,L^{-1}}
\]

The system is at equilibrium.

The solution is diluted by adding water until its volume triples. Temperature remains constant.

Assume both reaction directions are elementary steps, with rate laws:

\[
r_f=k_f[M]^2
\]

\[
r_r=k_r[N]
\]

Calculate:

a) The immediate concentrations of M and N.

b) The new reaction quotient.

c) The direction of the subsequent reaction.

d) The ratio of the reverse rate to the forward rate immediately after dilution.

Solution 3

The reaction shifts to the left, and the reverse reaction rate becomes three times the forward reaction rate immediately after dilution.

Part a: Calculate the immediate concentrations.

The volume triples, so all dissolved concentrations are divided by three.

\[
[M]’=\frac{0.80}{3}
\]

\[
[M]’=0.267\ \mathrm{mol\,L^{-1}}
\]

Similarly:

\[
[N]’=\frac{0.32}{3}
\]

\[
[N]’=0.107\ \mathrm{mol\,L^{-1}}
\]

These are the immediate concentrations, before any reaction occurs.

Part b: Calculate the new reaction quotient.

For this reaction:

\[
Q_c=\frac{[N]}{[M]^2}
\]

First, calculate the original equilibrium constant:

\[
K_c=\frac{0.32}{0.80^2}
\]

\[
K_c=0.50
\]

Immediately after dilution:

\[
Q_c=\frac{0.107}{0.267^2}
\]

\[
Q_c=1.50
\]

Since:

\[
Q_c>K_c
\]

the reverse reaction is favoured.

Part c: Identify the subsequent reaction.

The reverse reaction converts N into M.

Therefore, N decreases and M increases as the system approaches equilibrium.

Part d: Compare the reaction rates.

The forward rate law is:

\[
r_f=k_f[M]^2
\]

Since the concentration of M decreases to one-third of its original value:

\[
r_f’=\frac{1}{9}r_f
\]

The reverse rate law is:

\[
r_r=k_r[N]
\]

Since N also decreases to one-third of its original concentration:

\[
r_r’=\frac{1}{3}r_r
\]

Before dilution, the system was at equilibrium, so:

\[
r_f=r_r=r_0
\]

Therefore:

\[
r_f’=\frac{r_0}{9}
\]

\[
r_r’=\frac{r_0}{3}
\]

Taking the ratio:

\[
\frac{r_r’}{r_f’}
=
\frac{r_0/3}{r_0/9}
\]

\[
\boxed{\frac{r_r’}{r_f’}=3}
\]

The reverse reaction is three times faster than the forward reaction immediately after dilution.

This demonstrates why saying that both reaction rates decrease is not enough. You must compare how much each rate changes.

Question 4

Consider the equilibrium:

\[
\ce{2X(g) <=> Y(g)}
\]

Initially:

\[
[X]=1.00\ \mathrm{mol\,L^{-1}}
\]

\[
[Y]=2.00\ \mathrm{mol\,L^{-1}}
\]

The mixture is at equilibrium at constant temperature.

A selective membrane removes half of the Y present in the vessel. At the same time, a catalyst is introduced.

Explain:

a) The immediate concentration changes.

b) The new reaction quotient and subsequent direction of reaction.

c) Whether the catalyst changes the direction of the equilibrium shift or the value of \(K_c\).

d) What happens to the forward and reverse reaction rates.

Solution 4

The reaction shifts forwards, towards Y. The catalyst speeds up the approach to equilibrium but does not change the equilibrium position or the direction favoured by the disturbance.

Part a: Identify the immediate concentration changes.

The membrane selectively removes half of Y.

Therefore:

\[
[Y]’=1.00\ \mathrm{mol\,L^{-1}}
\]

X is not removed, so:

\[
[X]’=1.00\ \mathrm{mol\,L^{-1}}
\]

The catalyst does not cause an instantaneous change in the concentrations of X or Y.

Part b: Calculate the reaction quotient.

For this reaction:

\[
Q_c=\frac{[Y]}{[X]^2}
\]

Initially:

\[
K_c=\frac{2.00}{1.00^2}=2.00
\]

Immediately after removing Y:

\[
Q_c=\frac{1.00}{1.00^2}=1.00
\]

Since:

\[
Q_c<K_c
\]

the forward reaction is favoured.

X is consumed and Y is produced until the system reaches equilibrium again.

Part c: Explain the effect of the catalyst.

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

It increases the rates of both the forward and reverse reactions.

However, it does not change \(K_c\) at a given temperature.

The catalyst therefore does not cancel the effect of removing Y. The reaction still proceeds forwards because \(Q_c<K_c\).

Part d: Describe the rate changes.

The removal of Y disturbs the equilibrium and changes the balance between the forward and reverse rates.

The catalyst accelerates both reaction directions, allowing the system to reach its new equilibrium more quickly.

At the new equilibrium:

\[
r_f=r_r
\]

The important distinction is that removing Y determines the direction of the net reaction, while the catalyst affects how quickly equilibrium is restored.

Question 5

Three aqueous isomers, A, B, and C, can interconvert through two reversible reactions in the same well-mixed vessel:

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

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

The equilibrium constants are:

\[
K_{c,1}=1.00
\]

\[
K_{c,2}=4.00
\]

Initially, both reactions are at equilibrium, with:

\[
[A]=1.00,\quad [B]=1.00,\quad [C]=4.00
\]

All concentrations are in \(\mathrm{mol\,L^{-1}}\).

