Enthalpy, Heat Capacities & Thermochemistry

Table of Contents
1. Enthalpy (H)
2. Heat Capacity and Specific Heats
3. Solved Examples
4. Degrees of Freedom (DOF)
5. Law of Equipartition of Energy
6. Thermochemistry
7. Rules for Thermochemical Equations
8. Properties of Matter: Intensive and Extensive
9. Further Discussion on Enthalpy
10. Kirchhoff's Equation and Temperature Dependence of Enthalpy
11. Different Types of Enthalpies
12. Bomb Calorimeter
13. Other Types of Enthalpies
14. Born-Haber Cycle (brief)
15. Worked Numerical Example (Standard Enthalpy Calculation)
16. Final Notes and Important Points to Remember
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Enthalpy (H)

Enthalpy is a thermodynamic function defined as the sum of the internal energy and the product of pressure and volume of the system. It is denoted by H and is an extensive state function.

Mathematically,

The absolute value of enthalpy for a system cannot be determined; only changes in enthalpy can be measured experimentally. For a change between two states 1 and 2,

ΔH = H₂ - H₁ = ΔU + Δ(PV)

At constant pressure the heat exchanged with surroundings, q_p, equals the change in enthalpy:

q_p = ΔH

Sign convention:

• ΔH < 0 for exothermic reactions (heat is evolved).
• ΔH > 0 for endothermic reactions (heat is absorbed).

Derivation of ΔH = q_p

Start from the first law of thermodynamics:

ΔU = q + w

For pressure-volume work only, w = -P ΔV. At constant pressure, denote heat as q_p:

ΔU = q_p - P ΔV

Rearrange:

q_p = ΔU + P ΔV = (U₂ + P V₂) - (U₁ + P V₁)

Define enthalpy H = U + PV, then

q_p = H₂ - H₁ = ΔH

Since U, P and V are state functions, H is also a state function and ΔH is path independent for a given initial and final state at the same pressure (and temperature) conditions.

Relationship between ΔH and ΔU

General relation:

ΔH = ΔU + Δ(PV)

For reactions involving ideal gases at constant temperature and pressure the change in PV can be related to change in number of moles of gases.

Using PV = nRT, for reactants (A) and products (B):

P ΔV = (n_B - n_A) R T = Δn_g R T

Therefore

ΔH = ΔU + Δn_g R T

Note: At constant volume (ΔV = 0),

ΔU = q_V

and thus ΔH = ΔU + P ΔV reduces to ΔH = ΔU when ΔV = 0.

Heat Capacity and Specific Heats

Heat Capacity

Heat capacity (C) of a system is the amount of heat required to raise the temperature of that system by 1 K (or 1 °C).

For an infinitesimal heat d q producing a temperature change dT,

C = d q / dT

• SI unit: J K^-1.
• Heat capacity is an extensive property (depends on the amount of substance).

Molar Heat Capacity

Molar heat capacity is the amount of heat required to raise the temperature of 1 mole of a substance by 1 K.

SI unit: J mol^-1 K^-1.

Specific Heat Capacity

Specific heat capacity (c) is the heat required to raise the temperature of 1 unit mass of a substance by 1 K.

• m = mass of substance (in g or kg).
• c = specific heat capacity (J g^-1 K^-1 or J kg^-1 K^-1).
• ΔT = temperature change.

Heat Capacities at Constant Volume and Constant Pressure

• At constant volume: q_V = ΔU = C_V ΔT
• At constant pressure: q_p = ΔH = C_P ΔT

Relation between C_P and C_V for ideal gases

For one mole of an ideal gas, Δ(PV) = R ΔT, so

ΔH = ΔU + R ΔT

Dividing by ΔT gives

C_P = C_V + R

Hence

For ideal monatomic gas (classical equipartition):

C_V = (3/2) R, C_P = (5/2) R

Ratio γ = C_P / C_V depends on atomicity:

• Monoatomic: γ ≈ 1.66
• Diatomic: γ ≈ 1.40
• Triatomic: γ ≈ 1.33

Solved Examples

Example 1. Calculate the energy needed to raise the temperature of 10 g of iron from 25 degree Celsius to 500 degree Celsius if specific heat capacity of iron is 0.45 J g -1˚C -1.

Solution: Here change in temp.=500 - 25=475 ˚C

Total heat = mass × Specific Heat Capacity × change in temp

= m × C_s × ∆T

=10 × 0.45 × 475

=2137.5 Joules

Example 2. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.

Solution.

