How to Choose the Physical Root of an Equilibrium Quadratic

Your equilibrium quadratic gives two positive answers: \(x=0.0845\) and \(x=3.32\ \mathrm{mol\,L^{-1}}\). Both solve the equation. So which one describes the actual reaction?

Only the first. The larger root would consume more reactant than the mixture contains, producing negative equilibrium concentrations.

The physical root is the mathematically valid solution that also satisfies the chemical constraints of the reaction. To identify it, check the equilibrium concentrations, the amount of each substance available, and the direction of the reaction. Never choose a root simply because it is positive, negative, smaller, or larger.

There is an important twist: a negative root can sometimes be the correct answer. It depends on how you define \(x\).

01What makes an equilibrium root physical?

An equilibrium quadratic usually comes from an initial-change-equilibrium (ICE) table. The table tracks how the concentrations of reactants and products change as a reversible reaction approaches equilibrium.

Each row has a different job:

The relationship connecting the rows is:

\[ \boxed{E=I+C} \]

The change row is particularly important because it determines the expressions you eventually substitute into \(K_c\).

Understand what \(x\) represents

Consider the reversible reaction:

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

Suppose the reaction proceeds in the forward direction by an amount \(x\ \mathrm{mol\,L^{-1}}\).

One mole of A reacts with one mole of B to produce one mole of C. Therefore:

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

The ICE table becomes:

Here, \(a\), \(b\), and \(c\) are the initial concentrations in \(\mathrm{mol\,L^{-1}}\).

For this article, we’ll use a signed forward-reaction convention:

• \(x>0\): the net reaction proceeds forwards.
• \(x<0\): the net reaction proceeds backwards.
• \(x=0\): the mixture is already at equilibrium.

This means a negative value of \(x\) is not automatically a mistake.

The physical constraint: concentrations cannot be negative

Imagine starting with:

\[ [A]_0=0.80\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0.60\ \mathrm{mol\,L^{-1}} \]

\[ [C]_0=0.10\ \mathrm{mol\,L^{-1}} \]

The equilibrium concentrations are:

\[ [A]_{\mathrm{eq}}=0.80-x \]

\[ [B]_{\mathrm{eq}}=0.60-x \]

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

Every concentration must be zero or greater.

Therefore:

\[ 0.80-x\geq 0 \]

\[ 0.60-x\geq 0 \]

\[ 0.10+x\geq 0 \]

Combining these inequalities gives:

\[ \boxed{-0.10\leq x\leq 0.60} \]

This is the physically allowed interval for \(x\).

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

Notice that the upper limit is \(0.60\), not \(0.80\). B is the limiting reactant, so the reaction cannot consume more than the initial amount of B.

Check the reaction direction using Qc and Kc

There is another useful constraint: the reaction must move in the direction predicted by the reaction quotient.

The reaction quotient, \(Q_c\), uses the concentrations at the moment you examine the mixture. The equilibrium constant, \(K_c\), describes the equilibrium ratio at that temperature.

For:

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

the reaction quotient is:

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

Compare the initial \(Q_c\) with \(K_c\):

This gives you an independent check on the sign of your answer.

For example, if \(Q_c>K_c\), the reaction must move backwards. A positive forward extent would contradict that prediction.

If your initial mixture already contains products, remember to include them in the initial row. Their presence does not automatically mean they must decrease. The comparison between \(Q_c\) and \(K_c\) determines the direction.

For a more detailed explanation of that situation, see How to Set Up an ICE Table When Product Is Present Initially.

04A reliable method for choosing the physical root

When you reach a quadratic in an equilibrium calculation, use this sequence.

There are three common mistakes worth watching for.

Mistake 1: Always choosing the positive root.

A negative root can represent a perfectly valid reverse reaction when \(x\) is defined as forward progress.

Mistake 2: Always choosing the smaller root.

The smaller root often turns out to be physical, but that is a consequence of the available concentrations, not a general mathematical rule.

Mistake 3: Assuming that solving the quadratic finishes the question.

The quadratic gives mathematical solutions. You still need to check whether those solutions describe a possible chemical mixture.

One final check is useful: substitute your chosen root into the original equilibrium expression, not just the expanded quadratic. This confirms that the equilibrium constant is satisfied and that the original expression is defined.

The examples and practice questions below use illustrative data. Assume concentration-based equilibrium expressions, constant volume during each equilibration, and no competing reactions. The concentration changes are determined by the balanced equation.

