How to Set Up an ICE Table When Product Is Present Initially

A reaction mixture can already contain plenty of product and still produce more. It can also contain product and use some of it up. The amount of product present at the start does not tell you which direction the reaction will move.

This is the detail that catches students when setting up an ICE table. They see product in the initial mixture and assume its concentration must decrease. That is not necessarily true.

The correct approach is to compare the initial reaction quotient, \(Q_c\), with the equilibrium constant, \(K_c\). This tells you the direction of the net reaction and, therefore, the signs in your change row.

Once you know the direction, setting up the ICE table is straightforward. The harder part is solving the resulting equation without accepting a mathematically valid but physically impossible answer.

## 1. Setting up an ICE table when product is present initially

Consider the reversible reaction:

\[
\mathrm{A + B \rightleftharpoons C}
\]

Suppose a flask initially contains all three substances. You know the starting concentrations of A, B, and C, and you are given the value of \(K_c\).

Your job is to calculate their concentrations once equilibrium is established.

An ICE table has three rows:

- **Initial (I):** The concentrations before the mixture has shifted towards equilibrium.
- **Change (C):** The increases or decreases in concentration as the reaction moves towards equilibrium.
- **Equilibrium (E):** The final concentrations after the forward and reverse reaction rates become equal.

The important point is that the initial row includes every substance already in the mixture, including products.

[IMAGE: A reversible reaction in a sealed laboratory flask containing particles of A, B, and C from the beginning. Show A and B as separate reactant particles and C as a bonded product particle. Include a small, clearly labelled three-row ICE table beside the flask, with the initial row containing non-zero concentrations for all three substances.]

For example, suppose the initial concentrations are:

\[
[\mathrm A]_0=0.20\ \mathrm{mol\,L^{-1}}
\]

\[
[\mathrm B]_0=0.20\ \mathrm{mol\,L^{-1}}
\]

\[
[\mathrm C]_0=0.30\ \mathrm{mol\,L^{-1}}
\]

Notice that C is already present. You must enter \(0.30\ \mathrm{mol\,L^{-1}}\) in the initial row, not zero.

But what should you put in the change row?

You cannot decide that just by looking at the reaction equation. First, you need to establish which direction the reaction will move.

## 2. Decide which direction the reaction will move

The reaction quotient, \(Q_c\), measures the ratio of product concentrations to reactant concentrations at a particular moment.

For the reaction:

\[
\mathrm{A+B\rightleftharpoons C}
\]

the reaction quotient is:

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

It has the same mathematical form as \(K_c\), but there is an important difference:

- \(Q_c\) uses the concentrations at the time you are examining.
- \(K_c\) uses the equilibrium concentrations and is fixed for a particular reaction at a particular temperature.

Compare the initial reaction quotient with the equilibrium constant.

| Comparison | Net reaction direction | What happens to product C? |
|---|---|---|
| \(Q_c<K_c\) | Forward | C increases |
| \(Q_c>K_c\) | Reverse | C decreases |
| \(Q_c=K_c\) | Already at equilibrium | No net change |

[IMAGE: A clear three-panel comparison of reversible reaction mixtures. Panel one shows Qc less than Kc, with reactants converting into additional product. Panel two shows Qc greater than Kc, with product converting into reactants. Panel three shows Qc equal to Kc, with no net change. Use particle arrangements to show the direction of net change, not just arrows.]

Think of \(Q_c\) as a snapshot of the mixture and \(K_c\) as the ratio the mixture must reach at equilibrium.

If the initial ratio is too small, the forward reaction increases the ratio. If it is too large, the reverse reaction decreases it.

This gives you the sign convention for the change row.

### Choosing the signs in the change row

For clarity, we'll define \(x\) as a positive amount of reaction progress in the direction the mixture actually shifts.

For the general reaction:

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

the two possible change rows are:

| Row | Forward reaction | Reverse reaction |
|---|---|---|
| Initial | \(a,\ b,\ c\) | \(a,\ b,\ c\) |
| Change | \(-x,\ -x,\ +x\) | \(+x,\ +x,\ -x\) |
| Equilibrium | \(a-x,\ b-x,\ c+x\) | \(a+x,\ b+x,\ c-x\) |

Here, \(a\), \(b\), and \(c\) are the initial concentrations of A, B, and C, measured in M.

For reactions with coefficients other than one, multiply each change by its coefficient. For example:

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

The forward change row is \(-2x,+x\), while the reverse change row is \(+2x,-x\).

This is a direct consequence of the balanced equation. The coefficients determine the relative amounts consumed and produced.

If you need to revise how these coefficients work, see [How to Use Stoichiometric Coefficients in an ICE Table](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-use-stoichiometric-coefficients-in-the-change-row-of-an-ice-table/).

