Why Equilibrium Concentrations Cannot Be Negative

An equilibrium calculation gives two possible values for \(x\): \(0.10\ \mathrm{M}\) and \(0.35\ \mathrm{M}\). Both solve the quadratic equation. However, substituting the second value into the ICE table gives a reactant concentration of \(-0.30\ \mathrm{M}\).

Which answer should you keep?

Only \(x=0.10\ \mathrm{M}\). The other root is mathematically valid but chemically impossible.

Equilibrium concentrations cannot be negative because concentration is the amount of a substance divided by the volume it occupies. The amount of substance cannot be negative, and volume is positive. Therefore, a concentration must be zero or positive.

There is an important distinction, though: the value of \(x\) in an ICE table can be negative. A negative \(x\) can simply mean that the reaction has proceeded in the reverse direction.

Understanding this distinction will help you avoid rejecting valid solutions and accepting impossible ones.

01Why equilibrium concentrations must be non-negative

Concentration measures how much of a substance is present in a given volume.

It is calculated using:

\[ [X]=\frac{n(X)}{V} \]

where:

• \([X]\) is the concentration of substance \(X\), measured in \(\mathrm{mol\,L^{-1}}\), or M.
• \(n(X)\) is the amount of substance \(X\), measured in mol.
• \(V\) is the volume of the mixture, measured in L.

You can have no particles of a substance, giving a concentration of zero. You can also have a very small amount, giving a small positive concentration.

But you cannot have less than zero particles.

This physical restriction becomes important when solving equilibrium problems because algebra does not automatically know which answers are physically possible.

A quadratic equation can have two real roots, even when only one corresponds to a possible chemical mixture.

To identify the correct root, you need to understand what the rows of an ICE table represent.

02Understanding the initial, change, and equilibrium rows

An ICE table has three rows:

• Initial (I): The concentrations present before the reaction adjusts towards equilibrium.
• Change (C): The concentration changes caused by the net reaction.
• Equilibrium (E): The final concentrations after the reaction reaches equilibrium.

Consider the reaction:

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

Suppose a solution initially contains:

• \([A]=0.40\ \mathrm{M}\)
• \([B]=0.10\ \mathrm{M}\)

We will define \(x\) as the forward reaction extent, measured in M.

The balanced equation tells us that two units of A are consumed for every one unit of B produced.

Therefore:

• A decreases by \(2x\).
• B increases by \(x\).

Our ICE table becomes:

The equilibrium row is calculated by adding the initial and change rows.

For example:

\[ [A]_{\mathrm{eq}}=0.40-2x \]

This expression tells us something important. If \(x\) becomes too large, the calculated concentration of A becomes negative.

That cannot happen in a real solution.

03Finding the physically possible range of x

Before solving an equilibrium equation, you can determine the range of \(x\) that keeps every concentration non-negative.

Using our example:

\[ [A]_{\mathrm{eq}}=0.40-2x \]

Since concentration cannot be negative:

\[ 0.40-2x\geq 0 \]

Rearranging:

\[ x\leq 0.20 \]

Now consider B:

\[ [B]_{\mathrm{eq}}=0.10+x \]

Again, the concentration must be non-negative:

\[ 0.10+x\geq 0 \]

Therefore:

\[ x\geq -0.10 \]

Combining both restrictions gives:

\[ \boxed{-0.10\leq x\leq 0.20} \]

This is the physically possible range of \(x\).

Any mathematical root outside this interval must be rejected because it produces at least one negative concentration.

Notice that the range includes negative values of \(x\). This is completely acceptable.

A negative value of \(x\) simply means that the net reaction has proceeded backwards relative to the direction we initially defined as positive.

The general rule

For a reaction:

\[ \ce{aA + bB <=> cC + dD} \]

If we define positive \(x\) as the forward reaction, the change row is:

\[ -ax,\quad -bx,\quad +cx,\quad +dx \]

Every equilibrium concentration must satisfy:

\[ [\text{Species}]_{\mathrm{eq}}\geq 0 \]

You can apply this restriction to every species to find the allowable range of \(x\).

The physical range is the intersection of all these restrictions.

One small detail: zero is physically possible, but if a zero concentration makes a denominator in the \(K_c\) expression equal to zero, that boundary cannot be accepted as a solution for a finite equilibrium constant.

05A negative value of x can be physically correct

Here is the misconception that catches students out:

A negative concentration is impossible, but a negative reaction extent is not.

Think of \(x\) as a direction convention. If you choose forwards as positive, moving backwards gives a negative value. That does not mean the chemical substances have negative concentrations.

