8.3 Use formation enthalpies in reaction energy balances | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Chemical Reactions and Combustion

MEC E 340 Applied Thermodynamics — Study topic 8.3

This lesson considers a reacting mixture as either a closed system, with no mass crossing its boundary, or a steady-flow control volume, with reactants entering and products leaving. Number the reactants as state 1 and the products as state 2 when that helps track the process. We use supplied formation enthalpies at a stated reference condition and assume the reaction equation is balanced. Formation enthalpies account for the chemical contribution to enthalpy; they do not replace the first-law balance. Take heat into the system or control volume as positive. Unless a problem says otherwise, we will state when changes in kinetic and potential energy and work can be neglected.

What you will learn

Calculate reaction enthalpy from balanced reaction coefficients and supplied formation enthalpies.
Keep reaction basis, species phase, and energy units consistent.
Apply reaction enthalpy in a closed-system or steady-flow energy balance under stated assumptions.
Interpret the sign of heat transfer using a clearly stated sign convention.

1. Formation enthalpy gives the reaction enthalpy

The standard enthalpy of formation, written as

\Delta H_f^\circ

, is the enthalpy change for forming one mole of a substance from its elements in their standard reference forms at the stated reference condition. Its value depends on the substance and its phase, so match each reaction species to the phase in the supplied data.

An element in its standard reference form is assigned a standard formation enthalpy of zero. This is a reference convention, not a statement that the element contains no energy. Use the supplied values and do not substitute a value for a different phase.

For a balanced reaction, multiply each species’ formation enthalpy by its stoichiometric coefficient. Add the product terms and subtract the reactant terms. The result is the reaction enthalpy for the reaction as written. Multiplying the entire reaction by a factor multiplies its reaction enthalpy by the same factor.

\Delta H_{\mathrm{rxn}}^\circ=\sum_{\mathrm{products}}\nu_i\Delta H_{f,i}^\circ-\sum_{\mathrm{reactants}}\nu_i\Delta H_{f,i}^\circ

Balance atoms before calculating reaction enthalpy.
Use products minus reactants, with each term weighted by its coefficient.
Report the reaction basis and units, such as per mole of fuel reacted.

2. Choose the correct first-law balance

Before using a reaction enthalpy, identify the boundary. A closed-system balance has no mass-flow terms. A steady-flow control-volume balance includes enthalpy carried in and out. In both cases, state the sign convention and whether work, kinetic energy, and potential energy changes are negligible.

For a closed system, the first law is based on the change in total energy, which includes internal, kinetic, and potential energy. If the kinetic and potential energy changes are negligible, the balance relates heat and work to the change in internal energy. It does not generally equate heat transfer to a change in enthalpy.

A useful special case is a closed system undergoing a constant-pressure process with only pressure-volume work, negligible changes in kinetic and potential energy, and no other work. Under these conditions, heat transfer equals the system’s enthalpy change. A reaction enthalpy can represent that change when the reaction occurs at the stated reference condition and other enthalpy changes are negligible or accounted for.

For a steady-flow control volume, with negligible kinetic and potential energy changes and no shaft work, heat transfer equals the net enthalpy flow out minus the net enthalpy flow in. For reacting streams, form the stream enthalpies consistently: use formation enthalpies for chemical contributions and include sensible enthalpy increments when supplied or required. A reference-condition reaction enthalpy alone does not include a product stream’s temperature rise.

Q-W=\Delta U+\Delta KE+\Delta PE

Do not substitute enthalpy for internal energy without conditions that justify it.
For a steady-flow balance, compare total enthalpy flow out with total enthalpy flow in.
With heat positive inward, negative heat transfer means heat leaves the boundary.

3. A consistent solution method

First write and balance the reaction, including the phases specified by the problem. Identify whether the requested result is per mole of reaction, per mole of fuel, a total energy, or a rate. This prevents a correct per-mole result from being mistaken for a total heat transfer or power.

Next list the supplied formation enthalpies with their units, phases, and reaction coefficients. Calculate the product sum minus the reactant sum. If the problem gives sensible enthalpy increments, add them on the same basis before applying the energy balance.

Finally write the appropriate closed-system or steady-flow balance. Include any work and energy terms that the assumptions require. Convert a per-reaction result to a total amount or rate only after the reaction basis is clear. Check that atom balances hold, units agree, and the sign of heat transfer fits the stated convention.