[IMAGE: Two side-by-side annotated ICE tables for A + B ⇌ C. Show identical non-zero initial concentrations in both tables. In the forward table, A and B decrease while C increases. In the reverse table, A and B increase while C decreases. Include arrows between reactant and product particle arrangements to make the opposite directions clear.]

There is one more detail worth remembering: if \(Q_c=K_c\), the mixture is already at equilibrium. Its net concentration changes are zero. The forward and reverse reactions still occur, but at equal rates.

## 3. What an equilibrium experiment would actually measure

An ICE table is a mathematical model of what happens inside a reaction vessel. It is useful to connect that model to the measurements an experiment could produce.

Consider the equilibrium between nitrogen dioxide and dinitrogen tetroxide:

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

Nitrogen dioxide is brown, while dinitrogen tetroxide is colourless.

Imagine a sealed, transparent reaction cell held at a constant temperature. A light source sits on one side, and a detector sits on the opposite side. Light passes through the gas mixture, and the detector measures how much light is absorbed.

[IMAGE: An accurately arranged experimental apparatus for studying the NO2/N2O4 equilibrium. Show a sealed transparent gas cell inside a temperature-controlled holder, a light source on the left, and a light detector on the right. Show a brown gas mixture inside the cell, with labels for NO2 and N2O4. Include the light path through the cell and the detector reading absorbance. Do not show a numerical result.]

A calibration allows the absorbance to be related to the concentration of nitrogen dioxide.

What should happen if the reaction proceeds forwards?

- The concentration of \(\ce{NO2}\) decreases.
- The concentration of \(\ce{N2O4}\) increases.
- The brown colour becomes less intense.

The detector does not directly count molecules of \(\ce{N2O4}\). However, if the cell is sealed and there are no side reactions, conservation of nitrogen atoms lets us calculate its concentration from the measured \(\ce{NO2}\) concentration.

### A small experimental calculation

All the following measurements are illustrative, not results from a reported experiment.

Suppose the initial concentrations are:

\[
[\ce{NO2}]_0=0.30\ \mathrm{M}
\]

\[
[\ce{N2O4}]_0=0.10\ \mathrm{M}
\]

After equilibrium is reached, the calibrated detector gives:

\[
[\ce{NO2}]_{\mathrm{eq}}=0.10\ \mathrm{M}
\]

Each \(\ce{NO2}\) molecule contains one nitrogen atom, while each \(\ce{N2O4}\) molecule contains two.

The total nitrogen atom concentration must therefore remain constant:

\[
[\ce{NO2}]+2[\ce{N2O4}]=\text{constant}
\]

Initially:

\[
0.30+2(0.10)=0.50\ \mathrm{M}
\]

At equilibrium:

\[
0.10+2[\ce{N2O4}]_{\mathrm{eq}}=0.50
\]

Therefore:

\[
[\ce{N2O4}]_{\mathrm{eq}}=0.20\ \mathrm{M}
\]

The concentration of \(\ce{N2O4}\) has increased from \(0.10\) M to \(0.20\) M.

The reaction has proceeded forwards, even though product was present at the beginning.

This is exactly what an ICE table predicts when \(Q_c<K_c\).

There is also a useful experimental limitation here. An open vessel would be a poor choice because gas could escape, breaking the assumption that the total amount of nitrogen remains constant. A sealed cell controls that problem, while temperature control ensures that the equilibrium constant remains unchanged during the measurement.

## 4. Worked examples: From the initial mixture to equilibrium

The first example focuses on choosing the correct direction and building the table. The second takes the calculation further, including a reverse shift and a quadratic equation.

### Worked example: Product is present, but the reaction proceeds forwards

A sealed vessel initially contains \(0.40\ \mathrm{M}\) nitrogen dioxide and \(0.10\ \mathrm{M}\) dinitrogen tetroxide.

At this temperature, the equilibrium constant is \(K_c=5.0\ \mathrm{M^{-1}}\).

For the reaction:

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

Determine the direction of the net reaction and construct the ICE table.

[IMAGE: A sealed reaction vessel containing a mixture of brown NO2 gas and colourless N2O4 gas. Label the initial concentrations as 0.40 M NO2 and 0.10 M N2O4. Show the balanced reaction beside the vessel, without showing the predicted direction or equilibrium concentrations.]

**Step 1**: Calculate the initial reaction quotient.

For this reaction:

\[
Q_c=\frac{[\ce{N2O4}]}{[\ce{NO2}]^2}
\]

Substitute the initial concentrations:

\[
Q_c=\frac{0.10}{(0.40)^2}
\]

\[
Q_c=\frac{0.10}{0.16}
=0.625\ \mathrm{M^{-1}}
\]

**Step 2**: Compare \(Q_c\) with \(K_c\).

\[
0.625<5.0
\]

Therefore, \(Q_c<K_c\), so the reaction proceeds in the forward direction.

