How to Use Measured Percentage Conversion in an ICE Table

A rigid reactor starts with \(0.800\ \mathrm{mol\,L^{-1}}\) of carbon monoxide. After the mixture reaches equilibrium, a gas analysis shows that 40.0% of the original CO has converted.

How much CO remains? And how do you use that measurement to calculate the equilibrium concentrations of the other gases?

The key is that 40.0% conversion tells you how much of the original CO has reacted. It does not tell you the percentage change of every substance in the reaction.

First, calculate the amount of CO consumed:

\[
\Delta[\ce{CO}]_{\mathrm{consumed}}
=0.400\times0.800
=0.320\ \mathrm{mol\,L^{-1}}
\]

So \(0.320\ \mathrm{mol\,L^{-1}}\) of CO has reacted, leaving \(0.480\ \mathrm{mol\,L^{-1}}\).

You then use the balanced chemical equation to calculate how much every other substance has changed. These changes fill the C row of your initial-change-equilibrium (ICE) table.

01Convert percentage conversion into a change

Percentage conversion measures the fraction of an initial reactant that has been consumed.

The general formula is:

\[
\%\text{ conversion}=
\frac{\text{amount of reactant consumed}}
{\text{initial amount of reactant}}
\times100
\]

Rearranging gives:

\[
\text{amount consumed}
=
\frac{\%\text{ conversion}}{100}
\times\text{initial amount}
\]

You can use this formula with moles or concentrations.

If the reaction takes place at constant volume, the ratio of concentrations is equivalent to the ratio of amounts in moles. Therefore:

\[
\Delta[A]_{\mathrm{consumed}}
=
\frac{p}{100}[A]_i
\]

where:

• \(p\) is the measured percentage conversion of reactant A.
• \([A]_i\) is its initial concentration in \(\mathrm{mol\,L^{-1}}\).
• \(\Delta[A]_{\mathrm{consumed}}\) is the positive magnitude of the concentration decrease.

In the ICE table, however, the change for a reactant is written with a negative sign.

Watch the denominator. Percentage conversion is calculated using the initial amount of the particular reactant being measured. It is not calculated using the total mixture or the equilibrium amount.

For example, if 40% of an initial \(0.800\ \mathrm{M}\) reactant converts:

• \(0.320\ \mathrm{M}\) reacts.
• \(0.480\ \mathrm{M}\) remains.
• The equilibrium concentration is 60% of the initial concentration, not 40%.

02Use the measured conversion to complete an ICE table

An ICE table organises the concentrations of every substance involved in an equilibrium reaction.

The three rows have specific jobs:

The balanced chemical equation determines the relationship between the changes.

Consider:

\[
\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}
\]

The coefficients are all 1. This means that for every mole of CO consumed:

• One mole of \(\ce{H2O}\) is consumed.
• One mole of \(\ce{CO2}\) is produced.
• One mole of \(\ce{H2}\) is produced.

If \(x\) represents the concentration of CO consumed, the change row becomes:

The signs matter:

• Reactants decrease in a forward reaction, so their changes are negative.
• Products increase in a forward reaction, so their changes are positive.

The equilibrium row is simply the initial row plus the change row.

How the measurement is obtained experimentally

It helps to understand what the percentage conversion represents.

Imagine investigating the water-gas shift equilibrium:

\[
\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}
\]

An illustrative experimental design could use a rigid, sealed reactor containing a known mixture of the four gases. The reactor is maintained at a constant temperature, and a catalyst helps the system reach equilibrium more quickly.

Once the reaction has proceeded, a small gas sample passes through a sampling valve and a heated transfer line into a calibrated gas chromatograph.

The chromatograph separates the gases as they pass through its column. A detector records signals for the different components. Comparing these signals with calibration standards allows the remaining CO concentration to be determined.

What should the experiment show?

As the forward reaction proceeds, CO and steam decrease while carbon dioxide and hydrogen increase. Eventually, repeated measurements show that the gas composition is no longer changing overall.

This is evidence that equilibrium has been reached. The forward and reverse reactions are still occurring, but at equal rates.

There is an important experimental limitation. If the sampling line cools the gas, some steam may condense. That changes the sample composition before it reaches the analyser. A heated sampling line helps prevent this problem.

The temperature of the reactor must also remain constant because changing temperature changes the equilibrium constant.

This is an illustrative experimental design, not a description of a particular historical experiment.

04What changes when the measured reactant has a coefficient greater than 1?

This is where students often make a mistake.

