Why Removing Product Increases Further Product Formation

A reaction mixture is at equilibrium. You remove some of its product, seal the vessel again, and wait. Surprisingly, the reaction starts producing more product.

Why would a reaction make more of something you have just removed?

The answer is in the competition between the forward and reverse reaction rates. Removing a product can slow the reverse reaction without immediately changing the forward reaction. The forward reaction then becomes faster than the reverse reaction, producing more product until a new equilibrium is established.

This is also why chemical manufacturers can increase the total amount of product collected by continuously separating products from a reaction mixture.

There is an important qualification, though: **producing more product does not necessarily mean having more product inside the reaction vessel.** We need to distinguish the amount formed, the amount remaining in the reactor, and the amount collected outside it.

## 1. Why removing product changes the reaction rates

Consider a reversible reaction:

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

The forward reaction combines A and B to produce C. The reverse reaction breaks C apart to regenerate A and B.

At dynamic equilibrium, both reactions are still occurring, but their rates are equal:

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

There is no overall change in concentration because the forward reaction produces substances at exactly the rate that the reverse reaction consumes them.

Now imagine that you selectively remove some C from the mixture.

Predict what happens to each reaction rate immediately afterwards.

- The forward reaction initially has the same concentrations of A and B available, so its rate is unchanged.
- The reverse reaction has less C available, so its rate decreases.

Consequently:

\[
r_{\mathrm{forward}}>r_{\mathrm{reverse}}
\]

The forward reaction now occurs faster than the reverse reaction. There is a net conversion of A and B into C.

As C accumulates again, the reverse reaction speeds up. Eventually, the forward and reverse rates become equal, establishing a new equilibrium.

[IMAGE: Show a reversible reaction as two opposing molecular pathways. On the forward pathway, A and B molecules combine to form C. On the reverse pathway, C separates into A and B. Show several C molecules being selectively removed from the reaction mixture, leaving fewer available for the reverse pathway. Include arrows indicating the initially unchanged forward rate and reduced reverse rate.]

### A useful analogy: two competing conveyor belts

Imagine two conveyor belts moving objects between two rooms.

- The forward belt carries A and B into a machine that produces C.
- The reverse belt carries C back through a machine that separates it into A and B.

At equilibrium, both belts process objects at the same rate.

If someone removes some C before it reaches the reverse belt, fewer objects are available for the reverse process. The forward belt initially keeps operating at its original rate, so more C accumulates.

The analogy helps explain the imbalance, but real reaction rates depend on molecular collisions, concentrations, temperature, and the reaction mechanism. Molecules are not literally travelling along conveyor belts.

### The important role of the system boundary

Before explaining product removal, identify exactly what belongs to the reaction system.

Imagine a rigid reaction vessel connected to a selective outlet. The vessel contains A, B, and C. A separation mechanism allows C to leave and enter a collection container, while A and B remain in the reaction vessel.

For this explanation, the reaction system is the contents of the reaction vessel. The collection container is outside that boundary.

[IMAGE: Annotated apparatus diagram showing a rigid reaction vessel containing A, B, and C, connected through a selective product outlet to an external collection vessel. Draw a dashed system boundary around the reaction vessel only. Show C crossing the boundary into the receiver, while A and B remain inside. Label the reaction vessel, collection vessel, product outlet, and system boundary.]

When C crosses the boundary, its amount inside the reaction vessel decreases.

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 the reaction vessel volume in L.

If the vessel volume stays constant, selectively removing C decreases its concentration. The concentrations of A and B do not change immediately.

This is different from withdrawing an ordinary sample of the entire mixture. If you remove 10% of a well-mixed solution, you remove 10% of every dissolved species and 10% of the solution volume. The concentrations remaining in the vessel are unchanged.

The distinction is between removing a particular product and removing a representative sample of the whole mixture.

For a more detailed treatment of these immediate changes, see [Instantaneous Concentration Changes vs Equilibrium Shifts](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-distinguish-an-instantaneous-concentration-change-from-a-later-equilibrium-shift/).

## 2. Following the reaction after product removal

We can make the rate changes more precise using a simple kinetic model.