This means \(\ce{NO2}\) decreases and \(\ce{N2O4}\) increases.

Notice that the initial concentration of \(\ce{N2O4}\) was already \(0.10\) M. It still increases because the mixture needs to produce more product to reach the equilibrium ratio.

**Step 3**: Use the balanced equation to write the change row.

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

Two moles of nitrogen dioxide produce one mole of dinitrogen tetroxide.

Therefore:

- \(\ce{NO2}\) decreases by \(2x\).
- \(\ce{N2O4}\) increases by \(x\).

**Step 4**: Complete the ICE table.

|  | \(\ce{NO2}\) | \(\ce{N2O4}\) |
|---|---:|---:|
| Initial (M) | 0.40 | 0.10 |
| Change (M) | \(-2x\) | \(+x\) |
| Equilibrium (M) | \(0.40-2x\) | \(0.10+x\) |

The variable \(x\) represents the forward reaction progress in \(\mathrm{mol\,L^{-1}}\).

**Step 5**: Establish the physically possible range.

The concentration of nitrogen dioxide cannot become negative:

\[
0.40-2x\geq0
\]

Therefore:

\[
x\leq0.20\ \mathrm{M}
\]

Since the reaction moves forwards, \(x\geq0\).

The physically possible range is:

\[
0\leq x\leq0.20
\]

For a finite, non-zero equilibrium constant in this system, the equilibrium concentration of nitrogen dioxide must remain positive, so the upper boundary is not reached.

You now have the correct ICE table and the range in which a valid solution must lie.

### Worked example: Product is initially present and the reaction shifts backwards

A sealed vessel contains the following initial concentrations:

\[
[\mathrm A]_0=0.20\ \mathrm{M}
\]

\[
[\mathrm B]_0=0.30\ \mathrm{M}
\]

\[
[\mathrm C]_0=0.50\ \mathrm{M}
\]

The reaction is:

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

At this temperature, \(K_c=4.0\ \mathrm{M^{-1}}\).

Calculate the equilibrium concentrations of all three substances.

**Step 1**: Calculate \(Q_c\) to determine the direction.

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

Substitute the initial concentrations:

\[
Q_c=\frac{0.50}{(0.20)(0.30)}
\]

\[
Q_c=\frac{0.50}{0.06}
=8.33\ \mathrm{M^{-1}}
\]

Since:

\[
Q_c>K_c
\]

the reaction proceeds backwards.

A common mistake would be to see that C is present and simply write a negative change for it without checking the equilibrium constant. In this case, that prediction happens to be correct, but it needs to come from the comparison of \(Q_c\) and \(K_c\), not from the mere presence of product.

**Step 2**: Construct the ICE table.

Because the reaction moves backwards:

- A increases by \(x\).
- B increases by \(x\).
- C decreases by \(x\).

|  | A | B | C |
|---|---:|---:|---:|
| Initial (M) | 0.20 | 0.30 | 0.50 |
| Change (M) | \(+x\) | \(+x\) | \(-x\) |
| Equilibrium (M) | \(0.20+x\) | \(0.30+x\) | \(0.50-x\) |

**Step 3**: Write the equilibrium expression.

For the reaction:

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

the expression is:

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

Substitute the equilibrium row:

\[
4.0=\frac{0.50-x}{(0.20+x)(0.30+x)}
\]

**Step 4**: Expand and solve the equation.

Multiply both sides by the denominator:

\[
4(0.20+x)(0.30+x)=0.50-x
\]

Expand:

\[
4(0.06+0.50x+x^2)=0.50-x
\]

\[
0.24+2x+4x^2=0.50-x
\]

Rearrange:

\[
4x^2+3x-0.26=0
\]

Using the quadratic formula:

\[
x=\frac{-3\pm\sqrt{3^2-4(4)(-0.26)}}{2(4)}
\]

\[
x=\frac{-3\pm\sqrt{13.16}}{8}
\]

Therefore:

\[
x=0.0785\ \mathrm{M}
\]

or:

\[
x=-0.828\ \mathrm{M}
\]

**Step 5**: Reject the nonphysical root.

Remember that \(x\) was defined as a positive amount of reverse reaction progress.

Also, C cannot become negative:

\[
0.50-x\geq0
\]

Therefore:

\[
0\leq x\leq0.50
\]

The root \(x=-0.828\) is outside this range.

It would also give negative concentrations for A and B:

\[
[\mathrm A]=0.20-0.828=-0.628\ \mathrm{M}
\]

This is physically impossible.