Suppose the balanced equation is:

\[
\ce{2SO2(g) + O2(g) <=> 2SO3(g)}
\]

The coefficient of \(\ce{SO2}\) is 2, while the coefficient of \(\ce{O2}\) is 1.

For every two moles of \(\ce{SO2}\) consumed:

• One mole of \(\ce{O2}\) is consumed.
• Two moles of \(\ce{SO3}\) are produced.

The measured conversion still tells you the amount of the particular reactant consumed. You must then apply the coefficient ratios.

If you define \(x\) as the reaction extent in concentration units, the changes are:

\[
\begin{aligned}
\Delta[\ce{SO2}]&=-2x\\
\Delta[\ce{O2}]&=-x\\
\Delta[\ce{SO3}]&=+2x
\end{aligned}
\]

This follows directly from the balanced equation.

More generally, for:

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

if \(x\) represents the concentration of A consumed, the changes are:

\[
\begin{aligned}
\Delta[A]&=-x\\
\Delta[B]&=-\frac{b}{a}x\\
\Delta[C]&=+\frac{c}{a}x\\
\Delta[D]&=+\frac{d}{a}x
\end{aligned}
\]

This version is useful when the measured conversion is given for a reactant whose coefficient is not 1.

05How to reject nonphysical mathematical roots

A measured percentage conversion usually gives you the change directly. You do not need to solve a quadratic to obtain it.

However, a quadratic can appear when you use the equilibrium constant calculated from experimental data to predict the equilibrium concentrations of a different starting mixture.

When this happens, you must check every mathematical root against the equilibrium row.

For a physically possible equilibrium, no concentration can be negative.

For example, consider:

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

Suppose the forward reaction proceeds by an amount \(x\), with initial concentrations \(a\), \(b\), and \(c\).

The equilibrium row is:

The physical constraints are:

\[
a-x\geq0
\]

\[
b-x\geq0
\]

\[
c+x\geq0
\]

Therefore, the allowed range is:

\[
-c\leq x\leq\min(a,b)
\]

Any mathematical root outside this range must be rejected.

An important distinction: A negative \(x\) is not automatically impossible. If products are present initially, a negative value can represent a physically valid shift in the reverse direction. You must check the actual equilibrium concentrations, not simply reject every negative root.

06A reliable decision process

When a question gives you a measured percentage conversion, follow this sequence.

There are three additional conditions worth remembering.

The measurement must represent equilibrium. A conversion measured while the reaction is still progressing cannot automatically be used as the equilibrium row.

Conversion is not the same as yield. Conversion measures how much of a reactant disappears. If competing reactions consume that reactant, its conversion alone does not tell you how much of the desired product formed. The ordinary ICE method assumes the specified reaction is the only reaction being considered.

Use moles if the volume changes. Percentage conversion is fundamentally based on amounts of substance. If volume changes during the reaction, calculate the amount consumed in moles first, then convert to concentrations using the relevant volume.

07Questions and solutions

The following questions progress from straightforward conversion calculations to problems involving stoichiometry, experimental interpretation, and changes in reaction conditions.

Question 1

Consider the equilibrium:

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

A vessel initially contains:

• \([A]_i=0.500\ \mathrm{M}\)
• \([B]_i=0.800\ \mathrm{M}\)
• \([C]_i=0.100\ \mathrm{M}\)

At equilibrium, 20.0% of the initial A has converted.

Calculate the equilibrium concentrations of A, B, and C.

Solution 1

The equilibrium concentrations are:

\[
[A]_{\mathrm{eq}}=0.400\ \mathrm{M}
\]

\[
[B]_{\mathrm{eq}}=0.700\ \mathrm{M}
\]

\[
[C]_{\mathrm{eq}}=0.200\ \mathrm{M}
\]

The key is to calculate the amount of A consumed before applying the stoichiometric ratios.

Step 1: Calculate A consumed.

\[
\Delta[A]_{\mathrm{consumed}}
=0.200\times0.500
=0.100\ \mathrm{M}
\]

Step 2: Apply the coefficients.

All coefficients are 1, so B decreases by \(0.100\ \mathrm{M}\), and C increases by \(0.100\ \mathrm{M}\).

Step 3: Complete the ICE table.

Notice that 20.0% conversion of A does not mean 20.0% conversion of B. The changes are determined by the balanced equation, not by applying the measured percentage to every reactant.