For the illustrative reaction:

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

assume the forward and reverse steps are elementary, with rate laws:

\[
r_f=k_f[A][B]
\]

\[
r_r=k_r[C]
\]

Here:

- \(r_f\) is the forward reaction rate.
- \(r_r\) is the reverse reaction rate.
- \(k_f\) and \(k_r\) are the respective rate constants.
- Square brackets represent concentrations in \(\mathrm{mol\,L^{-1}}\).

These rate laws are assumptions for this particular model. A balanced chemical equation alone does not establish the rate law of a real reaction.

We will define the net reaction rate as positive in the forward direction:

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

Therefore:

| Rate comparison | Net reaction | Result |
|---|---|---|
| \(r_f>r_r\) | Forward | More C is formed |
| \(r_f<r_r\) | Reverse | C is consumed |
| \(r_f=r_r\) | Neither direction dominates | Dynamic equilibrium |

Immediately after selectively removing C:

- \([A]\) remains unchanged.
- \([B]\) remains unchanged.
- \([C]\) decreases.
- \(r_f\) initially remains unchanged.
- \(r_r\) decreases.

The net reaction rate becomes positive, meaning that C is being formed overall.

### What happens as equilibrium is restored?

The forward reaction consumes A and B and produces C.

As this happens:

1. The concentrations of A and B decrease.
2. The concentration of C increases.
3. The forward reaction rate decreases.
4. The reverse reaction rate increases.

Eventually:

\[
r_f=r_r
\]

The system has reached a new dynamic equilibrium.

[IMAGE: Draw a reaction-rate-versus-time graph for a reversible reaction after selective product removal. Before the disturbance, forward and reverse rates are equal at a non-zero value. At the instant of removal, show the reverse rate dropping vertically while the forward rate remains unchanged. Subsequently, show the forward rate decreasing and reverse rate increasing until both meet at a new equal, non-zero rate. Mark the instant of product removal and the new equilibrium.]

Notice that the new equilibrium does not require the two rates to return to their original values. They only need to become equal again.

This is the central mechanism: product removal creates an initial difference between the forward and reverse rates. The reaction then proceeds forwards until that difference disappears.

### An illustrative set of rate measurements

Suppose a rigid vessel initially contains an equilibrium mixture with:

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

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

The forward and reverse rates are initially both \(0.060\ \mathrm{mol\,L^{-1}\,s^{-1}}\).

A selective separation process removes some C. The following table shows illustrative model values, not measurements from an actual experiment.

| Stage | \([A]\) | \([B]\) | \([C]\) | \(r_f\) | \(r_r\) |
|---|---:|---:|---:|---:|---:|
| Initial equilibrium | 0.40 | 0.40 | 0.60 | 0.060 | 0.060 |
| Immediately after removal | 0.40 | 0.40 | 0.36 | 0.060 | 0.036 |
| During forward reaction | 0.38 | 0.38 | 0.38 | 0.054 | 0.038 |
| New equilibrium | - | - | - | Equal | Equal |

All concentrations are in \(\mathrm{mol\,L^{-1}}\), and rates are in \(\mathrm{mol\,L^{-1}\,s^{-1}}\). The intermediate row is illustrative rather than a complete calculated equilibrium trajectory.

The key observation is the immediate rate difference:

\[
r_{\mathrm{net}}=0.060-0.036
\]

\[
r_{\mathrm{net}}=0.024\ 
\mathrm{mol\,L^{-1}\,s^{-1}}
\]

The positive result confirms that the net reaction is proceeding forwards.

The concentrations in the final row are deliberately not specified by this table. Equal rates identify equilibrium, but calculating the final concentrations requires additional equilibrium information.

## 3. Does the product concentration increase above its original value?

Not necessarily. This is where students often confuse the amount of product formed with the amount remaining in the vessel.

Consider two separate events:

**Event 1: Product removal**

The amount of C in the vessel decreases immediately. Some C has crossed the system boundary and is now in the collection container.

**Event 2: The forward reaction**

The reaction produces additional C. Its concentration inside the vessel increases from its reduced value.

The second event partially replaces the product that was removed. It does not automatically restore the original concentration.

For example, suppose a vessel initially contains 0.50 mol of C. You remove 0.30 mol, leaving 0.20 mol inside.