\Delta H_{\mathrm{streams}}=\Delta H_{\mathrm{rxn}}^\circ+\Delta H_{\mathrm{sensible}}

Formation-enthalpy sums give a reaction enthalpy, not automatically heat transfer.
Include temperature-related enthalpy changes when the problem supplies them.
State assumptions and carry the basis through every calculation.

Worked example

Reaction enthalpy for methane combustion

For

\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)}

, use the supplied formation enthalpies in kJ/mol: methane,

-74.8

; oxygen,

0

; carbon dioxide,

-393.5

; and liquid water,

-285.8

. Find the standard reaction enthalpy per mole of methane reacted. Assume all data use the same standard reference condition.

Confirm the basis
The equation is balanced: one mole of methane reacts with two moles of oxygen. The requested basis is one mole of reaction as written, equivalent here to one mole of methane consumed.
Calculate the product sum
Multiply each product’s supplied formation enthalpy by its reaction coefficient. Use the liquid-water value because the products specify liquid water.
$-393.5+2(-285.8)=-965.1\ \mathrm{kJ/mol\ reaction}$
Subtract the reactant sum
Oxygen has zero formation enthalpy under the standard-element convention. Subtract the reactant sum from the product sum.
$\Delta H_{\mathrm{rxn}}^\circ=-965.1-[-74.8+2(0)]=-890.3\ \mathrm{kJ/mol\ CH_4}$

Answer: The standard reaction enthalpy is

-890.3\ \mathrm{kJ}

per mole of methane reacted.

Check: The negative sign means the product enthalpy sum is lower than the reactant sum at the stated reference condition. The result is per mole of methane, not a total energy for an unspecified amount.

Worked example

Heat transfer in a constant-pressure reacting system

A closed system undergoes one mole of the balanced reaction

\mathrm{H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)}

at the supplied reference condition. Supplied formation enthalpies are

0

,

0

, and

-241.8\ \mathrm{kJ/mol}

for hydrogen gas, oxygen gas, and water vapour, respectively. The system remains at constant pressure, has only pressure-volume work, and has negligible changes in kinetic and potential energy. Find the heat transfer, taking heat into the system as positive.

Find the reaction enthalpy
Use the balanced coefficients and the supplied gas-phase water value. The result is the enthalpy change for one mole of reaction at the stated reference condition.
$\Delta H_{\mathrm{rxn}}^\circ=-241.8-[0+\frac{1}{2}(0)]=-241.8\ \mathrm{kJ/mol\ reaction}$
Apply the stated closed-system conditions
For a closed system at constant pressure with only pressure-volume work, and negligible kinetic and potential energy changes, the first law gives heat transfer equal to enthalpy change. The supplied reference-condition reaction enthalpy is the applicable enthalpy change here.
$Q=\Delta H=-241.8\ \mathrm{kJ}$

Answer: The heat transfer is

-241.8\ \mathrm{kJ}

for one mole of reaction, so

241.8\ \mathrm{kJ}

leaves the system.

Check: The result depends on the stated constant-pressure and work assumptions. A rigid vessel would instead require an internal-energy change in the closed-system first-law balance; reaction enthalpy alone would not determine its heat transfer.

Worked example

Steady-flow combustor with a sensible enthalpy increment

A steady-flow control volume burns methane completely according to

\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)}

. Methane and oxygen enter at the reference condition at

1.00\ \mathrm{mol/s}

and

2.00\ \mathrm{mol/s}

. Products leave with a supplied total sensible enthalpy increment of

+80.0\ \mathrm{kJ}

per mole of methane reacted, relative to that reference condition. Supplied formation enthalpies in kJ/mol are methane,

-74.8

; oxygen,

0

; carbon dioxide,

-393.5

; and water vapour,

-241.8

. Neglect shaft work and changes in kinetic and potential energy. Find the heat-transfer rate, with heat into the control volume positive.

Steady-flow combustor

Schematic control volume with methane and oxygen entering and products leaving. Heat is shown outward; shaft work is neglected.