We therefore accept:

\[
x=0.0785\ \mathrm{M}
\]

**Step 6**: Calculate the equilibrium concentrations.

For A:

\[
[\mathrm A]_{\mathrm{eq}}=0.20+0.0785
\]

\[
=0.2785\ \mathrm{M}
\]

For B:

\[
[\mathrm B]_{\mathrm{eq}}=0.30+0.0785
\]

\[
=0.3785\ \mathrm{M}
\]

For C:

\[
[\mathrm C]_{\mathrm{eq}}=0.50-0.0785
\]

\[
=0.4215\ \mathrm{M}
\]

**Step 7**: Check the result against \(K_c\).

\[
K_c=\frac{0.4215}{(0.2785)(0.3785)}
\]

\[
K_c\approx4.00\ \mathrm{M^{-1}}
\]

This agrees with the given equilibrium constant.

The final concentrations are:

| Substance | Equilibrium concentration |
|---|---:|
| A | \(0.2785\ \mathrm{M}\) |
| B | \(0.3785\ \mathrm{M}\) |
| C | \(0.4215\ \mathrm{M}\) |

The important lesson is that the initial concentration of a product does not determine whether it increases or decreases. The comparison between \(Q_c\) and \(K_c\) determines that.

## 5. Common mistakes when product is present initially

There are a few traps worth checking before moving on.

| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Entering zero for the initial product concentration | The product is already in the mixture | Enter its actual starting concentration |
| Assuming product must decrease | The reaction may proceed forwards | Compare \(Q_c\) with \(K_c\) |
| Using the forward change row automatically | The reaction might shift backwards | Choose signs based on the net direction |
| Ignoring balanced-equation coefficients | Concentration changes must follow the mole ratio | Multiply \(x\) by each coefficient |
| Substituting initial concentrations into \(K_c\) | \(K_c\) describes equilibrium | Substitute the equilibrium row |
| Rejecting every negative value of \(x\) | A negative \(x\) can represent reverse reaction progress under a signed convention | Check the resulting concentrations |
| Accepting every quadratic root | Some roots give negative concentrations | Check the physical range before accepting a root |

### A note about negative values of x

In the worked examples, \(x\) was defined as a positive amount of reaction progress in the actual direction of the shift.

There is another valid convention. You can define \(x\) as positive in the forward direction of the balanced equation, regardless of which direction the reaction actually moves.

For example, for:

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

you would always write:

|  | \(\ce{NO2}\) | \(\ce{N2O4}\) |
|---|---:|---:|
| Change (M) | \(-2x\) | \(+x\) |
| Equilibrium (M) | \([\ce{NO2}]_0-2x\) | \([\ce{N2O4}]_0+x\) |

If the reaction moves backwards, the calculated value of \(x\) will be negative.

That is not automatically a problem. The actual concentrations must still be non-negative.

The distinction is between a negative reaction-progress variable and a negative concentration. Only the latter is physically impossible.

## Questions and solutions

The following questions test the method in increasingly unfamiliar situations. Pay particular attention to the initial conditions: they determine the starting point of every new ICE table.

### Question 1

A mixture contains the following gases at a constant temperature:

\[
[\ce{H2}]_0=0.20\ \mathrm{M}
\]

\[
[\ce{I2}]_0=0.20\ \mathrm{M}
\]

\[
[\ce{HI}]_0=0.10\ \mathrm{M}
\]

For the reaction:

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

the equilibrium constant is \(K_c=4.0\).

Calculate \(Q_c\), determine the direction of the net reaction, and construct the complete ICE table.

### Solution 1

The reaction proceeds forwards, and the change row is \(-x,-x,+2x\).

First, calculate the reaction quotient:

\[
Q_c=\frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}
\]

Substitute the initial concentrations:

\[
Q_c=\frac{(0.10)^2}{(0.20)(0.20)}
\]

\[
Q_c=\frac{0.01}{0.04}=0.25
\]

Since:

\[
0.25<4.0
\]

we know that \(Q_c<K_c\), so the net reaction proceeds forwards.

Hydrogen and iodine are consumed, while hydrogen iodide is produced.

The balanced equation has coefficients of \(1:1:2\), giving the following table:

|  | \(\ce{H2}\) | \(\ce{I2}\) | \(\ce{HI}\) |
|---|---:|---:|---:|
| Initial (M) | 0.20 | 0.20 | 0.10 |
| Change (M) | \(-x\) | \(-x\) | \(+2x\) |
| Equilibrium (M) | \(0.20-x\) | \(0.20-x\) | \(0.10+2x\) |

The important point is that hydrogen iodide already has a non-zero initial concentration. It still increases because the reaction moves forwards.

### Question 2

A sealed vessel contains nitrogen dioxide and dinitrogen tetroxide at a constant temperature.

The initial concentrations are:

\[
[\ce{NO2}]_0=0.30\ \mathrm{M}
\]

\[
[\ce{N2O4}]_0=0.12\ \mathrm{M}
\]

For the reaction:

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

the equilibrium constant is:

\[
K_c=2.0\ \mathrm{M^{-1}}
\]

Calculate both equilibrium concentrations. Show how you determine the correct quadratic root.