If the forward reaction subsequently forms another 0.10 mol of C, the vessel now contains:

\[
n(C)=0.20+0.10=0.30\ \mathrm{mol}
\]

The reaction has formed an additional 0.10 mol of product, but the vessel still contains less C than it did before the removal.

Meanwhile, the collection container holds the 0.30 mol that was withdrawn.

The distinction can be summarised as follows:

| Quantity | What it measures |
|---|---|
| Product in the vessel | C currently remaining in the reacting mixture |
| Product collected | C transferred outside the reaction vessel |
| Additional product formed | C produced by the net forward reaction after removal |

[IMAGE: Side-by-side comparison of two reaction vessels. In the first, show a single product withdrawal followed by partial re-formation of C and a new equilibrium. In the second, show repeated product withdrawal, with C continually entering an external collection vessel and the forward reaction continuing. Clearly distinguish product remaining inside from cumulative product collected.]

### Worked example: How much product is re-formed?

A rigid 1.00 L vessel initially contains an equilibrium mixture for:

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

The initial concentrations are:

\[
[P]=[Q]=[R]=0.50\ \mathrm{mol\,L^{-1}}
\]

A selective separation process removes 0.30 mol of R. The temperature and vessel volume remain constant.

Immediately after removal, determine the concentrations and explain what happens next.

**Step 1**: Calculate the immediate concentration change.

The amount removed is 0.30 mol, and the vessel volume is 1.00 L.

\[
\Delta[R]=\frac{0.30}{1.00}
=0.30\ \mathrm{mol\,L^{-1}}
\]

Therefore:

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

\[
[R]=0.50-0.30=0.20
\]

Only R changes immediately.

**Step 2**: Compare the forward and reverse rates.

The concentrations of P and Q are unchanged, so the forward rate initially remains unchanged.

However, the concentration of R has decreased. The reverse reaction therefore becomes slower.

Consequently:

\[
r_f>r_r
\]

The net reaction proceeds forwards, consuming P and Q and producing R.

**Step 3**: Explain the final result.

As R is produced, its concentration increases and the reverse reaction becomes faster.

At the same time, P and Q are consumed, reducing the forward reaction rate.

The reaction continues until:

\[
r_f=r_r
\]

Some of the removed R is therefore regenerated, although the final amount depends on the equilibrium constant.

The important point is that product removal initiates further product formation; it does not guarantee that the original concentration will be restored.

## 4. Calculating the new equilibrium after product removal

We can go one step further and calculate exactly how much product is regenerated.

For a reaction:

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

the equilibrium constant is:

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

Here, \(K_c\) describes the equilibrium composition at a particular temperature. Its value does not change simply because product is removed.

Suppose the initial equilibrium concentrations are:

\[
[P]=[Q]=[R]=0.50\ \mathrm{mol\,L^{-1}}
\]

The original equilibrium constant is:

\[
K_c=\frac{0.50}{0.50\times0.50}
\]

\[
K_c=2.0\ \mathrm{L\,mol^{-1}}
\]

After removing 0.30 mol L\(^{-1}\) of R, the immediate concentrations are:

\[
[P]=[Q]=0.50
\]

\[
[R]=0.20
\]

The forward reaction then proceeds.

Let \(x\) represent the concentration of R formed as the system moves towards its new equilibrium.

Because the balanced equation has a 1:1:1 mole ratio:

| Species | Immediately after removal | Change | New equilibrium |
|---|---:|---:|---:|
| P | 0.50 | \(-x\) | \(0.50-x\) |
| Q | 0.50 | \(-x\) | \(0.50-x\) |
| R | 0.20 | \(+x\) | \(0.20+x\) |

### Worked example: Calculate the new equilibrium concentrations

Using the information above, calculate the new equilibrium concentrations of P, Q, and R.

**Step 1**: Substitute the equilibrium concentrations into the \(K_c\) expression.

The temperature is unchanged, so \(K_c=2.0\ \mathrm{L\,mol^{-1}}\).

\[
2.0=\frac{0.20+x}{(0.50-x)^2}
\]

**Step 2**: Rearrange the equation.

\[
2.0(0.50-x)^2=0.20+x
\]

Expanding and collecting terms:

\[
2x^2-3x+0.30=0
\]

**Step 3**: Solve the quadratic equation.

Using the quadratic formula:

\[
x=\frac{3\pm\sqrt{9-4(2)(0.30)}}{4}
\]

\[
x=\frac{3\pm\sqrt{6.6}}{4}
\]

The two mathematical solutions are approximately:

\[
x=0.108 \quad \text{or} \quad x=1.392
\]

The second solution is physically impossible because it would make the concentrations of P and Q negative.