05Questions and solutions

Question 1

Consider the reaction:

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

Initially, the vessel contains:

\[ [A]_0=0.50\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0 \]

At constant temperature:

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

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

Calculate the equilibrium concentrations and identify the physical root.

Solution 1

The physical root is \(x=0.1351\ \mathrm{mol\,L^{-1}}\), giving \([A]_{\mathrm{eq}}=0.3649\ \mathrm{mol\,L^{-1}}\) and \([B]_{\mathrm{eq}}=0.2702\ \mathrm{mol\,L^{-1}}\).

Step 1: Construct the ICE table.

The balanced equation shows that one mole of A produces two moles of B.

Step 2: Establish the allowed interval.

For A:

\[ 0.50-x\geq0 \]

Therefore, \(x\leq0.50\).

For B:

\[ 2x\geq0 \]

Therefore, \(x\geq0\).

Combining these:

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

Step 3: Form the equilibrium equation.

For this reaction:

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

Substitute:

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

Rearrange:

\[ 4x^2=0.20(0.50-x) \]

\[ 4x^2+0.20x-0.10=0 \]

Step 4: Solve the quadratic.

\[ x=\frac{-0.20\pm\sqrt{0.20^2-4(4)(-0.10)}}{8} \]

The roots are:

\[ x_1=0.1351 \]

\[ x_2=-0.1851 \]

The second root is outside the allowed interval because it would give a negative concentration of B.

Therefore:

\[ x=0.1351\ \mathrm{mol\,L^{-1}} \]

Step 5: Calculate the equilibrium concentrations.

\[ [A]_{\mathrm{eq}}=0.50-0.1351 \]

\[ \boxed{[A]_{\mathrm{eq}}=0.3649\ \mathrm{mol\,L^{-1}}} \]

\[ [B]_{\mathrm{eq}}=2(0.1351) \]

\[ \boxed{[B]_{\mathrm{eq}}=0.2702\ \mathrm{mol\,L^{-1}}} \]

Substitution gives:

\[ K_c=\frac{0.2702^2}{0.3649} \approx0.20 \]

The result satisfies the equilibrium expression and all physical constraints.

Question 2

Consider the reversible reaction:

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

A vessel initially contains:

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

\[ [B]_0=0.10\ \mathrm{mol\,L^{-1}} \]

\[ [C]_0=0.0248\ \mathrm{mol\,L^{-1}} \]

At constant temperature:

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

A quadratic calculation produces two positive roots.

Determine both roots, identify the physical root, and calculate the equilibrium concentrations.

Solution 2

The physical root is \(x=0.020\ \mathrm{mol\,L^{-1}}\). The larger positive root, \(0.880\ \mathrm{mol\,L^{-1}}\), is impossible because it would make both reactant concentrations negative.

Step 1: Determine the direction of the reaction.

\[ Q_c=\frac{0.0248}{0.30\times0.10} \]

\[ Q_c=0.8267\ \mathrm{L\,mol^{-1}} \]

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

Therefore, \(x\) should be positive.

Step 2: Construct the ICE table.

Step 3: Establish the allowed interval.

The equilibrium concentrations must satisfy:

\[ 0.30-x\geq0 \]

\[ 0.10-x\geq0 \]

\[ 0.0248+x\geq0 \]

Therefore:

\[ \boxed{-0.0248\leq x\leq0.10} \]

Step 4: Form the quadratic.

\[ 2.00=\frac{0.0248+x} {(0.30-x)(0.10-x)} \]

Expanding:

\[ 2.00(0.030-0.40x+x^2)=0.0248+x \]

Rearranging:

\[ 2x^2-1.80x+0.0352=0 \]

Factorising:

\[ 2(x-0.020)(x-0.880)=0 \]

Hence:

\[ x_1=0.020 \]

\[ x_2=0.880 \]

Step 5: Reject the nonphysical root.

The allowed interval is:

\[ -0.0248\leq x\leq0.10 \]

The root \(0.020\) lies inside this interval.

The root \(0.880\) exceeds the maximum possible extent of \(0.10\).

It would produce:

\[ [A]_{\mathrm{eq}}=0.30-0.88=-0.58 \]

\[ [B]_{\mathrm{eq}}=0.10-0.88=-0.78 \]

Both are impossible.

Step 6: Calculate the final concentrations.