Question 2

Consider the equilibrium:

\[
\ce{2SO2(g) + O2(g) <=> 2SO3(g)}
\]

A vessel initially contains:

• \(1.20\ \mathrm{M}\) of \(\ce{SO2}\)
• \(0.700\ \mathrm{M}\) of \(\ce{O2}\)
• \(0.150\ \mathrm{M}\) of \(\ce{SO3}\)

Experimental measurements show that 30.0% of the initial oxygen has converted.

Calculate:

a) The equilibrium concentration of each gas.

b) The equilibrium constant \(K_c\), including its units under the concentration convention.

Solution 2

The equilibrium concentrations are:

\[
[\ce{SO2}]_{\mathrm{eq}}=0.780\ \mathrm{M}
\]

\[
[\ce{O2}]_{\mathrm{eq}}=0.490\ \mathrm{M}
\]

\[
[\ce{SO3}]_{\mathrm{eq}}=0.570\ \mathrm{M}
\]

The equilibrium constant is approximately:

\[
K_c=1.09\ \mathrm{L\,mol^{-1}}
\]

Step 1: Calculate the oxygen consumed.

\[
\Delta[\ce{O2}]_{\mathrm{consumed}}
=0.300\times0.700
=0.210\ \mathrm{M}
\]

Step 2: Use the stoichiometric ratios.

The balanced equation gives:

\[
2\ce{SO2}:1\ce{O2}:2\ce{SO3}
\]

Therefore, consuming \(0.210\ \mathrm{M}\) of oxygen consumes twice that concentration of sulfur dioxide:

\[
\Delta[\ce{SO2}]=-0.420\ \mathrm{M}
\]

The sulfur trioxide concentration increases by:

\[
\Delta[\ce{SO3}]=+0.420\ \mathrm{M}
\]

Step 3: Complete the ICE table.

Step 4: Calculate \(K_c\).

\[
K_c=
\frac{[\ce{SO3}]^2}
{[\ce{SO2}]^2[\ce{O2}]}
\]

Substitute:

\[
\begin{aligned}
K_c&=\frac{(0.570)^2}
{(0.780)^2(0.490)}\\
&=\frac{0.3249}{0.298116}\\
&=1.09\ \mathrm{L\,mol^{-1}}
\end{aligned}
\]

The important detail is that the measured conversion was for oxygen, whose coefficient is 1. The sulfur dioxide change must therefore be twice the oxygen change.

Question 3

Nitrogen reacts with hydrogen to form ammonia:

\[
\ce{N2(g) + 3H2(g) <=> 2NH3(g)}
\]

A rigid \(4.00\ \mathrm{L}\) vessel initially contains:

• \(1.20\ \mathrm{mol}\) of \(\ce{N2}\)
• \(4.50\ \mathrm{mol}\) of \(\ce{H2}\)
• \(0.200\ \mathrm{mol}\) of \(\ce{NH3}\)

At equilibrium, 25.0% of the initial nitrogen has converted.

Calculate:

a) The equilibrium number of moles of each gas.

b) The equilibrium concentration of each gas.

c) The value of \(K_c\).

Assume constant volume and no competing reactions.

Solution 3

The equilibrium amounts are:

• \(0.900\ \mathrm{mol}\) of \(\ce{N2}\)
• \(3.60\ \mathrm{mol}\) of \(\ce{H2}\)
• \(0.800\ \mathrm{mol}\) of \(\ce{NH3}\)

The equilibrium concentrations are:

• \(0.225\ \mathrm{M}\) of \(\ce{N2}\)
• \(0.900\ \mathrm{M}\) of \(\ce{H2}\)
• \(0.200\ \mathrm{M}\) of \(\ce{NH3}\)

The equilibrium constant is:

\[
K_c=0.244\ \mathrm{L^2\,mol^{-2}}
\]

Step 1: Calculate the nitrogen consumed.

\[
n(\ce{N2})_{\mathrm{consumed}}
=0.250\times1.20
=0.300\ \mathrm{mol}
\]

Step 2: Apply the mole ratios.

The balanced equation gives:

\[
1\ce{N2}:3\ce{H2}:2\ce{NH3}
\]

Therefore:

\[
n(\ce{H2})_{\mathrm{consumed}}
=3(0.300)
=0.900\ \mathrm{mol}
\]

\[
n(\ce{NH3})_{\mathrm{produced}}
=2(0.300)
=0.600\ \mathrm{mol}
\]

Step 3: Calculate the equilibrium amounts.

For nitrogen:

\[
1.20-0.300=0.900\ \mathrm{mol}
\]

For hydrogen:

\[
4.50-0.900=3.60\ \mathrm{mol}
\]

For ammonia:

\[
0.200+0.600=0.800\ \mathrm{mol}
\]

Step 4: Convert the equilibrium amounts into concentrations.

