Why Every Mole of Water Counts in Enthalpy Calculations
Learn how balanced-equation coefficients determine water's contribution to reaction enthalpy, how its physical state changes the result, and why reaction water must be distinguished from calorimeter water.
Burn 1 mole of ethanol completely. How many moles of water should you include when calculating its enthalpy of combustion?
Not one. Three.
The balanced equation is:
\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]
If you calculate the reaction enthalpy using only one mole of water, you leave out the energy contribution from the other two moles. Your answer will be wrong, even if every other part of your calculation is correct.
The rule is simple: every substance in a balanced thermochemical equation must be multiplied by its stoichiometric coefficient when calculating the total enthalpy change. A coefficient of 3 means three moles, so you must include three times that substance’s molar enthalpy contribution.
There is one further distinction worth understanding: water formed in a chemical reaction is not necessarily the same water that absorbs heat in a calorimetry experiment. These two quantities are used in different calculations.
01Why the coefficient in front of water matters
Think about what a balanced chemical equation actually tells you.
The equation for ethanol combustion tells us that:
- 1 mole of ethanol reacts.
- 3 moles of oxygen react.
- 2 moles of carbon dioxide form.
- 3 moles of water form.
Those coefficients describe the mole ratios of the reaction. They aren’t optional numbers added to make the equation balance.
Now consider the standard enthalpy of formation of water:
\[ \Delta_\mathrm{f}H^\circ [\ce{H2O(l)}]=-286\ \mathrm{kJ\,mol^{-1}} \]
This value represents the enthalpy change associated with forming one mole of liquid water from its elements in their standard states.
If your reaction produces three moles of water, the total contribution from water is:
\[ 3\times(-286)=-858\ \mathrm{kJ} \]
Using only \(-286\ \mathrm{kJ}\) would account for just one of the three moles.
The same principle applies to every other substance in the equation. If a reaction produces four moles of carbon dioxide, you must include four times its standard enthalpy of formation.
The equation used to calculate reaction enthalpy
When standard enthalpies of formation are supplied, use:
\[ \Delta_\mathrm{r}H^\circ = \sum \nu\Delta_\mathrm{f}H^\circ \text{ (products)} - \sum \nu\Delta_\mathrm{f}H^\circ \text{ (reactants)} \]
Here:
- \(\Delta_\mathrm{r}H^\circ\) is the standard enthalpy change of the reaction.
- \(\Delta_\mathrm{f}H^\circ\) is the standard enthalpy of formation of a substance, in kJ mol⁻¹.
- \(\nu\) is the stoichiometric coefficient from the balanced equation.
- The summation means you include every substance on the relevant side of the equation.
Notice that the equation has two separate operations:
- Multiply each formation enthalpy by its coefficient.
- Subtract the total for the reactants from the total for the products.
The order matters. Products come first, reactants second.
An element in its standard state has a standard enthalpy of formation of zero. For example, \(\ce{O2(g)}\) has a value of zero.
Worked example: Calculate the enthalpy of ethanol combustion
Calculate the standard enthalpy change for the complete combustion of liquid ethanol using the following rounded, illustrative standard enthalpies of formation.
| Substance | Standard enthalpy of formation (kJ mol⁻¹) |
|---|---|
| \(\ce{C2H5OH(l)}\) | -278 |
| \(\ce{CO2(g)}\) | -394 |
| \(\ce{H2O(l)}\) | -286 |
| \(\ce{O2(g)}\) | 0 |
Step 1
\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]
There are three moles of water, so the water contribution must be multiplied by three.
Step 2
\[ \begin{aligned} \text{Products} &=2(-394)+3(-286)\\ &=-788-858\\ &=-1646\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Notice that water contributes \(-858\ \mathrm{kJ\,mol^{-1}}\), not \(-286\).
Step 3
\[ \begin{aligned} \text{Reactants} &=1(-278)+3(0)\\ &=-278\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 4
\[ \begin{aligned} \Delta_\mathrm{r}H^\circ &=-1646-(-278)\\ &=-1368\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 5
The standard enthalpy change is:
\[ \boxed{-1368\ \mathrm{kJ\,mol^{-1}}} \]
One mole of ethanol releases 1368 kJ when it undergoes complete combustion under the specified standard conditions, with liquid water as the final water product.
What happens if you only include one mole of water?
Let’s see how much the mistake changes the answer.