Therefore:

\[
x=0.108\ \mathrm{mol\,L^{-1}}
\]

**Step 4**: Calculate the final concentrations.

\[
[P]=0.50-0.108
=0.392\ \mathrm{mol\,L^{-1}}
\]

\[
[Q]=0.50-0.108
=0.392\ \mathrm{mol\,L^{-1}}
\]

\[
[R]=0.20+0.108
=0.308\ \mathrm{mol\,L^{-1}}
\]

<badge color="success">New equilibrium</badge>

\[
\boxed{[P]=[Q]=0.392\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[R]=0.308\ \mathrm{mol\,L^{-1}}}
\]

The reaction has regenerated approximately 0.108 mol of R.

However, the final concentration of R is still below its original value of \(0.50\ \mathrm{mol\,L^{-1}}\). The reaction has only partially replaced the removed product.

This is exactly why you must distinguish the amount regenerated from the amount present before the disturbance.

05Why continuous product removal can increase total yield

The difference between removing product once and removing it continuously is important.

Consider two ways of operating a reaction.

A batch reaction with one-off product removal

1. Allow the reaction to reach equilibrium.
2. Remove some product.
3. Close the vessel.
4. Allow the forward reaction to regenerate some product.
5. Wait until the new equilibrium is reached.

The reaction produces some additional product, but eventually the forward and reverse rates become equal again.

A reaction with continuous product separation

1. Supply reactants to the reaction vessel.
2. Allow them to react.
3. Continuously separate product C from the reacting mixture.
4. Collect C outside the reaction vessel.
5. Maintain the reactant supply so that further reaction can occur.

Because C is continually removed, its concentration in the reacting mixture can be kept relatively low. This suppresses the reverse reaction and allows further net production of C.

This is the principle behind using product separation to improve the yield of a reversible reaction.

However, there is a limit. Removing C cannot produce an unlimited amount of product. Eventually, reactants become depleted, and the reaction is restricted by the available starting materials, reaction conditions, and the effectiveness of the separation process.

A constant product concentration does not mean production has stopped

Here is another subtle point.

Imagine a continuous reactor in which C is being formed and removed at the same time.

The amount of C inside the reactor might remain constant. Does that mean the chemical reaction has stopped?

No. It could mean that C is being formed at exactly the rate at which it leaves the reactor.

We can describe this using a simple material balance:

\[ \frac{\Delta n_C}{\Delta t} = \text{formation rate} – \text{removal rate} \]

Here, both rates are expressed as amounts per unit time, such as \(\mathrm{mol\\,min^{-1}}\).

If the amount of C inside the reactor stays constant:

\[ \text{formation rate} = \text{removal rate} \]

Both rates can be non-zero.

This is called a steady state. It is not necessarily chemical equilibrium.

The distinction is important:

• At chemical equilibrium, the forward and reverse reaction rates are equal.
• At steady state in an operating reactor, concentrations can remain constant because material entering and leaving balances the chemical changes.

The whole continuous reactor is an open system because matter crosses its boundary.

06When removing product does not increase product formation

The explanation so far assumes that removing the product actually decreases its concentration or partial pressure in the reacting mixture.

That assumption matters. Not every removal causes an equilibrium shift.

Case 1: Removing a reactant instead of a product

Consider:

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

Suppose you selectively remove A.

The forward reaction initially slows because less A is available. The reverse reaction initially remains unchanged because the concentration of C has not changed.

Therefore:

\[ r_{\mathrm{reverse}}>r_{\mathrm{forward}} \]

The net reaction proceeds backwards.

C is consumed, producing more A and B.

Notice the opposite result:

The reaction responds to which species has been removed, not simply to the fact that something has been taken away.

Case 2: Removing a pure solid product

There is another important exception in heterogeneous equilibria.