From the expression of heat (q),

q = m. c. ΔT

Where, c = molar heat capacity, m = mass of substance, ΔT = change in temperature

q = 1066.7 J

q = 1.07 kJ

Degrees of Freedom (DOF)

The number of independent coordinates required to describe the motion of a particle or system is called the degree of freedom.

• Total DOF for a molecule with N atoms = 3N.
• Translational DOF (f_t) = 3 for every molecule.
• Rotational DOF (f_r): 0 for monoatomic species, 2 for linear diatomic or linear polyatomic, 3 for nonlinear polyatomic molecules.
• Vibrational DOF (f_v) = 3N - f_t - f_r.

Examples and common cases:

• Monoatomic: Total DOF = 3 (only translational).
• Diatomic molecule: Total DOF = 6 = 3 translational + 2 rotational + 1 vibrational (vibrational mode becomes significant at higher temperatures).

Break up for diatomic (illustrative):

(i) 3 translational degrees of freedom.

(ii) 2 rotational degrees of freedom.

(iii) 1 vibrational degree of freedom.

Example. Find the total degree of freedom and break up as translational, rotational, or vibrational DOFs in following cases.

(i) O = C = O

(ii)

(iii) He

(iv) NH₃

Solution.

(i) CO₂: Total dof = 3 × 3 = 9. Translational = 3. Rotational = 2 (linear). Vibrational = 4.

(ii) SO₂: Total dof = 9. Translational = 3. Rotational = 3 (bent molecule). Vibrational = 3.

(iii) He: Total dof = 3 (monoatomic). Translational = 3.

(iv) NH₃: Total dof = 12. Translational = 3. Rotational = 3 (nonlinear). Vibrational = 6.

Law of Equipartition of Energy

According to the law of equipartition of energy, each quadratic degree of freedom contributes an average energy of 1/2 kT per molecule, where k is the Boltzmann constant (k = R/N_A).

• Each translational or rotational degree of freedom contributes ½ kT per molecule.
• Each vibrational degree of freedom contributes kT per molecule (because vibrational motion contains both kinetic and potential energy parts).

For n moles, multiply by N_A to obtain energy in terms of R and T.

On ignoring vibrational degrees of freedom for certain temperature ranges:

Thermochemistry

Thermochemistry deals with heat changes associated with chemical reactions and physical changes.

Definitions and Units

Specific heat (s): Amount of energy required to raise the temperature of 1 g of a substance by 1 °C. Unit: J g^-1 K^-1.

Heat capacity (m s): Heat required to raise temperature of a given amount of substance by 1 °C. Unit: J K^-1.

Total heat given to increase the temperature by ΔT:

q = m s ΔT

Molar heat capacity (C_m): Heat required to raise temperature of 1 mole by 1 K. Two important variants:

• Molar heat capacity at constant pressure, C_p,m.
• Molar heat capacity at constant volume, C_v,m.

Relation between C_p and C_v

and

C_p / C_v = γ (Poisson's ratio)

Rules for Thermochemical Equations

• Always state the physical state of reactants and products (s, l, g, aq).
• If A → B with ΔH = x kJ, then reversing the reaction gives B → A with ΔH = -x kJ.
• When two thermochemical equations are added algebraically, their enthalpy changes add algebraically.
• If an equation is multiplied by a factor, ΔH is multiplied by the same factor.

Properties of Matter: Intensive and Extensive

Intensive properties do not depend on the amount of substance (e.g., density, refractive index, specific heat).

Extensive properties depend on amount of substance (e.g., volume, enthalpy ΔH, internal energy U, number of moles).

Examples: ΔH, ΔS, ΔG, V, U, resistance, number of moles are extensive.

Note: The ratio of two extensive properties is intensive.

Further Discussion on Enthalpy

For an ideal gas:

H = U + PV = U + nRT

Hence enthalpy of an ideal gas depends solely on temperature (for ideal gas behaviour).

For an ideal gas reaction, a useful form is:

ΔH = ΔU + Δn C_p T (derived earlier; use Δn_g for gaseous changes)

Kirchhoff's Equation and Temperature Dependence of Enthalpy

Kirchhoff's equation relates the change in reaction enthalpy with temperature through heat capacities:

ΔH₂ - ΔH₁ = ΔC_p (T₂ - T₁)

where ΔC_p = Σ C_p(products) - Σ C_p(reactants).

If ΔC_p depends on temperature (e.g., ΔC_p = a T² + b T + c), integrate ΔC_p with respect to temperature to obtain the enthalpy change between T₁ and T₂.