[IMAGE: A sealed gas reaction vessel containing both NO2 and N2O4, with the initial concentrations 0.30 M and 0.12 M clearly labelled. Include the balanced equation. Show the vessel at its initial state only, without indicating the reaction direction or equilibrium values.]

### Solution 2

The equilibrium concentrations are approximately \(0.263\ \mathrm{M}\) for nitrogen dioxide and \(0.138\ \mathrm{M}\) for dinitrogen tetroxide.

First, calculate the initial reaction quotient:

\[
Q_c=\frac{0.12}{(0.30)^2}
\]

\[
Q_c=1.33\ \mathrm{M^{-1}}
\]

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

**Step 1**: Construct the ICE table.

The balanced equation has a \(2:1\) ratio.

|  | \(\ce{NO2}\) | \(\ce{N2O4}\) |
|---|---:|---:|
| Initial (M) | 0.30 | 0.12 |
| Change (M) | \(-2x\) | \(+x\) |
| Equilibrium (M) | \(0.30-2x\) | \(0.12+x\) |

**Step 2**: Substitute the equilibrium concentrations into \(K_c\).

\[
2.0=\frac{0.12+x}{(0.30-2x)^2}
\]

Rearrange:

\[
2.0(0.30-2x)^2=0.12+x
\]

Expand:

\[
0.18-2.4x+8x^2=0.12+x
\]

Collect terms:

\[
8x^2-3.4x+0.06=0
\]

**Step 3**: Solve the quadratic.

Using the quadratic formula:

\[
x=\frac{3.4\pm\sqrt{(3.4)^2-4(8)(0.06)}}{16}
\]

\[
x=\frac{3.4\pm\sqrt{9.64}}{16}
\]

This gives:

\[
x=0.0184\ \mathrm{M}
\]

or:

\[
x=0.407\ \mathrm{M}
\]

**Step 4**: Reject the nonphysical root.

The nitrogen dioxide concentration must remain non-negative:

\[
0.30-2x\geq0
\]

Therefore:

\[
x\leq0.15\ \mathrm{M}
\]

The root \(x=0.407\) is impossible because it exceeds the maximum amount of nitrogen dioxide available.

It would produce:

\[
[\ce{NO2}]=0.30-2(0.407)
\]

\[
=-0.514\ \mathrm{M}
\]

Reject this root.

The valid value is:

\[
x=0.0184\ \mathrm{M}
\]

**Step 5**: Calculate the equilibrium concentrations.

For nitrogen dioxide:

\[
[\ce{NO2}]_{\mathrm{eq}}
=0.30-2(0.0184)
\]

\[
=0.263\ \mathrm{M}
\]

For dinitrogen tetroxide:

\[
[\ce{N2O4}]_{\mathrm{eq}}
=0.12+0.0184
\]

\[
=0.138\ \mathrm{M}
\]

Both concentrations are positive, and substituting them into the equilibrium expression gives approximately \(2.0\ \mathrm{M^{-1}}\).

The second quadratic root was mathematically valid but chemically impossible. This is why checking the physical range is part of the calculation, not an optional final step.

### Question 3

The following reaction is exothermic:

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

At \(300\ \mathrm{K}\), a sealed vessel is at equilibrium with:

\[
[\mathrm A]=0.20\ \mathrm{M}
\]

\[
[\mathrm B]=0.20\ \mathrm{M}
\]

\[
[\mathrm C]=0.40\ \mathrm{M}
\]

The equilibrium constant at this temperature is \(10.0\ \mathrm{M^{-1}}\).

The vessel is heated to \(400\ \mathrm{K}\). The new equilibrium constant is \(4.0\ \mathrm{M^{-1}}\).

Assume the vessel remains at constant volume and the concentrations do not change instantaneously during heating.

Determine:

a) The reaction quotient immediately after heating.

b) The direction of the net reaction.

c) The correct ICE table for the new equilibrium.

d) The new equilibrium concentrations.

[IMAGE: A sealed reaction vessel containing A, B, and C, shown inside a temperature-controlled bath. Illustrate the initial equilibrium state at 300 K and the subsequent heating to 400 K. Keep the initial concentrations visible and do not show the predicted reaction direction.]

### Solution 3

The reaction shifts backwards, and the new equilibrium concentrations are approximately \(0.282\ \mathrm{M}\) A, \(0.282\ \mathrm{M}\) B, and \(0.318\ \mathrm{M}\) C.

**Step 1**: Calculate the reaction quotient immediately after heating.

Immediately after heating, the concentrations have not yet changed.

Therefore:

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

\[
Q_c=\frac{0.40}{(0.20)(0.20)}
\]

\[
Q_c=10.0\ \mathrm{M^{-1}}
\]

**Step 2**: Compare the new \(Q_c\) with the new \(K_c\).

This is where the temperature change matters.