Use:

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

For nitrogen:

\[
[\ce{N2}]_{\mathrm{eq}}
=\frac{0.900}{4.00}
=0.225\ \mathrm{M}
\]

For hydrogen:

\[
[\ce{H2}]_{\mathrm{eq}}
=\frac{3.60}{4.00}
=0.900\ \mathrm{M}
\]

For ammonia:

\[
[\ce{NH3}]_{\mathrm{eq}}
=\frac{0.800}{4.00}
=0.200\ \mathrm{M}
\]

Step 5: Calculate \(K_c\).

The equilibrium expression is:

\[
K_c=
\frac{[\ce{NH3}]^2}
{[\ce{N2}][\ce{H2}]^3}
\]

Substituting:

\[
\begin{aligned}
K_c&=\frac{(0.200)^2}
{(0.225)(0.900)^3}\\
&=\frac{0.0400}{0.164025}\\
&=0.244\ \mathrm{L^2\,mol^{-2}}
\end{aligned}
\]

The main challenge here is that the percentage conversion gives you an amount in moles, while the equilibrium expression requires concentrations.

Since the vessel volume is constant, convert all equilibrium amounts using the same \(4.00\ \mathrm{L}\) volume.

Question 4

A student investigates the equilibrium:

\[
\ce{2A(g) + B(g) <=> A2B(g)}
\]

The initial concentrations are:

• \([A]_i=0.600\ \mathrm{M}\)
• \([B]_i=0.500\ \mathrm{M}\)
• \([\ce{A2B}]_i=0.100\ \mathrm{M}\)

At equilibrium, 30.0% of the initial A has converted.

The student writes the following change row:

They conclude that the equilibrium concentration of \(\ce{A2B}\) is \(0.280\ \mathrm{M}\).

Identify the student’s error, calculate the correct equilibrium concentrations, and explain why the student’s answer violates conservation of atoms.

Solution 4

The correct equilibrium concentrations are:

\[
[A]_{\mathrm{eq}}=0.420\ \mathrm{M}
\]

\[
[B]_{\mathrm{eq}}=0.410\ \mathrm{M}
\]

\[
[\ce{A2B}]_{\mathrm{eq}}=0.190\ \mathrm{M}
\]

The student correctly calculated how much A was consumed but failed to apply the stoichiometric ratios to B and \(\ce{A2B}\).

Step 1: Calculate A consumed.

\[
\Delta[A]_{\mathrm{consumed}}
=0.300\times0.600
=0.180\ \mathrm{M}
\]

Step 2: Determine the reaction extent.

The balanced equation contains two A particles for every one B particle and one \(\ce{A2B}\) particle.

Therefore:

\[
2x=0.180
\]

\[
x=0.090\ \mathrm{M}
\]

The correct changes are:

• A decreases by \(0.180\ \mathrm{M}\).
• B decreases by \(0.090\ \mathrm{M}\).
• \(\ce{A2B}\) increases by \(0.090\ \mathrm{M}\).

Step 3: Complete the ICE table.

Step 4: Check conservation of atoms.

Each molecule of \(\ce{A2B}\) contains two A atoms.

Therefore, the total concentration of A atoms must remain constant:

\[
[A]+2[\ce{A2B}]
\]

Initially:

\[
0.600+2(0.100)=0.800\ \mathrm{M}
\]

At equilibrium:

\[
0.420+2(0.190)=0.800\ \mathrm{M}
\]

The total is conserved.

Now check the student’s result:

\[
0.420+2(0.280)=0.980\ \mathrm{M}
\]

The student has apparently created \(0.180\ \mathrm{M}\) of A atoms from nowhere.

That is impossible. Their error is not merely an incorrect ICE table entry; it violates the conservation of atoms.

Question 5

This final problem combines measured conversion, equilibrium constants, dilution, and physical root selection.

Consider the gaseous equilibrium:

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

Two separate experiments are conducted at the same temperature.

Experiment 1

A rigid \(1.00\ \mathrm{L}\) vessel initially contains:

• \(1.00\ \mathrm{mol}\) of A
• \(1.00\ \mathrm{mol}\) of B
• No C

At equilibrium, measurements show that 60.0% of the initial A has converted.

Experiment 2

A second rigid vessel contains exactly the same initial amounts:

• \(1.00\ \mathrm{mol}\) of A
• \(1.00\ \mathrm{mol}\) of B
• No C

However, this vessel has a volume of \(2.00\ \mathrm{L}\).