Different Types of Enthalpies

(i) Heat (Enthalpy) of Formation (ΔH_f)

Heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable reference states under standard conditions.

Example:

C(s, graphite) + O₂(g) ⇒ CO₂(g), ΔH = ΔH_f(CO₂)

Heat of formation of the most stable form of an element is taken as zero.

Standard Enthalpy (ΔH°)

Standard conditions: T = 298 K (25 °C), P = 1 atm, concentration = 1 M.

Standard enthalpy change of reaction can be computed from standard enthalpies of formation:

ΔH° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants)

Example 1. From the following data:

C_{(s, graphite)} + O_2(g) ⇒ CO_2(g) ΔH° = -393.5 kJ mol^-1

H₂(g) + 1/2 O₂(g) ⇒ H₂O(l) ΔH° = -286 kJ mol^-1

2 C₂H₆(g) + 7 O₂(g) ⇒ 4 CO₂(g) + 6 H₂O(l) ΔH° = -3120 kJ mol^-1

Calculate the standard enthalpy of formation of C₂H₆(g).

Solution.

Use ΔH°(reaction) = Σ ΔH_f°(products) - Σ ΔH_f°(reactants).

From given data ΔH_f°(CO₂) = -393.5 kJ mol^-1, ΔH_f°(H₂O) = -286 kJ mol^-1.

For the combustion reaction:

-3120 = 4 (-393.5) + 6 (-286) - 2 ΔH_f°(C₂H₆)

-3120 = -1574 - 1716 - 2 ΔH_f°(C₂H₆)

-3120 + 3290 = -2 ΔH_f°(C₂H₆)

170 = -2 ΔH_f°(C₂H₆)

ΔH_f°(C₂H₆) = -85 kJ mol^-1

(ii) Heat (Enthalpy) of Combustion

Heat of combustion is the enthalpy change when one mole of a substance burns completely in oxygen; it is usually exothermic (negative ΔH).

For example, C(s) + O₂ ⇒ CO₂ (g), the ΔH° equals ΔH_f°(CO₂) since ΔH_f° for C(graphite) and O₂(g) is zero.

Note: Most combustion reactions are exothermic, but some reactions between elements can be endothermic depending on bond making/breaking.

Example 2. The enthalpy change for the reaction

C₃H₈(g) + H₂(g) ⇒ C₂H₆(g) + CH₄(g) at 25ºC is -55.7 kJ mol^-1. Calculate the enthalpy of combustion of C₂H₆(g). The enthalpy of combustion of H₂, and CH₄ are -285.8 and -890.0 kJ mol^-1 respectively. Enthalpy of combustion of propane is -2220 kJ mol^-1.

Solution.

Use ΔH_r° = ΔH_c°(reactants) - ΔH_c°(products).

-55.7 = (-2220 - 285.8) - {ΔH_c°(C₂H₆) + (-890)}

Solve for ΔH_c°(C₂H₆):

ΔH_c°(C₂H₆) = -1560.1 kJ mol^-1

Problems Based on Both Heat of Combustion and Heat of Formation

Example 3. At 300 K, the standard enthalpies of formation of C₆H₅COOH(s), CO₂(g) and H₂O(l) are -408, -393 and -286 kJ mol^-1 respectively. Calculate the enthalpy of combustion of benzoic acid at

(i) constant pressure

(ii) constant volume

Solution.

Formation reaction: 7 C(s) + 3 H₂(g) + O₂(g) ⇒ C₆H₅COOH(s), ΔH° = -408 kJ ⇒ ΔH_f°(C₆H₅COOH) = -408 kJ.

Combustion: C₆H₅COOH(s) + 15/2 O₂(g) ⇒ 7 CO₂(g) + 3 H₂O(l)

ΔH_C°(C₆H₅COOH) = 7 ΔH_f°(CO₂) + 3 ΔH_f°(H₂O) - ΔH_f°(C₆H₅COOH)

= 7(-393) + 3(-286) - (-408)

ΔH_C° = -3201 kJ mol^-1 (constant pressure)

For constant volume, relate ΔH and ΔU by ΔH = ΔU + Δn RT.

Here Δn = (7 - 7.5) = -0.5 (gaseous moles change). Using R = 8.314 × 10^-3 kJ mol^-1 K^-1 and T = 300 K:

ΔU = ΔH - Δn R T = -3201 - (-0.5)(8.314 × 10^-3)(300)

ΔU ≈ -3199.75 kJ mol^-1 (constant volume)

Bond Energy

Bond energy (bond enthalpy) is defined for gaseous molecules as the enthalpy change for breaking one mole of a particular bond into isolated gaseous atoms.