Using \(x=0.020\):

\[ [A]_{\mathrm{eq}}=0.30-0.020 \]

\[ \boxed{[A]_{\mathrm{eq}}=0.28\ \mathrm{mol\,L^{-1}}} \]

\[ [B]_{\mathrm{eq}}=0.10-0.020 \]

\[ \boxed{[B]_{\mathrm{eq}}=0.08\ \mathrm{mol\,L^{-1}}} \]

\[ [C]_{\mathrm{eq}}=0.0248+0.020 \]

\[ \boxed{[C]_{\mathrm{eq}}=0.0448\ \mathrm{mol\,L^{-1}}} \]

Check:

\[ K_c=\frac{0.0448}{0.28\times0.08} \]

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

The smaller root is correct in this particular example because it satisfies the chemical constraints, not because smaller roots are always physical.

Question 3

A prepared aqueous mixture initially contains:

\[ [\mathrm{HA}]_0=0.100\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{H}^{+}]_0=0.0100\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{A}^{-}]_0=0.0100\ \mathrm{mol\,L^{-1}} \]

The weak acid dissociates according to:

\[ \ce{HA(aq) <=> H+(aq) + A-(aq)} \]

At this temperature:

\[ K_a=1.80\times10^{-5} \]

Assume constant volume and that any spectator ions do not participate in the equilibrium expression.

Using \(x\) as the signed forward dissociation extent:

a) Determine the direction of the net reaction.

b) Construct the ICE table and calculate both mathematical roots.

c) Identify the physical root and calculate the equilibrium concentrations.

Solution 3

The physical root is \(x=-0.00860\ \mathrm{mol\,L^{-1}}\). The reaction shifts backwards, consuming hydrogen ions and conjugate base ions to form more HA.

Step 1: Determine the initial reaction quotient.

For the acid dissociation:

\[ Q_a=\frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]} {[\mathrm{HA}]} \]

Substitute the initial concentrations:

\[ Q_a=\frac{0.0100\times0.0100}{0.100} \]

\[ Q_a=0.00100 \]

Since:

\[ Q_a>K_a \]

the reaction shifts backwards.

Because \(x\) is defined as positive forward dissociation, we expect a negative root.

Step 2: Construct the ICE table.

Step 3: Establish the allowed interval.

For HA:

\[ 0.100-x\geq0 \]

Therefore:

\[ x\leq0.100 \]

For hydrogen ions:

\[ 0.0100+x\geq0 \]

Therefore:

\[ x\geq-0.0100 \]

The conjugate base gives the same lower limit.

Thus:

\[ \boxed{-0.0100\leq x\leq0.100} \]

Step 4: Form the quadratic.

Substitute the equilibrium concentrations into \(K_a\):

\[ 1.80\times10^{-5} = \frac{(0.0100+x)^2}{0.100-x} \]

Cross-multiply:

\[ (0.0100+x)^2 = 1.80\times10^{-5}(0.100-x) \]

Expand:

\[ 0.000100+0.0200x+x^2 = 0.00000180-0.0000180x \]

Rearrange:

\[ x^2+0.020018x+0.00009820=0 \]

Step 5: Solve and test the roots.

Using the quadratic formula:

\[ x_1=-0.00860\ \mathrm{mol\,L^{-1}} \]

\[ x_2=-0.0114\ \mathrm{mol\,L^{-1}} \]

Both roots are negative, which is consistent with the reverse reaction.

However, the physical interval is:

\[ -0.0100\leq x\leq0.100 \]

The first root lies inside this interval.

The second root is too negative. It would produce negative concentrations of both \(\mathrm{H}^{+}\) and \(\mathrm{A}^{-}\).

Therefore:

\[ \boxed{x=-0.00860\ \mathrm{mol\,L^{-1}}} \]

Step 6: Calculate the equilibrium concentrations.

For HA:

\[ [\mathrm{HA}]_{\mathrm{eq}} =0.100-(-0.00860) \]

\[ \boxed{[\mathrm{HA}]_{\mathrm{eq}} =0.10860\ \mathrm{mol\,L^{-1}}} \]

For hydrogen ions:

\[ [\mathrm{H}^{+}]_{\mathrm{eq}} =0.0100-0.00860 \]

\[ \boxed{[\mathrm{H}^{+}]_{\mathrm{eq}} =0.00140\ \mathrm{mol\,L^{-1}}} \]

For the conjugate base:

\[ \boxed{[\mathrm{A}^{-}]_{\mathrm{eq}} =0.00140\ \mathrm{mol\,L^{-1}}} \]

The reverse shift increases the concentration of undissociated acid while reducing the concentrations of its ions.

The negative root is correct because it describes a physically possible reaction in the predicted direction.