The temperature is unchanged.

A student argues:

“The equilibrium constant is the same in both experiments, so the percentage conversion must also be 60.0%.”

Determine whether the student is correct.

Calculate the equilibrium percentage conversion in Experiment 2, reject any nonphysical mathematical roots, and explain the result using the relationship between volume and equilibrium position.

Assume ideal gas behaviour and that the specified reaction is the only reaction occurring.

Solution 5

The student is incorrect. The equilibrium conversion in Experiment 2 is approximately:

\[
\boxed{48.9\%}
\]

Although \(K_c\) remains constant at the same temperature, changing the initial concentrations changes the equilibrium composition.

Step 1: Calculate \(K_c\) using Experiment 1.

The initial concentrations in the \(1.00\ \mathrm{L}\) vessel are:

\[
[A]_i=[B]_i=1.00\ \mathrm{M}
\]

The measured conversion of A is 60.0%.

Therefore:

\[
x=0.600\times1.00
=0.600\ \mathrm{M}
\]

The ICE table is:

The equilibrium expression is:

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

Substitute:

\[
\begin{aligned}
K_c&=\frac{0.600}{(0.400)(0.400)}\\
&=3.75\ \mathrm{L\,mol^{-1}}
\end{aligned}
\]

This value applies to Experiment 2 because the temperature is unchanged.

Step 2: Calculate the initial concentrations in Experiment 2.

The vessel volume has doubled, but the initial amounts have not changed.

Using:

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

For A:

\[
[A]_i=\frac{1.00}{2.00}
=0.500\ \mathrm{M}
\]

For B:

\[
[B]_i=\frac{1.00}{2.00}
=0.500\ \mathrm{M}
\]

The initial concentration of C is zero.

Step 3: Construct the ICE table.

Let \(x\) represent the concentration of A and B consumed.

Step 4: Substitute into the equilibrium expression.

The equilibrium constant is still 3.75:

\[
3.75=
\frac{x}{(0.500-x)^2}
\]

Rearranging:

\[
3.75(0.500-x)^2=x
\]

Expand:

\[
3.75(0.250-1.00x+x^2)=x
\]

Collect terms:

\[
3.75x^2-4.75x+0.9375=0
\]

Step 5: Solve the quadratic and check the roots.

Using the quadratic formula:

\[
x=\frac{4.75\pm\sqrt{8.50}}{7.50}
\]

This gives:

\[
x=0.245\ \mathrm{M}
\]

or

\[
x=1.02\ \mathrm{M}
\]

The second root is physically impossible.

Both A and B initially have concentrations of only \(0.500\ \mathrm{M}\). Consuming \(1.02\ \mathrm{M}\) of either reactant would produce a negative equilibrium concentration.

Therefore, reject that root.

The valid reaction extent is:

\[
x=0.245\ \mathrm{M}
\]

Step 6: Calculate the percentage conversion.

The initial concentration of A in Experiment 2 is \(0.500\ \mathrm{M}\).

Therefore:

\[
\begin{aligned}
\%\text{ conversion}
&=\frac{0.245}{0.500}\times100\\
&=48.9\%
\end{aligned}
\]

The equilibrium conversion in Experiment 2 is 48.9%, not 60.0%.

Step 7: Explain why the conversion changed.

The forward reaction is:

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

There are two moles of gaseous reactants for every one mole of gaseous product.

Increasing the volume reduces the pressure and dilutes the mixture. The equilibrium responds by favouring the side with more gas particles: the reactants.

Consequently, less A converts into C.

The equilibrium constant does not change because the temperature is unchanged. However, the equilibrium concentrations do change to satisfy the same \(K_c\).

This is the important distinction:

The equilibrium constant is fixed at a given temperature, but percentage conversion depends on the initial conditions as well as the equilibrium constant.

08Where to go next

Using measured conversion is one way to fill an ICE table. The other common approach is to start with initial concentrations and use a supplied \(K_c\) to solve for the unknown reaction extent.

The two methods rely on exactly the same stoichiometric relationships.

For a broader explanation of building ICE tables and calculating equilibrium concentrations from \(K_c\), read ICE Tables for Equilibrium: Calculate Concentrations and Use Kc.

If your equilibrium question gives initial gas amounts in moles and a fixed container volume, continue with How to Calculate Gas Equilibrium at Constant Volume.

The next skill is recognising which information a question gives you: a measured conversion, an equilibrium constant, or an initial mixture. Once you identify that, you can choose the correct starting point for your ICE table.