The analogy has one limitation: \(x\) is a signed concentration change, not a physical distance. The actual concentrations still have to satisfy the non-negative restriction.

Let’s see what happens when the reaction genuinely shifts backwards.

06How to decide whether an equilibrium root is valid

When you solve a quadratic equation in an equilibrium problem, do not automatically choose the positive root or discard the negative one.

Use three checks.

The first check is the one most directly related to rejecting negative concentrations.

The second check prevents you from accepting a set of concentrations that cannot result from the stated reaction.

The third check confirms that the resulting mixture actually satisfies the equilibrium condition.

A useful habit is to calculate the allowable range of \(x\) before solving the quadratic. This gives you a clear test for every root as soon as it appears.

For a broader explanation of constructing the three ICE rows and forming the equilibrium expression, see ICE Tables for Equilibrium: Calculate Concentrations and Use Kc.

You can also use the HSC Chemistry Equilibrium Calculator to check numerical calculations. Remember that you still need to apply the physical restrictions yourself.

07Questions and solutions

Try each question before reading its solution. The questions progress from straightforward concentration restrictions to more demanding problems involving reaction direction, dilution, and alternative sign conventions.

Question 1

Consider the reaction:

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

A solution initially contains:

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

\[ [Q]=0.10\ \mathrm{M} \]

Let \(x\) represent the forward reaction extent.

(a) Write the initial, change, and equilibrium rows of an ICE table.

(b) Determine the physically possible range of \(x\).

(c) Decide whether each of the following values is physically possible:

• \(x=0.15\ \mathrm{M}\)
• \(x=0.30\ \mathrm{M}\)

Solution 1

Only \(x=0.15\ \mathrm{M}\) is physically possible. The value \(x=0.30\ \mathrm{M}\) produces a negative concentration of P.

(a) The ICE table is:

The coefficients are both 1, so P decreases by \(x\) while Q increases by \(x\).

(b) Both equilibrium concentrations must be non-negative.

For P:

\[ 0.25-x\geq0 \]

\[ x\leq0.25 \]

For Q:

\[ 0.10+x\geq0 \]

\[ x\geq-0.10 \]

Therefore:

\[ \boxed{-0.10\leq x\leq0.25} \]

(c) Test the two proposed values.

For \(x=0.15\):

\[ [P]_{\mathrm{eq}}=0.25-0.15=0.10\ \mathrm{M} \]

\[ [Q]_{\mathrm{eq}}=0.10+0.15=0.25\ \mathrm{M} \]

Both concentrations are positive, so this value is physically possible.

For \(x=0.30\):

\[ [P]_{\mathrm{eq}}=0.25-0.30 =-0.05\ \mathrm{M} \]

This is impossible.

The second value exceeds the maximum amount of P available to react.

Question 2

Consider the reaction:

\[ \ce{A(aq) + B(aq) <=> 2C(aq)} \]

A closed solution initially contains:

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

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

\[ [C]_0=0.10\ \mathrm{M} \]

A student reports the following equilibrium concentrations:

\[ [A]_{\mathrm{eq}}=0.10\ \mathrm{M} \]

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

\[ [C]_{\mathrm{eq}}=0.20\ \mathrm{M} \]

The equilibrium constant is \(K_c=2.0\).

The student argues that the answer must be correct because all concentrations are positive and the values reproduce \(K_c=2.0\).

Determine whether the reported equilibrium state is possible. Explain your reasoning using the stoichiometric coefficients.

Solution 2

The reported equilibrium state is impossible, even though all three concentrations are positive and the equilibrium constant is correct.

The problem is that the concentration changes do not satisfy the stoichiometric ratios in the balanced equation.

Step 1: Check the equilibrium constant.

For the reaction:

\[ K_c=\frac{[C]^2}{[A][B]} \]

Substituting the reported values:

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

\[ K_c=\frac{0.04}{0.02}=2.0 \]

So the reported concentrations satisfy the equilibrium expression.

Step 2: Check the change in A.

The initial concentration of A was \(0.30\ \mathrm{M}\), and its reported equilibrium concentration is \(0.10\ \mathrm{M}\).

Therefore:

\[ \Delta[A]=0.10-0.30=-0.20\ \mathrm{M} \]

This implies a forward reaction extent of:

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

Step 3: Check the change in B.

B should be consumed in the same ratio as A because both have a coefficient of 1.