Example: H₂(g) ⇒ 2 H(g), ΔH° = Σ H-H bond energy

In general, enthalpy of reaction in terms of bond energies is:

ΔH_reaction = Σ (bond energies of bonds broken - bond energies of bonds formed)

Example 4. Using bond enthalpy data, calculate ΔH for:

C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Solution.

ΔH_r° = Σ(bond energies of reactants) - Σ(bond energies of products)

Applying bond counts and given data, one obtains ΔH_r° ≈ -120.08 kJ mol^-1.

(iii) Heat of Hydrogenation

Heat of hydrogenation is the enthalpy change when one mole of H₂ is added to an unsaturated compound (usually exothermic).

Example: CH₂=CH₂ + H₂ ⇒ CH₃-CH₃, ΔH is heat of hydrogenation per mole of double bond.

Heat of hydrogenation depends on the stability of the unsaturated compound; more stable alkenes have smaller (less negative) heats of hydrogenation.

Trans isomer is generally more stable than cis, so heat of hydrogenation(trans) < heat of hydrogenation(cis).

Example 6. Find ΔH_f of HCl(g) if bond energies of H₂, Cl₂ and HCl are 104, 58, 103 kcal/mol respectively.

Solution.

H - Cl ⇒ H(g) + Cl(g)

Σ H-Cl = ΔH_f°H(g) + ΔH_f°Cl(g) - ΔH_f°HCl(g)

Calculation gives ΔH_f°(HCl(g)) = -22 kcal mol^-1.

(iv) Heat of Atomisation

Heat of atomisation is the enthalpy change when one mole of substance is converted into gaseous atoms. It is always positive (endothermic).

(v) Heat of Sublimation

Heat of sublimation is the enthalpy change when one mole of a solid converts directly to gaseous atoms or molecules. It is positive (endothermic).

Example 7. (Bond enthalpy based calculation for acetone formation)

Example 8. The enthalpy of combustion of acetylene is 312 kcal. If enthalpy of formation of CO₂ and H₂O are -94.38 and -68.38 kcal respectively, calculate C≡C bond enthalpy. Given enthalpy of atomisation of C = 150 kcal and H-H and C-H bond enthalpies are 103 kcal and 93.64 kcal respectively.

Solution.

From combustion data and bond atomisation considerations, one obtains C≡C bond enthalpy ≈ 160.86 kcal mol^-1.

Resonance Energy

Resonance energy is the energy difference between the resonance hybrid and the most stable contributing canonical (Lewis) structure. It's typically considered as the extra stability (negative ΔH) due to resonance.

Example 9. Given enthalpies of formation from gaseous atoms for ethane, ethylene and benzene are -2839.2, -2275.2 and -5506 kJ mol^-1 respectively. Bond enthalpy C-H = 410.87 kJ mol^-1. Calculate resonance energy of benzene.

Solution.

Using the bond-energy based enthalpies and comparing expected enthalpy for benzene from localized bonds with the actual enthalpy gives resonance energy ≈ -23.68 kJ mol^-1.

Bomb Calorimeter

Bomb calorimeter is used to measure the heat of combustion at constant volume. The total heat evolved increases the temperature of the calorimeter and its contents. If C is the heat capacity of the calorimeter assembly and ΔT is the observed temperature rise, then the heat released is C ΔT. Enthalpy of combustion per mole is obtained by dividing by the number of moles burned and correcting from ΔU to ΔH if required.

Example 10. When 1.0 g of fructose C₆H₁₂O₆(s) is burnt in oxygen in a bomb calorimeter, the temperature of the calorimeter water increases by 1.56 °C. If the heat capacity of the calorimeter and its contents is 10.0 kJ °C^-1, calculate the enthalpy of combustion of fructose at 298 K.

Solution.

Total heat evolved = C ΔT = 10.0 × 1.56 = 15.6 kJ for 1.0 g.

Convert to per mole of fructose to obtain ΔU (constant volume). If required, apply correction ΔH = ΔU + Δn RT to convert to enthalpy of combustion at constant pressure.

Other Types of Enthalpies

(vi) Heat of Solution

Heat of solution is the enthalpy change when 1 mole of a solute dissolves in a large excess of solvent. It may be exothermic or endothermic depending on solute-solvent interactions and hydration.

Examples: Dissolution of anhydrous CuSO₄ is exothermic (hydrate formation), whereas dissolution of NaCl is slightly endothermic.