If you incorrectly include just one mole of water, your products calculation becomes:
\[ \begin{aligned} \text{Products} &=2(-394)+1(-286)\\ &=-1074 \end{aligned} \]
Your calculated reaction enthalpy would then be:
\[ \Delta H=-1074-(-278) \]
\[ \Delta H=-796\ \mathrm{kJ\,mol^{-1}} \]
The correct answer was \(-1368\ \mathrm{kJ\,mol^{-1}}\).
The difference is:
\[ -1368-(-796)=-572\ \mathrm{kJ\,mol^{-1}} \]
That’s exactly the contribution from the two missing moles of water:
\[ 2\times(-286)=-572 \]
You haven’t made a small rounding error. You’ve left out a significant part of the reaction’s energy change.
02Why complete equations matter when using Hess’s law
The same principle becomes particularly important when manipulating thermochemical equations.
Hess’s law states that the total enthalpy change of a reaction depends on its initial and final states, not on the pathway taken between them.
This allows you to combine known thermochemical equations to calculate an unknown reaction enthalpy.
However, there are three rules you must follow:
| Manipulation | What happens to ΔH? |
|---|---|
| Reverse an equation | Change the sign of ΔH |
| Multiply an equation by a number | Multiply ΔH by the same number |
| Add two equations | Add their enthalpy changes |
Most importantly, you must manipulate the entire equation, not just the substance you’re interested in.
Worked example: Find an unknown enthalpy by cancelling water
Use the following illustrative thermochemical equations to calculate the enthalpy change for:
\[ \ce{CH4(g) + H2O(l) -> CO(g) + 3H2(g)} \]
Equation 1:
\[ \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \]
\[ \Delta H_1=-891\ \mathrm{kJ} \]
Equation 2:
\[ \ce{CO(g) + \frac{1}{2}O2(g) -> CO2(g)} \]
\[ \Delta H_2=-283\ \mathrm{kJ} \]
Equation 3:
\[ \ce{H2(g) + \frac{1}{2}O2(g) -> H2O(l)} \]
\[ \Delta H_3=-286\ \mathrm{kJ} \]
Step 1
Compare the supplied equations with the target reaction.
- Equation 1 already has methane on the reactant side.
- Equation 2 needs to be reversed so carbon dioxide becomes carbon monoxide.
- Equation 3 needs to be reversed and multiplied by three to produce three moles of hydrogen.
Step 2
Reversing a reaction changes the sign of its enthalpy.
\[ \ce{CO2(g) -> CO(g) + \frac{1}{2}O2(g)} \]
\[ \Delta H_2=+283\ \mathrm{kJ} \]
Step 3
The reversed equation is:
\[ \ce{H2O(l) -> H2(g) + \frac{1}{2}O2(g)} \]
Multiplying every coefficient by three gives:
\[ \ce{3H2O(l) -> 3H2(g) + \frac{3}{2}O2(g)} \]
Its enthalpy change becomes:
\[ \Delta H_3=3(+286)=+858\ \mathrm{kJ} \]
Notice that you now have three moles of water on the reactant side.
Step 4
The water terms are:
- Equation 1 produces 2 moles of water.
- The reversed, tripled equation 3 consumes 3 moles of water.
Therefore, two moles of water cancel, leaving one mole on the reactant side.
Carbon dioxide and oxygen also cancel.
The resulting equation is:
\[ \ce{CH4(g) + H2O(l) -> CO(g) + 3H2(g)} \]
Step 5
\[ \begin{aligned} \Delta H &=-891+283+858\\ &=+250\ \mathrm{kJ} \end{aligned} \]
The final answer is:
\[ \boxed{\Delta H=+250\ \mathrm{kJ}} \]
The reaction is endothermic and absorbs 250 kJ for the reaction as written.
The important detail is that two of the three moles of water cancel. One mole remains.
If you had counted only one mole of water in equation 1, your cancellation would have been incorrect, and your final chemical equation would not represent the target reaction.
Syllabus note: Hess’s law and simple enthalpy cycles are part of the Chemistry Stage 6 (2017) syllabus, particularly Module 4: Drivers of Reactions. NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028, with its first HSC examination in 2029. Students should check the content for their particular cohort rather than assume every Hess-cycle technique is required under the newer syllabus.
You can read the Chemistry Stage 6 (2017) syllabus information and the Chemistry 11-12 Syllabus (2025) implementation details.