Therefore, its equilibrium concentration should be:

\[ [B]_{\mathrm{eq}}=0.30-0.20 \]

\[ [B]_{\mathrm{eq}}=0.10\ \mathrm{M} \]

The reported value was \(0.20\ \mathrm{M}\), so it does not agree.

Step 4: Check the change in C.

The balanced equation produces two units of C for every unit of A consumed.

Therefore:

\[ [C]_{\mathrm{eq}}=0.10+2(0.20) \]

\[ [C]_{\mathrm{eq}}=0.50\ \mathrm{M} \]

The reported value was only \(0.20\ \mathrm{M}\).

The three reported concentrations cannot be produced by one reaction extent.

This question exposes an important limitation of checking concentrations individually. Non-negative concentrations are necessary, but they are not sufficient. The concentrations must also obey the balanced equation.

Question 3

An aqueous system is initially at equilibrium for the reaction:

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

At a fixed temperature:

\[ K_c=5.0\ \mathrm{M^{-1}} \]

The initial equilibrium concentrations are:

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

\[ [B]=0.60\ \mathrm{M} \]

\[ [C]=0.60\ \mathrm{M} \]

The solution is diluted by adding solvent, doubling its volume. Assume that the temperature remains constant and no reaction occurs during the dilution itself.

Calculate the new equilibrium concentrations.

Solution 3

The new equilibrium concentrations are approximately:

\[ \boxed{ \begin{aligned} [A]_{\mathrm{eq}}&=0.1464\ \mathrm{M}\\ [B]_{\mathrm{eq}}&=0.3464\ \mathrm{M}\\ [C]_{\mathrm{eq}}&=0.2536\ \mathrm{M} \end{aligned} } \]

The reaction shifts in reverse after dilution.

Step 1: Calculate the concentrations immediately after dilution.

Doubling the volume halves the concentration of every dissolved substance, because the amounts of substance have not changed during the dilution.

Therefore:

\[ [A]_0=0.10\ \mathrm{M} \]

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

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

These are the new initial concentrations for the ICE table.

Step 2: Determine the reaction direction.

Calculate the new reaction quotient:

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

Substituting the concentrations immediately after dilution:

\[ Q_c=\frac{0.30}{(0.10)(0.30)} \]

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

Since:

\[ Q_c>K_c \]

the reaction shifts backwards, towards A and B.

Step 3: Construct the new ICE table.

Define positive \(x\) as the forward reaction extent.

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

Step 4: Write the physical restrictions.

For A:

\[ 0.10-x\geq0 \]

\[ x\leq0.10 \]

For B:

\[ 0.30-x\geq0 \]

\[ x\leq0.30 \]

For C:

\[ 0.30+x\geq0 \]

\[ x\geq-0.30 \]

Therefore:

\[ -0.30\leq x\leq0.10 \]

Step 5: Substitute into the equilibrium expression.

The equilibrium constant remains \(5.0\) because the temperature has not changed.

\[ 5.0= \frac{0.30+x} {(0.10-x)(0.30-x)} \]

Rearranging:

\[ 0.30+x=5(0.10-x)(0.30-x) \]

Expanding:

\[ 0.30+x=0.15-2x+5x^2 \]

Therefore:

\[ 5x^2-3x-0.15=0 \]

Using the quadratic formula:

\[ x=\frac{3\pm\sqrt{12}}{10} \]

The two roots are:

\[ x=-0.0464\ \mathrm{M} \]

and

\[ x=0.6464\ \mathrm{M} \]

Only the negative root is within the physical range.

Step 6: Calculate the new equilibrium concentrations.

For A:

\[ [A]_{\mathrm{eq}}=0.10-(-0.0464) \]

\[ [A]_{\mathrm{eq}}=0.1464\ \mathrm{M} \]

For B:

\[ [B]_{\mathrm{eq}}=0.30-(-0.0464) \]

\[ [B]_{\mathrm{eq}}=0.3464\ \mathrm{M} \]

For C:

\[ [C]_{\mathrm{eq}}=0.30+(-0.0464) \]

\[ [C]_{\mathrm{eq}}=0.2536\ \mathrm{M} \]

All concentrations are positive, and the equilibrium expression gives \(K_c\approx5.0\).

The key lesson is that the initial row must represent the mixture immediately after the disturbance. Using the concentrations from before dilution would give the wrong equilibrium calculation.

Question 4

Consider the reaction:

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

A solution initially contains:

\[ [A]=0.00\ \mathrm{M} \]

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

\[ [C]=0.30\ \mathrm{M} \]

At this temperature:

\[ K_c=5.0\ \mathrm{M^{-1}} \]

Let \(x\) represent the forward reaction extent.

