Chemical Thermodynamics · Lesson 3 of 10

Enthalpy, Heat Capacity, and Calorimetry

“Connect measurable temperature changes with internal energy and enthalpy.”

Learning Objectives

• Derive enthalpy and state the conditions for heat to equal an energy-function change. • Calculate the difference between reaction enthalpy and internal-energy change. • Distinguish extensive, intensive, specific, and molar properties. • Use heat capacities and explain the ideal-gas molar heat-capacity difference. • Interpret constant-volume and constant-pressure calorimetry.

Why chemists introduce enthalpy

In a rigid vessel, heat supplied to a closed system changes its internal energy if no other work occurs. Most reactions, however, occur under approximately constant atmospheric pressure. If a gas forms or is consumed, the system can expand or contract while heat is transferred. Enthalpy is a state function that makes the constant-pressure energy account easier to use.

Assume the only work is pressure–volume work and the opposing pressure equals the constant pressure p at the equilibrium endpoints. The first law becomes $Δ U = q_{p} - p (V_{2} - V_{1})$ . Moving the volume term to the other side gives $q_{p} = (U_{2} + p V_{2}) - (U_{1} + p V_{1})$ . This combination suggests defining a new state property.

EnthalpyLaTeX

H = U + p V, Δ H = Δ U + Δ (p V)

H is enthalpy, U internal energy, p pressure, and V volume. H is a state function because U, p, and V are state functions.

At constant pressure, $Δ (p V) = p Δ V$ . The derivation then gives $q_{p} = Δ H$ . A negative enthalpy change means the system releases heat at constant pressure: it is exothermic. A positive change means it absorbs heat: it is endothermic. These equalities assume only pressure–volume work; electrical or other additional work must be included separately.

Experimental condition	Measured heat equals	Required qualification
Constant volume	$q_{V} = Δ U$	No non-volume work
Constant pressure	$q_{p} = Δ H$	Only pressure–volume work
General change	$Δ H = Δ U + Δ (p V)$	This is the defining state-function relation

Common mistake

Constant volume does not by itself imply $Δ H = Δ U$ . Pressure may change, so $Δ (p V)$ can be nonzero. The two changes are equal when the endpoint pV term is unchanged; they are often approximately equal for changes involving only condensed phases with negligible pV changes.

Counting gaseous moles

For ideal gaseous reactants and products compared at the same temperature, each gas contributes $n R T$ to pV. Solids and liquids usually make a much smaller pV contribution. Subtracting the gas contributions at the two endpoints gives a useful reaction correction. Only gaseous species are counted.

Enthalpy and internal energy of a gas reactionLaTeX

Δ_{r} H = Δ_{r} U + Δ n_{g} R T

Δn_g is the stoichiometric number of moles of gaseous products minus gaseous reactants per reaction as written. The relation assumes ideal gases and negligible condensed-phase pV contributions.

Example — Vaporisation changes the gas amount

Problem
One mole of liquid water becomes ideal water vapour at 373 K. The enthalpy change is 41.00 kJ. Find the internal-energy change.

1.The initial gas amount is zero and the final gas amount is one mole, so the gas contribution increases by $R T$ .
2.Calculate $R T = (8.314) (373) = 3101 J mo l^{- 1} \approx 3.10 kJ mo l^{- 1}$ .
3.Rearrange the relation: $Δ U = Δ H - R T = 41.00 - 3.10 = 37.90 kJ$ for the stated one mole.
4.The heat absorbed at constant pressure exceeds the increase in internal energy because the expanding vapour also does work.

Example — A reaction with no net gaseous-mole change

Problem
Compare $Δ H$ and $Δ U$ for $C (s) + O_{2} (g) \to C O_{2} (g)$ , assuming ideal gases.

1.Count only gases: there is one mole of gas on each side.
2.Thus $Δ n_{g} = 1 - 1 = 0$ . The solid carbon is not included in this count.
3.The gas correction $Δ n_{g} R T$ is zero, so $Δ H \approx Δ U$ under the stated approximation.
4.This is an equality of changes for this reaction, not a statement that the functions H and U are identical.

Size-dependent properties and heat capacity

Doubling a uniform sample doubles its mass, volume, internal energy, and enthalpy, but does not double its temperature or pressure. This distinction helps us compare samples of different sizes. Imagine partitioning a gas into two equal parts: each has half the amount and volume, while its temperature remains the same.

Definition

Extensive and intensive properties

An extensive property depends on the amount of matter, such as mass, volume, U, H, or total heat capacity. An intensive property does not scale with sample amount, such as temperature, pressure, or density. Dividing an extensive property by amount gives a molar property.

Partitioning a uniform sample— Halving the amount halves extensive properties. The intensive properties remain unchanged.

Heat capacity C tells us how much heat is needed for a specified temperature increase under the chosen process conditions. A large heat capacity means the same heat produces a smaller temperature rise. Total heat capacity scales with sample size. Molar heat capacity $C_{m} = C / n$ refers to one mole; specific heat capacity $c = C / m$ refers to unit mass.

Heat and temperature riseLaTeX

q = C Δ T = m c Δ T = n C_{m} Δ T

Use this form when heat capacity is approximately constant over the temperature interval and no phase change occurs. C has units J K−1, c J g−1 K−1 when mass is in grams, and C_m J mol−1 K−1.