For more practice with reaction pathways and combustion data, see How to Build a Hess Cycle from Combustion Data.
03The physical state of water also changes the answer
There is another detail that can catch students out.
Water can appear as either a liquid or a gas in a thermochemical equation. These are different physical states, and they have different standard enthalpies of formation.
Consider the approximate values:
| Substance | Standard enthalpy of formation |
|---|---|
| \(\ce{H2O(l)}\) | -286 kJ mol⁻¹ |
| \(\ce{H2O(g)}\) | -242 kJ mol⁻¹ |
Why are the values different?
Because converting liquid water into water vapour requires energy. The enthalpy of vaporisation is approximately:
\[ \Delta H_\mathrm{vap}=+44\ \mathrm{kJ\,mol^{-1}} \]
That means liquid water has a lower enthalpy than water vapour under the same reference conditions.
Now return to ethanol combustion.
If the products are liquid water, the reaction enthalpy is:
\[ \Delta H=-1368\ \mathrm{kJ\,mol^{-1}} \]
If the products are water vapour, you must use the formation enthalpy of gaseous water instead.
\[ \begin{aligned} \Delta H &=2(-394)+3(-242)-(-278)\\ &=-788-726+278\\ &=-1236\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
The difference is:
\[ 3\times44=132\ \mathrm{kJ\,mol^{-1}} \]
That difference exists because three moles of water are produced per mole of ethanol.
The key lesson is that you must account for both:
- How many moles of water the equation produces.
- Whether that water is liquid or gaseous.
Using the correct coefficient but the wrong physical state still gives an incorrect result.
For standard combustion calculations at 298 K, liquid water is commonly the specified product. However, always follow the physical state given in the question.
04Don’t confuse reaction water with calorimeter water
So far, we’ve discussed water as a chemical product. But what if you’re calculating enthalpy using experimental calorimetry?
This introduces a second use of water.
In a typical school experiment, a fuel burns beneath a container of water. The heat released by combustion warms the water, and you measure its temperature change.
The energy absorbed by the water is calculated using:
\[ q=mc\Delta T \]
Where:
- \(q\) is the heat absorbed, in joules.
- \(m\) is the mass of the water being heated, in grams.
- \(c\) is the specific heat capacity, in J g⁻¹ °C⁻¹.
- \(\Delta T\) is the temperature change, in °C.
Here, \(m\) refers to the actual water or solution being heated in the calorimeter.
It does not automatically include the water produced by the chemical reaction.
For example, burning ethanol produces three moles of water per mole of ethanol. But if those products escape from the flame and do not enter the beaker, you must not add their mass to the water already in the beaker.
These are two different quantities:
| Water in the reaction | Water in the calorimeter |
|---|---|
| Determined by the balanced equation | Determined by the experimental setup |
| Counted using stoichiometric coefficients | Measured using mass |
| Used in thermochemical equations | Used in \(q=mc\Delta T\) |
| Represents a chemical product or reactant | Represents the material absorbing heat |
A measured enthalpy is not the same as a model estimate
You should also distinguish between two ways of obtaining an enthalpy change.
1. A calculated model estimate
Using standard enthalpies of formation, you calculate the enthalpy change for a complete reaction under standard conditions.
For ethanol combustion, we calculated:
\[ \Delta H_\mathrm{model}=-1368\ \mathrm{kJ\,mol^{-1}} \]
This assumes the specified reaction occurs completely and that the products have the stated physical states.
2. An experimental estimate
In calorimetry, you measure a temperature change and use it to estimate the energy transferred.
The result can differ substantially from the calculated value because:
- Some heat escapes into the surrounding air.
- The container and thermometer absorb heat.
- Combustion may be incomplete.
- Some fuel may evaporate without burning.
- The experimental conditions may differ from standard conditions.
Strictly speaking, the temperature change is measured. The heat transferred and molar enthalpy are calculated from that measurement.
Worked example: Estimate an experimental enthalpy of combustion
A student burns 0.460 g of ethanol beneath a container holding 200.0 g of water.
The water temperature increases by 6.50 °C.
Given:
- Specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹.
- Molar mass of ethanol = 46.0 g mol⁻¹.
- All the ethanol is consumed.
- Ignore the heat capacity of the container and use the heat gained by the water as the experimental estimate of heat released.
Calculate the apparent experimental molar enthalpy of combustion and compare it with the model estimate of \(-1368\ \mathrm{kJ\,mol^{-1}}\).