The original equilibrium constant was \(10.0\ \mathrm{M^{-1}}\), but the new equilibrium constant is only \(4.0\ \mathrm{M^{-1}}\).

We must use the new value:

\[
Q_c=10.0>K_c=4.0
\]

Therefore, the reaction shifts backwards.

**Step 3**: Construct the new ICE table.

The reverse reaction produces A and B while consuming C.

|  | A | B | C |
|---|---:|---:|---:|
| Initial (M) | 0.20 | 0.20 | 0.40 |
| Change (M) | \(+x\) | \(+x\) | \(-x\) |
| Equilibrium (M) | \(0.20+x\) | \(0.20+x\) | \(0.40-x\) |

Notice that the initial row is the mixture immediately after heating, not a new mixture with zero product.

**Step 4**: Substitute into the new equilibrium expression.

\[
4.0=\frac{0.40-x}{(0.20+x)^2}
\]

Rearrange:

\[
4(0.20+x)^2=0.40-x
\]

Expand:

\[
4(0.04+0.40x+x^2)=0.40-x
\]

\[
0.16+1.6x+4x^2=0.40-x
\]

Therefore:

\[
4x^2+2.6x-0.24=0
\]

Solving:

\[
x=\frac{-2.6\pm\sqrt{(2.6)^2-4(4)(-0.24)}}{8}
\]

\[
x=\frac{-2.6\pm\sqrt{10.6}}{8}
\]

The physically valid root is:

\[
x=0.0820\ \mathrm{M}
\]

The other root would produce negative concentrations for A and B.

**Step 5**: Calculate the new equilibrium concentrations.

For A:

\[
[\mathrm A]_{\mathrm{eq}}=0.20+0.0820
=0.282\ \mathrm{M}
\]

For B:

\[
[\mathrm B]_{\mathrm{eq}}=0.20+0.0820
=0.282\ \mathrm{M}
\]

For C:

\[
[\mathrm C]_{\mathrm{eq}}=0.40-0.0820
=0.318\ \mathrm{M}
\]

The final mixture satisfies the new equilibrium constant.

The tempting mistake is to use the original \(K_c=10.0\) and conclude that the mixture is still at equilibrium. But temperature changes can change \(K_c\). The concentrations initially remain the same, so \(Q_c\) stays at 10.0 while the equilibrium target changes.

### Question 4

Consider the equilibrium:

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

A vessel initially contains:

\[
[\ce{NO2}]_0=0.10\ \mathrm{M}
\]

\[
[\ce{N2O4}]_0=0.40\ \mathrm{M}
\]

At this temperature:

\[
K_c=10.0\ \mathrm{M^{-1}}
\]

A student defines \(x\) as positive in the forward direction, as written in the balanced equation.

They obtain two roots when solving the equilibrium equation.

Calculate both roots, identify the physically valid one, and explain why a negative value of \(x\) can be accepted.

[IMAGE: A sealed vessel containing a mixture of NO2 and N2O4, with initial concentrations of 0.10 M and 0.40 M respectively. Show the balanced equation and a clear initial-state particle arrangement. Do not show reaction arrows or calculated equilibrium concentrations.]

### Solution 4

The physically valid root is approximately \(x=-0.0443\ \mathrm{M}\). The negative value is valid because \(x\) was defined as positive in the forward direction, while the actual reaction proceeds backwards.

**Step 1**: Calculate the initial reaction quotient.

\[
Q_c=\frac{0.40}{(0.10)^2}
\]

\[
Q_c=40.0\ \mathrm{M^{-1}}
\]

Since:

\[
Q_c>K_c
\]

the reaction proceeds backwards.

**Step 2**: Construct the ICE table using the student's sign convention.

The student has defined \(x\) as positive in the forward direction.

Therefore, the change row must remain:

\[
-2x,\quad +x
\]

The complete table is:

|  | \(\ce{NO2}\) | \(\ce{N2O4}\) |
|---|---:|---:|
| Initial (M) | 0.10 | 0.40 |
| Change (M) | \(-2x\) | \(+x\) |
| Equilibrium (M) | \(0.10-2x\) | \(0.40+x\) |

Because the reaction actually moves backwards, we expect \(x\) to be negative.