Consider the decomposition of calcium carbonate:

\[ \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \]

Suppose some calcium oxide, \(\ce{CaO(s)}\), is removed while some solid calcium oxide remains.

Does this necessarily shift the equilibrium forwards?

No.

The equilibrium expression is:

\[ K_c=[\ce{CO2}] \]

The pure solids do not appear in the equilibrium expression because their activities are constant.

Removing some solid \(\ce{CaO}\), while the solid phase remains present, does not change the equilibrium concentration of carbon dioxide.

This is different from removing a gaseous or dissolved product whose concentration or partial pressure changes.

For HSC equilibrium questions, remember to distinguish between:

• A product whose concentration or partial pressure can change when removed.
• A pure solid or liquid whose activity remains constant while that phase is present.

A quick decision process

Before predicting what happens when a product is removed, work through these questions:

For a concentration-time graph, remember that product removal produces an immediate decrease in product concentration, followed by a gradual increase as the forward reaction regenerates it.

You can explore the graph behaviour in How to Interpret a Concentration-Time Graph Approaching Equilibrium.

03Questions and solutions

The following questions progress from a straightforward rate comparison to problems involving system boundaries, continuous production, and coupled equilibria.

Question 1

A rigid 2.00 L vessel contains an equilibrium mixture undergoing:

\[ \ce{X(aq) + Y(aq) <=> Z(aq)} \]

A selective separation process removes 0.20 mol of Z without changing the vessel volume or temperature.

Explain:

a) The immediate concentration changes.

b) The effect on the forward and reverse reaction rates.

c) The direction of the subsequent net reaction.

Solution 1

The reaction proceeds forwards after the selective removal of Z. The reverse reaction initially slows, while the forward reaction initially remains unchanged.

Part a: Immediate concentration changes

The concentration change of Z is:

\[ \Delta[Z]=\frac{0.20}{2.00} \]

\[ \Delta[Z]=0.10\ \mathrm{mol\\,L^{-1}} \]

Therefore, the concentration of Z immediately decreases by \(0.10\ \mathrm{mol\\,L^{-1}}\).

The concentrations of X and Y remain unchanged because neither substance has crossed the system boundary.

Part b: Reaction rates

The forward reaction rate initially remains unchanged because the concentrations of X and Y are unchanged.

The reverse reaction rate decreases because less Z is available to decompose into X and Y.

Part c: Net reaction

Therefore:

\[ r_f>r_r \]

The net reaction proceeds forwards.

X and Y are consumed, while Z is regenerated. The reaction continues until the forward and reverse rates become equal again.

Question 2

A rigid vessel contains the following equilibrium:

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

At a particular temperature, the forward and reverse reaction rates are initially both:

\[ 0.050\ \mathrm{mol\\,L^{-1}\\,s^{-1}} \]

The concentrations are:

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

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

For this question, the rate laws are:

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

\[ r_r=k_r[N] \]

A selective separation process reduces the concentration of N to \(0.12\ \mathrm{mol\\,L^{-1}}\), without changing the volume or temperature.

Calculate:

a) The reverse rate constant, \(k_r\).

b) The immediate reverse reaction rate after removal.

c) The immediate net reaction rate and its direction.

d) Explain why the forward rate constant does not change.

Solution 2

The net reaction proceeds forwards at an immediate rate of \(0.020\ \mathrm{mol\\,L^{-1}\\,s^{-1}}\). The reduction in N decreases the reverse reaction rate, while the forward rate remains unchanged.

Part a: Calculate the reverse rate constant

The reverse rate law is:

\[ r_r=k_r[N] \]

Rearranging:

\[ k_r=\frac{r_r}{[N]} \]

Substituting the initial equilibrium values:

\[ k_r=\frac{0.050}{0.20} \]

\[ \boxed{k_r=0.25\ \mathrm{s^{-1}}} \]

Part b: Calculate the new reverse rate

Immediately after removal:

\[ [N]=0.12\ \mathrm{mol\\,L^{-1}} \]

The reverse rate is:

\[ r_r=0.25\times0.12 \]

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

Part c: Calculate the net reaction rate

The forward rate initially remains:

\[ r_f=0.050\ \mathrm{mol\\,L^{-1}\\,s^{-1}} \]

Using the positive-forward sign convention:

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

\[ r_{\mathrm{net}}=0.050-0.030 \]

\[ \boxed{r_{\mathrm{net}}=+0.020\ \mathrm{mol\\,L^{-1}\\,s^{-1}}} \]

The positive sign indicates a net forward reaction.