Step 1
\[ q=mc\Delta T \]
Substitute the supplied values:
\[ \begin{aligned} q&=200.0\times4.18\times6.50\\ &=5434\ \mathrm{J}\\ &=5.43\ \mathrm{kJ} \end{aligned} \]
The water absorbs 5.43 kJ.
Step 2
The reaction releases heat while the water absorbs it.
Therefore:
\[ q_\mathrm{reaction}\approx-5.43\ \mathrm{kJ} \]
The negative sign indicates an exothermic reaction.
Step 3
\[ n=\frac{m}{M} \]
\[ n=\frac{0.460}{46.0} =0.0100\ \mathrm{mol} \]
Step 4
\[ \begin{aligned} \Delta H_\mathrm{exp} &=\frac{-5.43}{0.0100}\\ &=-543\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 5
| Method | Enthalpy change |
|---|---|
| Formation enthalpy calculation | -1368 kJ mol⁻¹ |
| Experimental calorimetry estimate | -543 kJ mol⁻¹ |
The experimental estimate is much less exothermic.
This does not mean the formation enthalpy of water should have been counted only once. It means the simple experimental setup has not accounted for all the energy released by combustion.
The water’s temperature increase only tells you how much energy it gained. Some energy also heated the container and surrounding air.
The two calculations describe different ways of estimating the reaction’s energy change. Don’t confuse their limitations with an error in stoichiometric coefficients.
05A reliable method for enthalpy calculations
When you encounter water in an enthalpy question, follow this process.
| Step | What to do |
|---|---|
| 1. Balance | Write the complete balanced chemical equation. |
| 2. Count | Identify the coefficient of every water molecule. |
| 3. Check states | Determine whether water is liquid or gaseous. |
| 4. Multiply | Multiply each standard formation enthalpy by its coefficient. |
| 5. Subtract | Calculate products minus reactants. |
| 6. Check units | State the result in kJ for the equation as written, or kJ mol⁻¹ for the appropriate molar basis. |
For Hess’s law, also remember to reverse the sign when reversing an equation and scale the enthalpy whenever you multiply or divide the entire equation.
For calorimetry, use the actual mass of the water or solution being heated, and divide the inferred reaction heat by the moles that actually reacted.
If you need a refresher on converting a measured mass into moles, see Moles from Mass and Molar Mass: HSC Chemistry Guide.
06Questions and solutions
Try these questions without looking at the solutions first. They progress from a straightforward formation enthalpy calculation to problems combining reaction stoichiometry, experimental measurements, and physical states.
Question 1
Hydrogen combusts according to the equation:
\[ \ce{2H2(g) + O2(g) -> 2H2O(l)} \]
The standard enthalpy of formation of liquid water is \(-286\ \mathrm{kJ\,mol^{-1}}\). The standard enthalpies of formation of hydrogen and oxygen are both zero.
Calculate:
a) The enthalpy change for the reaction as written.
b) The enthalpy change per mole of hydrogen consumed.
Solution 1
The reaction enthalpy is \(-572\ \mathrm{kJ}\) as written, equivalent to \(-286\ \mathrm{kJ}\) per mole of hydrogen consumed.
Step 1: Identify the number of moles of water formed.
The coefficient of water is 2, so two moles of water form.
Step 2: Calculate the product contribution.
\[ 2\times(-286)=-572\ \mathrm{kJ} \]
Step 3: Calculate the reactant contribution.
Both reactants are elements in their standard states.
\[ 2(0)+1(0)=0 \]
Step 4: Subtract reactants from products.
\[ \Delta H=-572-0 \]
\[ \boxed{\Delta H=-572\ \mathrm{kJ}} \]
Step 5: Calculate the enthalpy change per mole of hydrogen.
The reaction consumes two moles of hydrogen.
\[ \frac{-572}{2} =-286\ \mathrm{kJ\,mol^{-1}} \]
The important distinction is that \(-572\ \mathrm{kJ}\) applies to the complete equation, which produces two moles of water. The molar value is \(-286\ \mathrm{kJ}\) per mole of hydrogen consumed.
Question 2
Propane undergoes complete combustion according to:
\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \]
The following rounded standard enthalpies of formation are supplied:
| Substance | ΔfH° (kJ mol⁻¹) |
|---|---|
| \(\ce{C3H8(g)}\) | -104 |
| \(\ce{CO2(g)}\) | -394 |
| \(\ce{H2O(l)}\) | -286 |
| \(\ce{O2(g)}\) | 0 |
a) Calculate the standard enthalpy change for the complete combustion of one mole of propane.
b) A student calculates the products contribution using three moles of carbon dioxide but only one mole of water. Calculate the incorrect enthalpy result and explain the size of the error.