Example — Heating aluminium

Problem
Find the heat required to raise 60.0 g of aluminium from 35 to 55 degrees Celsius. Its molar heat capacity is $24 J mo l^{- 1} K^{- 1}$ and molar mass is $27.0 g mo l^{- 1}$ .

1.Convert mass to amount: $n = 60.0/27.0 = 2.22 mol$ .
2.The temperature increase is $55 - 35 = 20 K$ .
3.Use molar rather than mass-specific capacity: $q = n C_{m} Δ T = (60.0/27.0) (24) (20) = 1067 J$ .
4.The required heat is approximately $1.07 kJ$ , positive because aluminium absorbs it.

A gas has different heat capacities at constant volume and at constant pressure. At constant volume the energy supplied changes U. At constant pressure some energy is also needed for expansion. For one mole of ideal gas, $H = U + R T$ , so $Δ H = Δ U + R Δ T$ . Substituting $C_{p, m} Δ T$ and $C_{V, m} Δ T$ and dividing by $Δ T$ gives the molar relationship.

Ideal-gas heat capacitiesLaTeX

C_{p, m} - C_{V, m} = R

For an n-mole sample, the total-capacity difference is C_p − C_V = nR. The simple difference R applies to molar capacities.

Measuring a reaction’s heat

A calorimeter turns a heat transfer into a measurable temperature change. The reaction mixture and the surrounding calorimeter must be distinguished: if the reaction loses heat, the calorimeter gains it. For an approximately insulated assembly, their heat changes add to zero. Its calibrated effective heat capacity can include the vessel, water, and other parts.

Calorimeter energy balanceLaTeX

q_{reaction} = - q_{cal} = - C_{cal} Δ T

A rising calorimeter temperature means positive q_cal and negative reaction heat. At constant volume this gives reaction ΔU; at constant pressure it gives ΔH under the usual work assumptions.

Constant-volume and constant-pressure measurements— In a bomb the rigid wall prevents boundary work. A cup arrangement permits the volume response associated with atmospheric pressure.

In a bomb calorimeter a combustible sample burns in oxygen inside a strong sealed vessel immersed in water. The bomb volume stays fixed. In a constant-pressure cup calorimeter the reaction occurs under atmospheric pressure, often in an insulated cup with a thermometer. Neither apparatus determines absolute U or H; it measures an energy change for a stated amount and process.

Example — Reading a bomb-calorimeter dataset

Problem
A supplied calculation dataset burns 1.00 g of graphite in excess oxygen. A calorimeter of effective heat capacity $20.7 kJ K^{- 1}$ warms from 298 to 299 K. Calculate the energy change implied by these data.

1.The calorimeter temperature change is $+ 1.00 K$ , so $q_{cal} = (20.7) (1.00) = + 20.7 kJ$ .
2.The reacting sample therefore has $q_{V} = - 20.7 kJ$ . With no other work, this is its $Δ U$ .
3.Using $12.0 g mo l^{- 1}$ for carbon, the inferred molar change is $(- 20.7) (12.0/1.00) = - 248.4 kJ mo l^{- 1}$ .
4.For carbon to carbon dioxide, $Δ n_{g} = 0$ , so the inferred molar enthalpy change is approximately the same. This result belongs to the supplied dataset; it is not a replacement for independently tabulated reference combustion data.

Quick check

When does heat absorbed at constant pressure equal the enthalpy change?

Quick check

Which quantity is extensive?

Quick check

For one mole of an ideal gas, which heat-capacity relationship is correct?

Quick check

A calorimeter warms by 2 K and has effective heat capacity 5 kJ/K. What heat did the reaction release?

Quick check

For ideal gases in $N_{2} + 3 H_{2} \to 2N H_{3}$ , what is the gas-mole correction to $Δ U$ ?

Quick check

Why must a molar heat capacity not be multiplied directly by mass in grams?

Practice Problems

Derive $q_{p} = Δ H$ and state the assumptions used.
For $N H_{2} CN (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ , calculate $Δ H$ at 298 K if $Δ U = - 742.7 kJ mo l^{- 1}$ .
A 100 g sample has specific heat capacity $0.90 J g^{- 1} K^{- 1}$ . Find the heat required for a 15 K rise.
A constant-volume calorimeter of heat capacity 8.0 kJ/K warms by 0.50 K when 0.010 mol of sample reacts. Find the molar internal-energy change.
Classify mass, density, enthalpy, temperature, molar volume, and total heat capacity as extensive or intensive.
Explain why $Δ V = 0$ alone does not prove that $Δ H = Δ U$ .

Key Takeaways

Enthalpy is $H = U + p V$ ; the work conditions determine whether measured heat equals ΔU or ΔH. For ideal-gas reactions, the gaseous-mole change supplies the Δn_gRT correction. Use total, mass-specific, or molar heat capacity with a matching sample measure. Calorimeter heat and reaction heat have opposite signs; the apparatus fixes whether the result measures ΔU or ΔH.

Previous · Lesson 2

Pressure–Volume Work and Gas Expansion

Next · Lesson 4

Reaction Enthalpy, Standard States, and Phase Changes

Why chemists introduce enthalpy

Counting gaseous moles

Size-dependent properties and heat capacity

Measuring a reaction’s heat

Quiz

Practice Problems

Key Takeaways