**Step 3**: Substitute into the equilibrium expression.

\[
10=\frac{0.40+x}{(0.10-2x)^2}
\]

Rearrange:

\[
10(0.10-2x)^2=0.40+x
\]

Expand:

\[
10(0.01-0.40x+4x^2)=0.40+x
\]

\[
0.10-4x+40x^2=0.40+x
\]

Collect terms:

\[
40x^2-5x-0.30=0
\]

**Step 4**: Solve the quadratic.

\[
x=\frac{5\pm\sqrt{(-5)^2-4(40)(-0.30)}}{80}
\]

\[
x=\frac{5\pm\sqrt{73}}{80}
\]

Therefore:

\[
x=0.169\ \mathrm{M}
\]

or:

\[
x=-0.0443\ \mathrm{M}
\]

**Step 5**: Check both roots against the concentration limits.

For nitrogen dioxide:

\[
0.10-2x\geq0
\]

Therefore:

\[
x\leq0.050
\]

For dinitrogen tetroxide:

\[
0.40+x\geq0
\]

Therefore:

\[
x\geq-0.40
\]

The physically possible range is:

\[
-0.40\leq x\leq0.050
\]

The positive root, \(x=0.169\), is outside this range.

It would produce:

\[
[\ce{NO2}]=0.10-2(0.169)
\]

\[
=-0.238\ \mathrm{M}
\]

This is impossible.

The negative root, \(x=-0.0443\), is within the physical range and must be accepted.

**Step 6**: Interpret the negative root.

For nitrogen dioxide:

\[
[\ce{NO2}]_{\mathrm{eq}}
=0.10-2(-0.0443)
\]

\[
=0.189\ \mathrm{M}
\]

For dinitrogen tetroxide:

\[
[\ce{N2O4}]_{\mathrm{eq}}
=0.40+(-0.0443)
\]

\[
=0.356\ \mathrm{M}
\]

Nitrogen dioxide has increased, while dinitrogen tetroxide has decreased.

That is exactly what we expect from the reverse reaction.

The important distinction is that a negative \(x\) does not necessarily mean a negative concentration. The signs in the change row depend on how you define \(x\). Always judge the final concentrations against physical constraints.

### Question 5

Two separate vessels contain mixtures of A, B, and C. Both vessels are at equilibrium at the same temperature for the reaction:

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

The equilibrium constant is:

\[
K_c=10.0\ \mathrm{M^{-1}}
\]

Their concentrations are:

| Substance | Vessel 1 | Vessel 2 |
|---|---:|---:|
| A | \(0.10\ \mathrm{M}\) | \(0.40\ \mathrm{M}\) |
| B | \(0.40\ \mathrm{M}\) | \(0.10\ \mathrm{M}\) |
| C | \(0.40\ \mathrm{M}\) | \(0.40\ \mathrm{M}\) |

Equal volumes from the two vessels are poured into a third vessel. Assume the volumes are additive, the temperature remains constant, and no reaction occurs during the transfer.

A student argues:

"Both original mixtures were at equilibrium, and the temperature hasn't changed. Therefore, the combined mixture must also be at equilibrium."

Is the student correct?

Calculate the initial reaction quotient after mixing, determine the reaction direction, and calculate the new equilibrium concentrations.

[IMAGE: A conceptual chemistry diagram showing two equal-volume sealed vessels being poured into a third vessel. Label vessel 1 with A = 0.10 M, B = 0.40 M, C = 0.40 M. Label vessel 2 with A = 0.40 M, B = 0.10 M, C = 0.40 M. Show the third vessel as the combined mixture, but do not supply its calculated concentrations or reaction direction.]

### Solution 5

The student is incorrect. The combined mixture is not initially at equilibrium. It has \(Q_c=6.4\ \mathrm{M^{-1}}\), which is less than \(K_c=10.0\ \mathrm{M^{-1}}\), so the reaction proceeds forwards.

This question tests something more subtle than simply comparing \(Q_c\) and \(K_c\). The key is understanding what happens to concentrations when two different mixtures are combined.

**Step 1**: Confirm that both original mixtures are individually at equilibrium.

For vessel 1:

\[
Q_c=\frac{0.40}{(0.10)(0.40)}
\]

\[
Q_c=10.0\ \mathrm{M^{-1}}
\]

For vessel 2:

\[
Q_c=\frac{0.40}{(0.40)(0.10)}
\]

\[
Q_c=10.0\ \mathrm{M^{-1}}
\]

Both mixtures are individually at equilibrium.

So far, the student's reasoning seems sensible.

**Step 2**: Calculate the concentrations after mixing.

Because equal volumes are combined, each final concentration is the average of the two original concentrations.

For A:

\[
[\mathrm A]_0=\frac{0.10+0.40}{2}
\]

\[
=0.25\ \mathrm{M}
\]

For B:

\[
[\mathrm B]_0=\frac{0.40+0.10}{2}
\]

\[
=0.25\ \mathrm{M}
\]

For C:

\[
[\mathrm C]_0=\frac{0.40+0.40}{2}
\]

\[
=0.40\ \mathrm{M}
\]

The new mixture initially contains:

\[
[\mathrm A]_0=0.25\ \mathrm{M}
\]

\[
[\mathrm B]_0=0.25\ \mathrm{M}
\]

\[
[\mathrm C]_0=0.40\ \mathrm{M}
\]

**Step 3**: Calculate the reaction quotient of the combined mixture.

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

Substitute:

\[
Q_c=\frac{0.40}{(0.25)(0.25)}
\]

\[
Q_c=\frac{0.40}{0.0625}
\]

\[
Q_c=6.4\ \mathrm{M^{-1}}
\]

Now compare:

\[
6.4<10.0
\]

Therefore:

\[
Q_c<K_c
\]

The combined mixture shifts forwards.