The vessel has a fixed volume of 1.00 L. An additional 3.00 mol of B is injected without changing the volume or temperature.

Assume:

• A, B, and C are isomers of the same molecule.
• No other reactions occur.
• The total concentration \([A]+[B]+[C]\) is conserved after injection.
• Both equilibria must be satisfied simultaneously at the final equilibrium.

A student claims:

“Adding B will cause the second reaction to shift right because B is a reactant. Therefore, the entire system shifts towards C.”

Evaluate this claim.

Determine:

a) The immediate concentrations after injection.

b) The initial direction of each of the two reactions after the disturbance.

c) The final equilibrium concentrations of A, B, and C.

d) Why describing the entire system as simply shifting right is misleading.

Solution 5

The student’s claim is incomplete. The first reaction shifts left, while the second reaction shifts right. Both reactions consume B, and the final concentrations must satisfy both equilibrium constants simultaneously.

This is a coupled equilibrium problem: the same species participates in two different reversible reactions.

Part a: Calculate the immediate concentrations.

The initial concentration of B is:

\[
[B]=1.00\ \mathrm{mol\,L^{-1}}
\]

Adding 3.00 mol of B to a 1.00 L vessel increases its concentration by:

\[
\Delta[B]=\frac{3.00}{1.00}
=3.00\ \mathrm{mol\,L^{-1}}
\]

Therefore, immediately after injection:

\[
[A]=1.00
\]

\[
[B]=4.00
\]

\[
[C]=4.00
\]

Only B experiences an instantaneous concentration increase.

Part b: Determine the direction of each reaction.

This is where the student’s reasoning fails.

You must calculate the reaction quotient for each equilibrium separately.

For the first reaction:

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

The reaction quotient is:

\[
Q_{c,1}=\frac{[B]}{[A]}
\]

Immediately after injection:

\[
Q_{c,1}=\frac{4.00}{1.00}=4.00
\]

Since:

\[
Q_{c,1}>K_{c,1}
\]

the first reaction proceeds in reverse.

Therefore, B is converted into A.

For the second reaction:

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

The reaction quotient is:

\[
Q_{c,2}=\frac{[C]}{[B]}
\]

Immediately after injection:

\[
Q_{c,2}=\frac{4.00}{4.00}=1.00
\]

Since:

\[
Q_{c,2}<K_{c,2}
\]

the second reaction proceeds forwards.

Therefore, B is converted into C.

Both reactions consume B, but they produce different substances.

The first reaction produces A, while the second produces C.

Part c: Calculate the final equilibrium concentrations.

At the new equilibrium, both equilibrium conditions must be satisfied.

For the first reaction:

\[
K_{c,1}=\frac{[B]}{[A]}=1.00
\]

Therefore:

\[
[B]=[A]
\]

For the second reaction:

\[
K_{c,2}=\frac{[C]}{[B]}=4.00
\]

Therefore:

\[
[C]=4[B]
\]

Now apply conservation of the total concentration.

Initially, the total concentration was:

\[
[A]+[B]+[C]=1+1+4=6
\]

Adding 3.00 mol of B to the 1.00 L vessel increases the total concentration to:

\[
[A]+[B]+[C]=9.00
\]

This total remains constant during the subsequent reactions because A, B, and C are isomers of the same molecule.

Let the final concentration of B be \(x\).

Then:

\[
[A]=x
\]

\[
[B]=x
\]

\[
[C]=4x
\]

Substituting into the conservation equation:

\[
x+x+4x=9.00
\]

\[
6x=9.00
\]

\[
x=1.50
\]

Therefore:

\[
\boxed{[A]=1.50\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[B]=1.50\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[C]=6.00\ \mathrm{mol\,L^{-1}}}
\]

Check the two equilibrium conditions:

\[
K_{c,1}=\frac{1.50}{1.50}=1.00
\]

\[
K_{c,2}=\frac{6.00}{1.50}=4.00
\]

Both conditions are satisfied.

Part d: Evaluate the student’s claim.

The student correctly recognises that adding B encourages the second reaction to proceed forwards.

However, B is also a product of the first reaction.

Increasing its concentration therefore encourages the first reaction to proceed backwards.

The two reactions respond simultaneously:

• Reaction 1: B is consumed to produce A.
• Reaction 2: B is consumed to produce C.

The mistake is treating a coupled system as if it contains only one equilibrium.

There is no single direction in which the entire reaction network shifts. Each reversible reaction has its own reaction quotient, equilibrium constant, and net reaction direction.

This is why checking each equilibrium separately is essential when a species participates in multiple reactions.

09The final distinction to remember

When answering an equilibrium question, keep three things separate:

1. The disturbance: What physically entered, left, or changed volume?
2. The immediate concentration change: What are the concentrations before any appreciable chemical reaction occurs?
3. The subsequent equilibrium shift: Which reaction direction becomes faster, and how does the system respond?

At constant temperature, use the immediate concentrations to calculate \(Q_c\), then compare it with \(K_c\).

If temperature also changes, there is another step: the equilibrium constant itself may change. You must establish the appropriate new \(K_c\) before deciding the net reaction direction.

Getting these distinctions right gives you a much stronger method for tackling equilibrium questions involving simultaneous changes, dilution, catalysts, and competing reactions.