More N is formed, while M is consumed.

Part d: Explain the rate constant

The forward rate constant, \(k_f\), depends on temperature and the reaction mechanism.

Neither has changed.

Removing N changes its concentration, not the forward rate constant. Since the concentrations of M also remain unchanged, the forward reaction rate is initially unchanged.

This distinction is important: a change in reaction rate does not necessarily mean that the rate constant has changed.

Question 3

A 1.00 L vessel contains an aqueous equilibrium mixture:

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

Its initial concentrations are:

\[ [A]=0.20,\quad [B]=0.30,\quad [C]=0.40 \]

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

A student performs two separate operations on identical equilibrium mixtures.

Experiment A: The student uses a pipette to withdraw 0.10 L of the well-mixed solution.

Experiment B: A selective separation process removes 0.10 mol of C without removing any A or B or changing the solution volume.

The student argues:

“Both experiments remove some C, so both must shift the equilibrium forwards.”

Evaluate this claim. For each experiment, determine the immediate concentrations and explain whether a subsequent net reaction occurs.

Solution 3

The student is incorrect. Experiment A does not cause an equilibrium shift, whereas Experiment B causes a net forward reaction. The difference is whether the concentrations change.

Experiment A: Withdrawing a representative sample

The original volume is 1.00 L, and the withdrawn volume is 0.10 L.

Therefore, 10% of the solution is removed.

The remaining volume is:

\[ V=1.00-0.10=0.90\ \mathrm{L} \]

The remaining amounts of each species are:

\[ n(A)=0.20\times0.90=0.18\ \mathrm{mol} \]

\[ n(B)=0.30\times0.90=0.27\ \mathrm{mol} \]

\[ n(C)=0.40\times0.90=0.36\ \mathrm{mol} \]

Calculate the remaining concentrations:

\[ [A]=\frac{0.18}{0.90}=0.20\ \mathrm{mol\\,L^{-1}} \]

\[ [B]=\frac{0.27}{0.90}=0.30\ \mathrm{mol\\,L^{-1}} \]

\[ [C]=\frac{0.36}{0.90}=0.40\ \mathrm{mol\\,L^{-1}} \]

All three concentrations remain unchanged.

The forward and reverse reaction rates therefore remain equal. No net reaction occurs.

The student has confused removing some product with selectively removing product.

Experiment B: Selectively removing C

The original volume remains 1.00 L.

The amount of C removed is 0.10 mol, so:

\[ \Delta[C]=\frac{0.10}{1.00} =0.10\ \mathrm{mol\\,L^{-1}} \]

Immediately after removal:

\[ [A]=0.20 \]

\[ [B]=0.30 \]

\[ [C]=0.40-0.10=0.30 \]

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

The forward reaction rate initially remains unchanged because A and B are unchanged.

The reverse reaction rate decreases because the concentration of C has decreased.

Therefore:

\[ r_f>r_r \]

The net reaction proceeds forwards, consuming A and B and regenerating C.

The two experiments produce different outcomes because they involve different changes to the system boundary.

Question 4

An industrial-style continuous reactor carries out the reversible reaction:

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

Reactants A and B are continuously supplied to the reaction zone. Product C is continuously removed through a separation system.

Assume:

• The reaction-zone volume remains constant at 1.00 L.
• No C enters with the reactant feed.
• There are no side reactions.
• The concentration of C remains constant at \(0.10\ \mathrm{mol\\,L^{-1}}\).
• The collection system receives 0.48 mol of C over 10.0 minutes.

A student concludes:

“The concentration of C is constant, so the reaction must be at equilibrium and no more product is being formed.”

a) Calculate the average rate at which C is collected.

b) Determine the net amount of C formed inside the reaction zone over the 10.0 minutes.

c) Explain why the student’s conclusion is incorrect.

d) Explain how continuous product removal can allow greater cumulative production than a closed batch reaction.