Solution 2
The correct enthalpy change is \(-2222\ \mathrm{kJ\,mol^{-1}}\) of propane. Counting only one mole of water gives an incorrect result of \(-1364\ \mathrm{kJ\,mol^{-1}}\).
Step 1: Calculate the total enthalpy contribution from the products.
There are three moles of carbon dioxide and four moles of water.
\[ \begin{aligned} \text{Products} &=3(-394)+4(-286)\\ &=-1182-1144\\ &=-2326\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 2: Calculate the reactants contribution.
\[ \begin{aligned} \text{Reactants} &=1(-104)+5(0)\\ &=-104\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 3: Calculate the reaction enthalpy.
\[ \begin{aligned} \Delta H &=-2326-(-104)\\ &=\boxed{-2222\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]
Step 4: Calculate the student’s incorrect result.
The student uses only one mole of water:
\[ \begin{aligned} \text{Incorrect products} &=3(-394)+1(-286)\\ &=-1468 \end{aligned} \]
Therefore:
\[ \begin{aligned} \Delta H_\mathrm{incorrect} &=-1468-(-104)\\ &=-1364\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Step 5: Explain the difference.
The student has omitted three moles of water.
The missing enthalpy contribution is:
\[ 3\times(-286)=-858\ \mathrm{kJ\,mol^{-1}} \]
This is exactly the difference between the correct and incorrect answers.
The tempting approach is to treat water as a single product regardless of its coefficient. However, four moles of water are formed for every mole of propane consumed. All four must be included.
Question 3
Propane undergoes complete combustion according to:
\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \]
The enthalpy change for the reaction as written is \(-2222\ \mathrm{kJ}\).
A reaction mixture contains:
- 0.400 mol of propane.
- 1.50 mol of oxygen.
Assume the reaction proceeds as far as the available reactants allow.
Calculate:
a) The limiting reactant.
b) The number of moles of propane consumed.
c) The number of moles of water produced.
d) The total heat released.
e) The amount of propane remaining after the reaction.
Solution 3
Oxygen is the limiting reactant. The reaction consumes 0.300 mol of propane, produces 1.20 mol of water, and releases approximately 667 kJ.
Step 1: Identify the limiting reactant.
The balanced equation requires five moles of oxygen for every mole of propane.
To completely consume 0.400 mol of propane, the oxygen required would be:
\[ n(\ce{O2})=0.400\times5 \]
\[ n(\ce{O2})=2.00\ \mathrm{mol} \]
However, only 1.50 mol of oxygen is available.
Therefore, oxygen is the limiting reactant.
Step 2: Calculate the amount of propane that reacts.
From the mole ratio:
\[ \frac{n(\ce{C3H8})}{n(\ce{O2})} =\frac{1}{5} \]
\[ n(\ce{C3H8})=\frac{1.50}{5} \]
\[ \boxed{n(\ce{C3H8})=0.300\ \mathrm{mol}} \]
Step 3: Calculate the amount of water produced.
The balanced equation produces four moles of water per mole of propane.
\[ n(\ce{H2O})=0.300\times4 \]
\[ \boxed{n(\ce{H2O})=1.20\ \mathrm{mol}} \]
Step 4: Calculate the heat released.
The enthalpy change of \(-2222\ \mathrm{kJ}\) applies to one mole of reaction as written.
Only 0.300 mol of the reaction occurs.
\[ \begin{aligned} \Delta H &=0.300\times(-2222)\\ &=-666.6\ \mathrm{kJ} \end{aligned} \]
Therefore:
\[ \boxed{667\ \mathrm{kJ}\text{ released}} \]
Step 5: Calculate the propane remaining.
Initially, there were 0.400 mol of propane, and 0.300 mol reacted.
\[ n_\mathrm{remaining}=0.400-0.300 \]
\[ \boxed{0.100\ \mathrm{mol}} \]
The tempting approach is to use all 0.400 mol of propane when calculating the heat released and water produced. But the oxygen supply prevents all the propane from reacting.
The enthalpy calculation must be based on the actual reaction extent, not simply the amount of fuel initially present.