Calculate the equilibrium concentrations, explaining why the absence of A initially does not mean that its equilibrium concentration must remain zero.

Solution 4

The equilibrium concentrations are approximately:

\[ \boxed{ \begin{aligned} [A]_{\mathrm{eq}}&=0.0873\ \mathrm{M}\\ [B]_{\mathrm{eq}}&=0.4873\ \mathrm{M}\\ [C]_{\mathrm{eq}}&=0.2127\ \mathrm{M} \end{aligned} } \]

The reaction proceeds backwards, producing A and B from C.

Step 1: Identify the possible reaction direction.

There is initially no A, so the forward reaction cannot proceed at the start.

However, C is present and can decompose to produce A and B.

The net reaction must initially proceed in the reverse direction.

Step 2: Construct the ICE table.

We still define positive \(x\) as the forward reaction extent.

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

Step 3: Determine the physically possible range.

For A:

\[ -x\geq0 \]

\[ x\leq0 \]

For B:

\[ 0.40-x\geq0 \]

\[ x\leq0.40 \]

For C:

\[ 0.30+x\geq0 \]

\[ x\geq-0.30 \]

Combining:

\[ \boxed{-0.30\leq x\leq0} \]

This is a useful example because the initial concentration of A is zero. Any positive forward extent would immediately produce a negative concentration of A.

Step 4: Substitute into the equilibrium expression.

For this reaction:

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

Substituting the equilibrium expressions:

\[ 5.0= \frac{0.30+x}{(-x)(0.40-x)} \]

Rearranging:

\[ 5(-x)(0.40-x)=0.30+x \]

Expanding:

\[ 5x^2-2x=0.30+x \]

Therefore:

\[ 5x^2-3x-0.30=0 \]

Using the quadratic formula:

\[ x=\frac{3\pm\sqrt{15}}{10} \]

The two roots are:

\[ x=-0.0873\ \mathrm{M} \]

and

\[ x=0.6873\ \mathrm{M} \]

The positive root is outside the physically possible range.

Step 5: Calculate the equilibrium concentrations.

Using \(x=-0.0873\):

For A:

\[ [A]_{\mathrm{eq}}=-(-0.0873) \]

\[ [A]_{\mathrm{eq}}=0.0873\ \mathrm{M} \]

For B:

\[ [B]_{\mathrm{eq}}=0.40-(-0.0873) \]

\[ [B]_{\mathrm{eq}}=0.4873\ \mathrm{M} \]

For C:

\[ [C]_{\mathrm{eq}}=0.30+(-0.0873) \]

\[ [C]_{\mathrm{eq}}=0.2127\ \mathrm{M} \]

Substituting these concentrations into \(K_c\) gives approximately \(5.0\ \mathrm{M^{-1}}\).

The absence of A in the initial mixture does not prevent it from appearing at equilibrium. The reverse reaction produces it.

Question 5

A closed solution contains:

\[ [A]_0=0.10\ \mathrm{M} \]

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

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

The reaction is:

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

At a fixed temperature:

\[ K_c=2.5\ \mathrm{M^{-1}} \]

Two students calculate the equilibrium concentrations using different reaction directions.

• Student 1 uses the reaction as written and defines positive \(x\) as the forward direction.
• Student 2 reverses the equation to \(\ce{C(aq) <=> A(aq) + B(aq)}\) and defines positive \(y\) as the reverse direction.

Student 1 obtains a negative root for \(x\) and initially rejects it.

Student 2 obtains a positive root for \(y\) and accepts it.

Are their calculations describing different equilibrium states, or can both approaches be correct?

Calculate the equilibrium concentrations using both reaction directions. Explain the relationship between \(x\), \(y\), and the two equilibrium constants.

Solution 5

Both approaches can describe exactly the same physical equilibrium state.

The crucial relationship is:

\[ \boxed{y=-x} \]

Reversing the reaction changes the sign convention for the reaction extent. It also changes the equilibrium constant to its reciprocal.

Step 1: Solve using the forward reaction.

Student 1 uses:

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

The ICE table is:

The physical restrictions are:

\[ 0.10-x\geq0 \]

\[ 0.30-x\geq0 \]

\[ 0.20+x\geq0 \]

Therefore:

\[ -0.20\leq x\leq0.10 \]

Step 2: Substitute into the forward equilibrium expression.