Solution 4

The reactor produces C continuously, even though its concentration remains constant. The product formation rate balances the product removal rate, creating a steady state rather than necessarily establishing chemical equilibrium.

Part a: Average collection rate

The average rate of collection is:

\[ \text{Rate}=\frac{\text{amount collected}}{\text{time}} \]

Substituting:

\[ \text{Rate}=\frac{0.48}{10.0} \]

\[ \boxed{0.048\ \mathrm{mol\\,min^{-1}}} \]

Part b: Net amount of C formed

The concentration of C remains constant, and the reaction-zone volume remains constant.

Therefore, the amount of C inside the reactor also remains constant.

Using the material balance:

\[ \Delta n_C=n_{\mathrm{formed}}-n_{\mathrm{removed}} \]

Since \(\Delta n_C=0\):

\[ 0=n_{\mathrm{formed}}-0.48 \]

Therefore:

\[ \boxed{n_{\mathrm{formed}}=0.48\ \mathrm{mol}} \]

A total of 0.48 mol of C has been formed during the 10.0 minutes.

The average net formation rate is \(0.048\ \mathrm{mol\\,min^{-1}}\).

Part c: Explain why the student is incorrect

A constant concentration only means that there is no net accumulation of C inside the reactor.

It does not mean that C is not being formed.

In this case, the net formation of C is balanced by its removal. The forward reaction is producing C while the separation system continuously removes it.

This is a steady state in an open system, not necessarily dynamic equilibrium.

At chemical equilibrium, the forward and reverse reaction rates would be equal. Here, the net chemical production of C is positive.

Part d: Explain the effect of continuous product removal

Continuously removing C keeps its concentration relatively low in the reaction zone.

This reduces the reverse reaction rate, allowing the forward reaction to continue producing C.

More reactants can consequently be converted into product, and the collection tank can accumulate a greater total amount of C than would be produced in a comparable closed batch reaction.

The advantage is in the cumulative amount collected, not necessarily the concentration of C inside the reactor.

Question 5

A reaction system contains three dissolved substances, A, B, and C, which are isomers of the same molecular formula.

Two reversible reactions occur:

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

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

The reactions have the following equilibrium constants at a fixed temperature:

\[ K_{c,1}=\frac{[B]}{[A]}=2.0 \]

\[ K_{c,2}=\frac{[C]}{[B]}=3.0 \]

A rigid 1.00 L vessel initially contains the system at equilibrium, with:

\[ [A]=0.10\ \mathrm{mol\\,L^{-1}} \]

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

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

A selective separation process removes 0.30 mol of C without changing the volume or temperature.

Assume there are no other reactions, and the total amount of A, B, and C inside the vessel is conserved during the subsequent reactions.

A student makes the following prediction:

“Removing C makes both reactions proceed forwards immediately. The system will therefore rapidly regenerate C until its original concentration is restored.”

Evaluate the prediction.

Determine:

a) Which reaction initially has unequal forward and reverse rates immediately after removal.

b) Why the other reaction does not necessarily shift immediately.

c) How the two reactions subsequently interact.

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

e) Whether the final concentration of C returns to its original value, and how much C has been regenerated.

Solution 5

The student’s prediction is partly correct but contains two important errors: the first reaction does not shift immediately, and the final concentration of C does not return to its original value. Both reactions eventually proceed forwards, but the second reaction responds first and causes the first reaction to become unbalanced.

Part a: Identify the initially affected reaction

Immediately after removing 0.30 mol of C from the 1.00 L vessel:

\[ [A]=0.10\ \mathrm{mol\\,L^{-1}} \]

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

\[ [C]=0.60-0.30=0.30\ \mathrm{mol\\,L^{-1}} \]

Consider the second reaction:

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

Its reaction quotient immediately after removal is:

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

\[ Q_{c,2}=\frac{0.30}{0.20}=1.5 \]

Since:

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

the forward reaction is favoured.

The forward reaction converts B into C.

Part b: Why the first reaction does not immediately shift

Now consider:

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

Its reaction quotient is:

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

Immediately after removal:

\[ Q_{c,1}=\frac{0.20}{0.10}=2.0 \]

This is still equal to its equilibrium constant:

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

Therefore, the first reaction initially remains at equilibrium.