**Step 4**: Construct the new ICE table.

The forward reaction consumes A and B and produces C.

|  | A | B | C |
|---|---:|---:|---:|
| Initial (M) | 0.25 | 0.25 | 0.40 |
| Change (M) | \(-x\) | \(-x\) | \(+x\) |
| Equilibrium (M) | \(0.25-x\) | \(0.25-x\) | \(0.40+x\) |

**Step 5**: Substitute into the equilibrium expression.

\[
10.0=\frac{0.40+x}{(0.25-x)^2}
\]

Rearrange:

\[
10(0.25-x)^2=0.40+x
\]

Expand:

\[
10(0.0625-0.50x+x^2)=0.40+x
\]

\[
0.625-5x+10x^2=0.40+x
\]

Collect terms:

\[
10x^2-6x+0.225=0
\]

**Step 6**: Solve and reject the nonphysical root.

Using the quadratic formula:

\[
x=\frac{6\pm\sqrt{36-4(10)(0.225)}}{20}
\]

\[
x=\frac{6\pm\sqrt{27}}{20}
\]

Therefore:

\[
x=0.0402\ \mathrm{M}
\]

or:

\[
x=0.560\ \mathrm{M}
\]

A and B initially have concentrations of \(0.25\ \mathrm{M}\), so:

\[
0.25-x\geq0
\]

This requires:

\[
x\leq0.25
\]

The root \(x=0.560\) would give negative concentrations for A and B. Reject it.

The valid root is:

\[
x=0.0402\ \mathrm{M}
\]

**Step 7**: Calculate the final equilibrium concentrations.

For A:

\[
[\mathrm A]_{\mathrm{eq}}
=0.25-0.0402
\]

\[
=0.210\ \mathrm{M}
\]

For B:

\[
[\mathrm B]_{\mathrm{eq}}
=0.25-0.0402
\]

\[
=0.210\ \mathrm{M}
\]

For C:

\[
[\mathrm C]_{\mathrm{eq}}
=0.40+0.0402
\]

\[
=0.440\ \mathrm{M}
\]

The final concentrations are approximately:

| Substance | New equilibrium concentration |
|---|---:|
| A | \(0.210\ \mathrm{M}\) |
| B | \(0.210\ \mathrm{M}\) |
| C | \(0.440\ \mathrm{M}\) |

**Why did mixing two equilibrium mixtures disturb equilibrium?**

This is the crucial part.

The equilibrium constant is the same in both original vessels, but the reaction quotient is calculated from the concentrations in the actual mixture.

When we combine the vessels, we average the individual concentrations. However, the product of two average concentrations is not necessarily the average of their products.

In this example:

\[
\frac{0.10+0.40}{2}
\times
\frac{0.40+0.10}{2}
=0.0625
\]

But the average of the original concentration products is:

\[
\frac{(0.10)(0.40)+(0.40)(0.10)}{2}
=0.0400
\]

The product of the averages is different.

That is why the combined mixture does not automatically preserve the original equilibrium condition.

The tempting assumption was that mixing two equilibrium systems at the same temperature must preserve equilibrium. The actual constraint is that equilibrium depends on the concentration ratio in the combined vessel, not on the history of the individual mixtures.

## 6. What to remember when product is present initially

When you encounter an ICE table with products already in the starting mixture, use this order:

1. Write the balanced equation.
2. Enter every initial concentration, including products.
3. Calculate \(Q_c\) using the initial concentrations.
4. Compare \(Q_c\) with \(K_c\) to determine the net reaction direction.
5. Use the balanced equation to construct the change row.
6. Add the initial and change rows to obtain the equilibrium row.
7. Substitute the equilibrium concentrations into the \(K_c\) expression.
8. Solve the equation and check every root against the physical concentration limits.
9. Substitute your final concentrations back into \(K_c\) to verify the result.

The most important distinction is between the initial mixture and the equilibrium mixture. A product can be present in both, but its concentration does not have to move in the same direction in every situation.

For the broader method of constructing ICE tables and calculating \(K_c\), see [ICE Tables for Equilibrium: Calculate Concentrations and Use Kc](https://tutorgum.com/learn/resources/ice-tables-for-equilibrium-from-initial-amounts-to-keq/).

Once you can handle product already being present, the next useful skill is rebuilding the initial row after a change in concentration or volume. The reaction quotient gives you the new direction before you start the next equilibrium calculation.