Calculate the reference-condition reaction enthalpy
Use the gas-phase water value and the balanced coefficients. The result is per mole of methane reacted.
$[-393.5+2(-241.8)]-[-74.8+2(0)]=-802.3\ \mathrm{kJ/mol\ CH_4}$
Include the supplied sensible increment
The reactants enter at the reference condition, so their sensible increments are zero on this basis. Add the given product increment to obtain the net stream enthalpy change per mole of methane.
$\Delta h_{\mathrm{streams}}=-802.3+80.0=-722.3\ \mathrm{kJ/mol\ CH_4}$
Apply the steady-flow balance
With no shaft work and negligible kinetic and potential energy changes, heat-transfer rate equals net enthalpy flow out minus net enthalpy flow in. The methane reaction rate is $1.00\ \mathrm{mol/s}$ , so multiplying the per-mole change gives kJ/s, equivalent to kW.
$\dot Q=(1.00\ \mathrm{mol/s})(-722.3\ \mathrm{kJ/mol})=-722.3\ \mathrm{kW}$

Answer: The heat-transfer rate is

-722.3\ \mathrm{kW}

, meaning

722.3\ \mathrm{kW}

of heat leaves the control volume.

Check: The sensible increment makes the stream enthalpy change less negative but does not reverse its sign. The calculated heat rate is therefore outward under the stated convention.

Common mistakes and how to avoid them

Adding reactant and product formation enthalpies instead of taking their difference.

Correction: Calculate the coefficient-weighted product sum minus the coefficient-weighted reactant sum.

Using a formation enthalpy for a different phase from the species in the reaction.

Correction: Match each species’ phase to the supplied formation-enthalpy data.

Treating reaction enthalpy as heat transfer without checking the system and process.

Correction: Write the applicable first-law balance and state the assumptions that connect heat transfer to enthalpy change.

Using a closed-system enthalpy balance for a rigid vessel.

Correction: A closed-system balance uses changes in total energy. For a rigid vessel with negligible kinetic and potential energy changes, relate heat and work to internal-energy change; do not substitute reaction enthalpy without a justified relation.

Reporting a per-mole result as a total energy or a power.

Correction: Keep the reaction basis visible and multiply by the reaction amount or rate when needed.

Lesson summary

Balance the reaction and identify the phase of every species.
Calculate reaction enthalpy from coefficient-weighted formation enthalpies: products minus reactants.
Keep chemical and sensible enthalpy contributions on a consistent basis.
Use the correct closed-system or steady-flow first-law balance and state its assumptions.
Check signs, units, reaction basis, and physical interpretation.

Check your understanding

Question 1

At the stated reference condition, what does a negative reaction enthalpy indicate?

The products have lower enthalpy than the reactants.
Every possible process must absorb heat.
The products must be at a lower temperature.
The reaction equation is unbalanced.

Show answer and explanation

The products have lower enthalpy than the reactants.

Reaction enthalpy is the product enthalpy sum minus the reactant enthalpy sum at the stated reference condition. A negative value means the product sum is lower; heat transfer in a particular process still depends on its energy balance.

Question 2

If every coefficient in a balanced reaction is multiplied by three, how does the reaction enthalpy for the equation as written change?

It is divided by three.
It is multiplied by three.
It becomes zero.
It is unchanged.

Show answer and explanation

It is multiplied by three.

The coefficient-weighted formation-enthalpy sums scale by the same factor as the reaction, so the reaction enthalpy also triples.

Question 3

A steady-flow control volume has a net stream enthalpy flow out minus in of

-25\ \mathrm{kW}

. There is no shaft work, and kinetic and potential energy changes are negligible. What is the heat-transfer rate if heat into the control volume is positive?

$+25\ \mathrm{kW}$
$-25\ \mathrm{kW}$
$0\ \mathrm{kW}$
It cannot be determined from the stated balance.

Show answer and explanation

-25\ \mathrm{kW}

Under the stated steady-flow assumptions, heat-transfer rate equals net enthalpy flow out minus in. The negative sign indicates heat leaves the control volume.

Key terms

Standard enthalpy of formation: Enthalpy change for forming one mole of a substance from its elements in their standard reference forms at the stated reference condition.
Reaction enthalpy: The coefficient-weighted product formation-enthalpy sum minus the corresponding reactant sum.
Sensible enthalpy increment: The enthalpy change associated with a temperature difference from the chosen reference condition.
Reaction basis: The specified amount of reaction used to express a result, such as one mole of fuel consumed or a stated reaction rate.

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8.3 · Use formation enthalpies in reaction energy balances

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 8.3

What you will learn

1. Formation enthalpy gives the reaction enthalpy

2. Choose the correct first-law balance

3. A consistent solution method

Reaction enthalpy for methane combustion

Heat transfer in a constant-pressure reacting system

Steady-flow combustor with a sensible enthalpy increment

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review