Removing C has not directly changed the concentrations of A or B.

This is the first trap in the question. It is tempting to assume that removing the final product immediately pushes every reaction in a connected sequence forwards. However, each reaction responds to the concentrations of the substances involved in that particular reaction.

Part c: Explain how the two reactions interact

The second reaction initially proceeds forwards:

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

This consumes B.

As the concentration of B decreases while A initially remains unchanged, the reaction quotient for the first reaction decreases:

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

It becomes smaller than \(K_{c,1}\).

Consequently, the first reaction also proceeds forwards:

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

The first reaction replenishes B, which can then react to form more C.

Both reactions therefore eventually proceed forwards.

The overall effect is the conversion of A into C through the intermediate B.

Part d: Calculate the new equilibrium concentrations

The total concentration initially inside the vessel is:

\[ [A]+[B]+[C]=0.10+0.20+0.60 \]

\[ =0.90\ \mathrm{mol\\,L^{-1}} \]

After removing 0.30 mol of C from the 1.00 L vessel, the total concentration becomes:

\[ 0.90-0.30=0.60\ \mathrm{mol\\,L^{-1}} \]

The total remains constant during the subsequent reactions because A, B, and C are isomers, and both reactions have a 1:1 stoichiometric ratio.

At the new equilibrium:

\[ [B]=2[A] \]

and:

\[ [C]=3[B] \]

Substituting the first relationship into the second:

\[ [C]=3(2[A]) \]

\[ [C]=6[A] \]

Therefore, the final concentrations have the ratio:

\[ [A]:[B]:[C]=1:2:6 \]

There are nine concentration units altogether.

Since their total is \(0.60\ \mathrm{mol\\,L^{-1}}\):

\[ [A]=\frac{0.60}{9} \]

\[ [A]=0.0667\ \mathrm{mol\\,L^{-1}} \]

Then:

\[ [B]=2(0.0667) \]

\[ [B]=0.133\ \mathrm{mol\\,L^{-1}} \]

And:

\[ [C]=6(0.0667) \]

\[ [C]=0.400\ \mathrm{mol\\,L^{-1}} \]

Therefore:

\[ \boxed{[A]=0.0667\ \mathrm{mol\\,L^{-1}}} \]

\[ \boxed{[B]=0.133\ \mathrm{mol\\,L^{-1}}} \]

\[ \boxed{[C]=0.400\ \mathrm{mol\\,L^{-1}}} \]

Both equilibrium conditions are satisfied:

\[ K_{c,1}=\frac{0.133}{0.0667}\approx2.0 \]

\[ K_{c,2}=\frac{0.400}{0.133}\approx3.0 \]

Part e: Evaluate the final amount of C

Immediately after removal, the vessel contains 0.30 mol of C.

At the new equilibrium, it contains 0.40 mol.

Therefore, the amount regenerated is:

\[ n(C)_{\mathrm{formed}}=0.40-0.30 \]

\[ \boxed{n(C)_{\mathrm{formed}}=0.10\ \mathrm{mol}} \]

The final concentration of C is still lower than its original value of \(0.60\ \mathrm{mol\\,L^{-1}}\).

However, the collection vessel contains the 0.30 mol of C removed earlier, while the reaction vessel contains another 0.40 mol.

The total amount of C across both vessels is:

\[ n(C)_{\mathrm{total}}=0.30+0.40 \]

\[ \boxed{n(C)_{\mathrm{total}}=0.70\ \mathrm{mol}} \]

Initially, there was only 0.60 mol of C.

The system has therefore produced a net additional 0.10 mol of C, even though the concentration of C remaining inside the reaction vessel is lower than it was originally.

This is the key conceptual distinction: removing a product can drive further product formation, but the final concentration inside the reactor, the amount regenerated, and the cumulative amount collected are three different quantities.

04Where to go next

Once you understand how product removal changes the forward and reverse rates, the next useful skill is interpreting how those rates appear on concentration-time graphs.

In particular, learn to distinguish an instantaneous decrease caused by removing a substance from the gradual concentration changes caused by the subsequent chemical reaction.

That distinction is essential when explaining equilibrium disturbances in HSC Chemistry, especially when a question combines selective separation, changing concentrations, and the establishment of a new equilibrium.