Question 4
An open calorimetry experiment is performed to estimate the enthalpy of combustion of ethanol.
A student supplies 0.500 g of ethanol to a spirit burner. Independent analysis confirms that only 0.400 g actually combusts; the remaining 0.100 g is recovered unreacted.
The burner heats 150.0 g of water, causing its temperature to increase by 5.00 °C.
Use the following information:
- Specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹.
- Molar mass of ethanol = 46.0 g mol⁻¹.
- Molar mass of water = 18.0 g mol⁻¹.
- The combustion products escape without entering the beaker.
- Ignore the container’s heat capacity.
a) Calculate the heat absorbed by the water.
b) Calculate the apparent experimental molar enthalpy of combustion using the amount of ethanol actually consumed.
c) Calculate the number of moles and mass of water produced.
d) Explain why the mass of product water must not be added to the mass of water in the beaker when using \(q=mc\Delta T\).
e) Explain why the experimental result may differ from the calculated standard enthalpy of combustion.
Solution 4
The apparent experimental molar enthalpy of combustion is approximately \(-361\ \mathrm{kJ\,mol^{-1}}\). The reaction produces 0.0261 mol of water, but this product water must not be added to the beaker’s 150.0 g of water.
Step 1: Calculate the heat absorbed by the beaker water.
\[ q=mc\Delta T \]
\[ \begin{aligned} q&=150.0\times4.18\times5.00\\ &=3135\ \mathrm{J}\\ &=3.135\ \mathrm{kJ} \end{aligned} \]
The water absorbs 3.135 kJ.
Step 2: Calculate the moles of ethanol actually consumed.
Only 0.400 g reacted.
\[ n=\frac{m}{M} \]
\[ n=\frac{0.400}{46.0} \]
\[ n=0.00870\ \mathrm{mol} \]
Step 3: Estimate the molar enthalpy change.
The water gains heat, so the reaction loses heat.
Using the simplified water-only calorimetry estimate:
\[ q_\mathrm{reaction}\approx-3.135\ \mathrm{kJ} \]
Divide by the actual moles of ethanol consumed:
\[ \begin{aligned} \Delta H_\mathrm{exp} &=\frac{-3.135}{0.00870}\\ &\approx-360.5\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]
Therefore:
\[ \boxed{\Delta H_\mathrm{exp}\approx-361\ \mathrm{kJ\,mol^{-1}}} \]
Step 4: Calculate the moles of water produced.
The balanced combustion equation is:
\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]
Three moles of water form per mole of ethanol.
\[ n(\ce{H2O})=3\times0.00870 \]
\[ \boxed{n(\ce{H2O})=0.0261\ \mathrm{mol}} \]
Step 5: Calculate the mass of product water.
\[ m=nM \]
\[ m=0.0261\times18.0 \]
\[ \boxed{m(\ce{H2O})=0.470\ \mathrm{g}} \]
Step 6: Explain which mass belongs in the calorimetry equation.
The correct mass is 150.0 g because that is the water physically heated in the beaker.
The 0.470 g of product water is part of the combustion products. Under the stated experimental conditions, it escapes without entering the beaker.
Adding it to the 150.0 g would confuse the chemical reaction’s stoichiometry with the calorimeter’s physical contents.
There is another important experimental detail. Using the original 0.500 g sample instead of the 0.400 g that actually reacted would give an incorrect number of moles in the denominator.
Step 7: Explain the difference between experimental and model values.
The calculated standard enthalpy of combustion of ethanol is approximately \(-1368\ \mathrm{kJ\,mol^{-1}}\) when liquid water is the final product.
The experimental estimate is much less negative because the simple open setup does not capture all the energy released.
Heat is transferred to the surroundings and apparatus, and incomplete combustion may also reduce the measured energy.
The difference is an experimental limitation, not a reason to omit water from the balanced equation.
Question 5
A chemistry student compares two calorimetry setups used to investigate the complete combustion of ethanol.
Both setups burn exactly 0.0200 mol of ethanol.
The balanced equation, when the water product is liquid, is:
\[ \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \]
Its standard enthalpy change is:
\[ \Delta H_\mathrm{liquid} =-1368\ \mathrm{kJ\,mol^{-1}} \]
The two experimental arrangements differ:
- Setup A: All the product water leaves as water vapour. No energy from subsequent condensation is captured.
- Setup B: All the product water condenses into liquid, and the heat released during condensation is captured.