Question 4

Consider the reaction:

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

Initially:

\[ [A]_0=0.60\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0.1392\ \mathrm{mol\,L^{-1}} \]

At constant temperature:

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

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

A student argues that the maximum possible forward extent is \(0.60\ \mathrm{mol\,L^{-1}}\), because this is the initial concentration of A.

a) Explain why this reasoning is incorrect.

b) Construct the ICE table and calculate the two quadratic roots.

c) Identify the physical root and calculate the equilibrium concentrations.

Solution 4

The physical root is \(x=0.120\ \mathrm{mol\,L^{-1}}\). The maximum forward extent is \(0.300\ \mathrm{mol\,L^{-1}}\), not \(0.600\), because each reaction event consumes two A particles.

Step 1: Explain the stoichiometric constraint.

The balanced equation is:

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

For every \(x\) amount of forward reaction:

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

Therefore, the equilibrium concentration of A is:

\[ [A]_{\mathrm{eq}}=0.60-2x \]

For A to remain non-negative:

\[ 0.60-2x\geq0 \]

Rearranging:

\[ x\leq0.30 \]

The maximum forward extent is therefore \(0.30\ \mathrm{mol\,L^{-1}}\).

The student has forgotten the coefficient of 2 in the balanced equation.

Step 2: Determine the reaction direction.

The equilibrium expression is:

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

The initial reaction quotient is:

\[ Q_c=\frac{0.1392}{0.60^2} \]

\[ Q_c=0.3867\ \mathrm{L\,mol^{-1}} \]

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

We expect \(x>0\).

Step 3: Construct the ICE table.

The allowed interval is:

\[ \boxed{-0.1392\leq x\leq0.300} \]

Step 4: Form the quadratic.

Substitute into the equilibrium expression:

\[ 2.00=\frac{0.1392+x}{(0.60-2x)^2} \]

Cross-multiply:

\[ 2.00(0.60-2x)^2=0.1392+x \]

Expand:

\[ 2.00(0.36-2.40x+4x^2)=0.1392+x \]

Rearrange:

\[ \boxed{8x^2-5.80x+0.5808=0} \]

Step 5: Solve and identify the physical root.

Using the quadratic formula:

\[ x=\frac{5.80\pm\sqrt{15.0544}}{16} \]

The two roots are:

\[ x_1=0.120 \]

\[ x_2=0.605 \]

The allowed interval ends at \(0.300\).

Therefore, \(x=0.605\) is impossible.

It would give:

\[ [A]_{\mathrm{eq}}=0.60-2(0.605) \]

\[ [A]_{\mathrm{eq}}=-0.61\ \mathrm{mol\,L^{-1}} \]

The physical root is:

\[ \boxed{x=0.120\ \mathrm{mol\,L^{-1}}} \]

Step 6: Calculate the equilibrium concentrations.

For A:

\[ [A]_{\mathrm{eq}}=0.60-2(0.120) \]

\[ \boxed{[A]_{\mathrm{eq}}=0.36\ \mathrm{mol\,L^{-1}}} \]

For B:

\[ [B]_{\mathrm{eq}}=0.1392+0.120 \]

\[ \boxed{[B]_{\mathrm{eq}}=0.2592\ \mathrm{mol\,L^{-1}}} \]

Check:

\[ K_c=\frac{0.2592}{0.36^2} \]

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

The result satisfies the equilibrium expression and the stoichiometric constraints.

Question 5

A sealed, rigid vessel contains the following equilibrium mixture at temperature \(T_1\):

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

The equilibrium concentrations are:

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

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

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

At \(T_1\), the equilibrium constant is \(K_{c,1}\).

The vessel is then rapidly brought to a different temperature, \(T_2\), without changing its volume or removing any gas. At the new temperature:

\[ K_{c,2}=0.75\ \mathrm{L\,mol^{-1}} \]

Assume the concentrations do not change during the instantaneous temperature adjustment. The system is then allowed to re-establish equilibrium at \(T_2\).

A student argues:

“The mixture was already at equilibrium before the temperature change, so the reaction extent must remain \(x=0\). Also, any negative root from the new quadratic must be rejected.”

Evaluate both claims.

Calculate the new equilibrium concentrations, showing why the physical root is the one it is.

Solution 5

The student’s two claims are incorrect. The temperature change alters the equilibrium constant, so the original mixture is no longer at equilibrium. The correct physical root is \(x=-0.0272\ \mathrm{mol\,L^{-1}}\), which represents a reverse reaction.