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

\[ 2.5= \frac{0.20+x} {(0.10-x)(0.30-x)} \]

Rearranging:

\[ 0.20+x=2.5(0.10-x)(0.30-x) \]

Expanding:

\[ 0.20+x=0.075-1.0x+2.5x^2 \]

Therefore:

\[ 2.5x^2-2x-0.125=0 \]

Solving:

\[ x=\frac{4\pm\sqrt{21}}{10} \]

The two roots are:

\[ x=-0.0583\ \mathrm{M} \]

and

\[ x=0.8583\ \mathrm{M} \]

Only the negative root lies within the physically possible range.

Student 1 should therefore keep:

\[ x=-0.0583\ \mathrm{M} \]

Step 3: Calculate the equilibrium concentrations.

For A:

\[ [A]_{\mathrm{eq}}=0.10-(-0.0583) \]

\[ [A]_{\mathrm{eq}}=0.1583\ \mathrm{M} \]

For B:

\[ [B]_{\mathrm{eq}}=0.30-(-0.0583) \]

\[ [B]_{\mathrm{eq}}=0.3583\ \mathrm{M} \]

For C:

\[ [C]_{\mathrm{eq}}=0.20+(-0.0583) \]

\[ [C]_{\mathrm{eq}}=0.1417\ \mathrm{M} \]

All three concentrations are positive.

Step 4: Reverse the reaction and its equilibrium constant.

Student 2 writes:

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

When an equilibrium equation is reversed, its equilibrium constant becomes the reciprocal of the original constant.

Therefore:

\[ K'_c=\frac{1}{K_c} \]

\[ K'_c=\frac{1}{2.5} \]

\[ K'_c=0.40\ \mathrm{M} \]

Notice that the units have also been inverted.

Step 5: Construct the reverse ICE table.

Student 2 defines positive \(y\) as the reverse reaction extent.

Therefore:

The physical restrictions give:

\[ -0.10\leq y\leq0.20 \]

Step 6: Solve using the reverse equilibrium expression.

For the reversed equation:

\[ K'_c=\frac{[A][B]}{[C]} \]

Substitute:

\[ 0.40= \frac{(0.10+y)(0.30+y)} {0.20-y} \]

Rearranging:

\[ 0.40(0.20-y) =(0.10+y)(0.30+y) \]

Expanding:

\[ 0.08-0.40y =0.03+0.40y+y^2 \]

Therefore:

\[ y^2+0.80y-0.05=0 \]

Using the quadratic formula:

\[ y=\frac{-0.80\pm\sqrt{0.84}}{2} \]

The roots are:

\[ y=0.0583\ \mathrm{M} \]

and

\[ y=-0.8583\ \mathrm{M} \]

Only the positive root is within the physical range.

Student 2 therefore keeps:

\[ y=0.0583\ \mathrm{M} \]

Step 7: Compare the two results.

Student 1 obtained:

\[ x=-0.0583\ \mathrm{M} \]

Student 2 obtained:

\[ y=+0.0583\ \mathrm{M} \]

Therefore:

\[ \boxed{y=-x} \]

Both students calculate the same equilibrium concentrations:

\[ \boxed{ \begin{aligned} [A]_{\mathrm{eq}}&=0.1583\ \mathrm{M}\\ [B]_{\mathrm{eq}}&=0.3583\ \mathrm{M}\\ [C]_{\mathrm{eq}}&=0.1417\ \mathrm{M} \end{aligned} } \]

The negative value of \(x\) was never a problem. It simply represented a reverse reaction relative to the equation Student 1 originally wrote.

The deeper lesson is that the sign of a reaction extent depends on your chosen direction. The physical equilibrium concentrations do not.

08What to remember in your HSC Chemistry calculations

When an equilibrium calculation produces multiple roots, keep these distinctions clear:

The most important rule is:

Never reject a root simply because \(x\) is negative. Reject it if it produces a physically impossible concentration or fails the other equilibrium conditions.

A reliable approach is to:

1. Write the initial row using the actual starting concentrations.
2. Use the balanced equation to construct the change row.
3. Add the rows to obtain the equilibrium expressions.
4. Find the allowable range of \(x\) by requiring every concentration to be non-negative.
5. Solve the \(K_c\) equation.
6. Keep only roots within the allowable range.
7. Substitute the remaining root into the original equilibrium expression and check the concentrations.

Once you can distinguish a negative reaction extent from a negative concentration, the next useful step is learning how to handle changes to an equilibrium mixture. In particular, adding or removing a reactant can change the initial row and reaction direction, so getting those two decisions right is essential before solving the next ICE table.