The enthalpy of vaporisation of water is:
\[ \Delta H_\mathrm{vap} =+44.0\ \mathrm{kJ\,mol^{-1}} \]
Assume complete combustion, no other energy losses, and complete capture of the relevant heat in each setup.
A student argues:
“Both setups burn the same amount of ethanol, so they must have exactly the same enthalpy change. The physical state of the water shouldn’t affect the answer because the chemical reaction is identical.”
Evaluate this claim.
Calculate:
a) The enthalpy change per mole of ethanol when the product water remains gaseous.
b) The heat released in Setup A.
c) The heat released in Setup B.
d) The difference between the heat captured in the two setups.
Explain why the student’s initial prediction is incorrect.
Solution 5
The student’s claim is incorrect. Setup B releases an additional 2.64 kJ for the 0.0200 mol sample because three moles of water are produced per mole of ethanol, and all three moles release energy when they condense.
Step 1: Determine the number of moles of water produced.
The balanced equation produces three moles of water per mole of ethanol.
For 0.0200 mol of ethanol:
\[ n(\ce{H2O})=3\times0.0200 \]
\[ n(\ce{H2O})=0.0600\ \mathrm{mol} \]
This is the first important detail. We cannot treat the reaction as producing only one mole of water per mole of ethanol.
Step 2: Calculate the enthalpy change when water remains gaseous.
The supplied enthalpy change of \(-1368\ \mathrm{kJ\,mol^{-1}}\) includes liquid water as the final product.
To leave water as a gas instead, we must supply the energy required to vaporise all three moles.
For one mole of ethanol:
\[ \Delta H_\mathrm{vap,total} =3\times44.0 \]
\[ =132\ \mathrm{kJ\,mol^{-1}} \]
Therefore, the combustion reaction ending in water vapour has an enthalpy change of:
\[ \begin{aligned} \Delta H_\mathrm{gas} &=-1368+132\\ &=\boxed{-1236\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]
Notice the positive correction. Vaporisation requires energy, so the reaction becomes less exothermic.
Step 3: Calculate the heat released in Setup A.
Setup A produces gaseous water.
The amount of ethanol burned is 0.0200 mol.
\[ \begin{aligned} q_A &=0.0200\times(-1236)\\ &=-24.72\ \mathrm{kJ} \end{aligned} \]
Therefore, Setup A releases:
\[ \boxed{24.72\ \mathrm{kJ}} \]
Step 4: Calculate the heat released in Setup B.
Setup B captures the heat from the condensation of all the product water.
Its final product is liquid water, so use the supplied liquid-water combustion enthalpy.
\[ \begin{aligned} q_B &=0.0200\times(-1368)\\ &=-27.36\ \mathrm{kJ} \end{aligned} \]
Therefore, Setup B releases:
\[ \boxed{27.36\ \mathrm{kJ}} \]
Step 5: Calculate the difference between the two setups.
\[ \begin{aligned} \Delta q &=27.36-24.72\\ &=\boxed{2.64\ \mathrm{kJ}} \end{aligned} \]
We can independently verify this using the amount of water produced.
\[ \begin{aligned} q_\mathrm{condensation} &=n\Delta H_\mathrm{vap}\\ &=0.0600\times44.0\\ &=2.64\ \mathrm{kJ} \end{aligned} \]
Both methods give the same result.
Step 6: Explain why the student’s prediction fails.
The student correctly identifies that both setups burn the same amount of ethanol. However, they overlook the final physical state of the water.
In Setup A, the products remain gaseous. Some energy remains associated with the water vapour.
In Setup B, the water condenses. This releases additional energy into the surroundings.
The two setups therefore have different final thermodynamic states, so their measured heat transfers are different.
There is an important boundary distinction here: if water vapour from Setup A eventually condenses somewhere outside the calorimeter, that additional heat is released outside the measured system. It does not become part of Setup A’s calorimeter reading.
The most tempting mistake is to focus only on the amount of fuel burned. A complete enthalpy calculation must account for the amount of every substance and its physical state.
The main principle to remember is that a balanced thermochemical equation describes the entire reaction, not just one molecule of fuel or one molecule of water.
Whenever you calculate an enthalpy change, include every stoichiometric coefficient, use the correct physical states, and keep experimental heat measurements separate from theoretical reaction enthalpies. Once those three habits are established, formation enthalpy calculations and Hess’s law become much more reliable.