This question requires you to distinguish between an instantaneous change in temperature and the subsequent chemical response.

Step 1: Calculate the original equilibrium constant.

At \(T_1\), the mixture is at equilibrium.

Therefore:

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

Substitute:

\[ K_{c,1}=\frac{0.10}{0.40\times0.20} \]

\[ \boxed{K_{c,1}=1.25\ \mathrm{L\,mol^{-1}}} \]

Step 2: Determine what happens immediately after the temperature change.

The vessel is rigid, so its volume remains constant.

No gas has been added or removed, and the chemical composition cannot change instantaneously.

Therefore, immediately after the temperature change:

\[ [A]=0.40 \]

\[ [B]=0.20 \]

\[ [C]=0.10 \]

However, the equilibrium constant has changed to:

\[ K_{c,2}=0.75\ \mathrm{L\,mol^{-1}} \]

The reaction quotient immediately after the temperature change is still:

\[ Q_c=\frac{0.10}{0.40\times0.20} \]

\[ Q_c=1.25\ \mathrm{L\,mol^{-1}} \]

Now compare:

\[ Q_c=1.25>K_{c,2}=0.75 \]

The reaction must shift backwards.

The original composition is no longer an equilibrium composition at \(T_2\).

Step 3: Construct a new ICE table.

Use the concentrations immediately after the temperature change as the new initial row.

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

Since the reaction moves backwards, we expect \(x<0\).

Step 4: Establish the physically allowed interval.

For A:

\[ 0.40-x\geq0 \]

Therefore:

\[ x\leq0.40 \]

For B:

\[ 0.20-x\geq0 \]

Therefore:

\[ x\leq0.20 \]

For C:

\[ 0.10+x\geq0 \]

Therefore:

\[ x\geq-0.10 \]

Combining the constraints:

\[ \boxed{-0.10\leq x\leq0.20} \]

Step 5: Form the new equilibrium quadratic.

Use the new equilibrium constant, not the old one.

\[ 0.75= \frac{0.10+x}{(0.40-x)(0.20-x)} \]

Expand the denominator:

\[ 0.75(0.08-0.60x+x^2)=0.10+x \]

Rearrange:

\[ \boxed{0.75x^2-1.45x-0.04=0} \]

Step 6: Solve the quadratic.

Using the quadratic formula:

\[ x= \frac{1.45\pm \sqrt{1.45^2-4(0.75)(-0.04)}} {1.50} \]

The two roots are approximately:

\[ x_1=-0.0272\ \mathrm{mol\,L^{-1}} \]

\[ x_2=1.961\ \mathrm{mol\,L^{-1}} \]

Now apply the allowed interval:

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

The first root is inside the interval.

The second root is much too large and would produce negative concentrations of A and B.

Therefore:

\[ \boxed{x=-0.0272\ \mathrm{mol\,L^{-1}}} \]

The negative value is physically correct because the reaction shifts backwards.

Step 7: Calculate the new equilibrium concentrations.

For A:

\[ [A]_{\mathrm{eq}}=0.40-(-0.0272) \]

\[ \boxed{[A]_{\mathrm{eq}}=0.4272\ \mathrm{mol\,L^{-1}}} \]

For B:

\[ [B]_{\mathrm{eq}}=0.20-(-0.0272) \]

\[ \boxed{[B]_{\mathrm{eq}}=0.2272\ \mathrm{mol\,L^{-1}}} \]

For C:

\[ [C]_{\mathrm{eq}}=0.10+(-0.0272) \]

\[ \boxed{[C]_{\mathrm{eq}}=0.0728\ \mathrm{mol\,L^{-1}}} \]

Check the new equilibrium constant:

\[ K_{c,2}= \frac{0.0728}{0.4272\times0.2272} \]

\[ \boxed{K_{c,2}\approx0.75\ \mathrm{L\,mol^{-1}}} \]

The new equilibrium composition satisfies the new equilibrium constant.

The important conceptual point is that an equilibrium mixture at one temperature is not necessarily an equilibrium mixture at another temperature. The concentrations initially remain unchanged, but the equilibrium constant can change.

This is why the original value of \(x=0\) cannot simply be carried over.

Once you can select the physical root confidently, the next useful skill is setting up the initial row correctly when a mixture is prepared from measured moles, mixed solutions, or an existing equilibrium mixture.

For the complete gas-equilibrium calculation process, including converting initial amounts into concentrations and constructing ICE tables, see How to Calculate Gas Equilibrium at Constant Volume.