(vii) Heat of Dilution

Heat of dilution is the enthalpy change when the concentration of a solution changes (e.g., dilution of a solution from one concentration to another). It is the difference between integral heats of solution for two solvent amounts.

(viii) Heat of Hydration

Heat of hydration is the enthalpy change when one mole of an anhydrous salt forms its hydrated form. This process is often exothermic.

Example: CuSO₄(s) + 5 H₂O ⇒ CuSO₄·5H₂O(s)

Example 11. Heat of solution of CuSO₄(s) and CuSO₄·5H₂O are 15.9 and 19.3 kJ mol^-1 respectively. Find the heat of hydration of CuSO₄(s).

Solution.

Reversing and combining appropriate solution/hydration equations yields heat of hydration = -35.2 kJ mol^-1 (exothermic).

(ix) Heat of Neutralisation

Heat of neutralisation is the enthalpy change when one equivalent of an acid is neutralised by one equivalent of a base in dilute aqueous solution. For strong acid + strong base the value is approximately constant (≈ -57 kJ per mole of H⁺ neutralised to form H₂O).

Example:

HCl(aq) + NaOH(aq) ⇒ NaCl(aq) + H₂O(l), ΔH ≈ -57 kJ mol^-1 (per mole of H⁺ neutralised).

Example 12. 100 mL of 0.5 M H₂SO₄ is neutralised by 200 mL of 0.2 M NH₄OH in a constant pressure calorimeter and temperature rises by 1.4 °C. Heat capacity of the calorimeter is 1.5 kJ °C^-1. Given: HCl + NaOH ⇒ NaCl + H₂O 57 kJ; CH₃COOH + NH₄OH ⇒ CH₃COONH₄ + H₂O 48.1 kJ. Which statements are correct?

(A) Enthalpy of neutralization of HCl vs NH₄OH is -52.5 kJ/mol

(B) Enthalpy of dissociation (ionisation) of NH₄OH is 4.5 kJ/mol

(C) Enthalpy of dissociation of CH₃COOH is 4.6 kJ/mol

(D) ΔH for 2 H₂O(l) ⇒ 2 H⁺(aq) + 2 OH^-(aq) is 114 kJ

Solution.

Total heat evolved = C ΔT = 1.5 × 1.4 = 2.1 kJ.

Milliequivalents: 100 mL × 0.5 M = 50 mmol equiv; 200 mL × 0.2 M = 40 mmol equiv. NH₄OH is limiting → 0.04 mol equiv produces 2.1 kJ.

Heat per mole for neutralisation = -52.5 kJ mol^-1 → (A) true.

Using -57 + x = -52.5 gives x = 4.5 kJ (enthalpy of dissociation of NH₄OH) → (B) true.

From relations x + y = 8.9 and x = 4.5, y ≈ 4.4 kJ → (C) approximately true (4.4 ≈ 4.6 given rounding); question statement uses 4.6 so depending on rounding (C) close.

ΔH for 2 H₂O ⇒ 2 H⁺ + 2 OH^- is +114 kJ (reverse of neutralisation) → (D) true.

Hence A, B and D are correct in the worked solution.

Born-Haber Cycle (brief)

• Ionisation energy: Energy required to remove an electron from a gaseous atom.
• Electron affinity: Energy released when an extra electron is added to a gaseous atom.
• Lattice energy: Energy released when gaseous ions combine to form one mole of an ionic solid.

Worked Numerical Example (Standard Enthalpy Calculation)

Example 13. Calculate the standard enthalpy change for the reaction CO₂(g) + H₂(g) → CO(g) + H₂O (g) given that ΔH_f⁰ for CO₂(g), CO(g) and H₂O(g) as -393.5, -110.5 and -241.8 KJ/mol respectively.

(A) -31.2 KJ

(B) -21.2 KJ

(C) -11.2 KJ

(D) 41.2 KJ

Ans. (D)

Solution.

ΔH° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants)

ΔH° = [-241.8 - 110.5] - [-393.5 + 0]

= -352.3 + 393.5 = 41.2 kJ

Final Notes and Important Points to Remember

• Enthalpy is a state function - ΔH depends only on initial and final states, not the path.
• At constant pressure q_p = ΔH; at constant volume q_V = ΔU.
• Mayer's relation for ideal gases: C_P - C_V = R.
• Use ΔH° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants) to compute standard enthalpies of reaction.
• Hess's law: Enthalpy changes add for stepwise reactions - useful to compute enthalpies from known reactions.
• Kirchhoff's equation provides temperature dependence of reaction enthalpies when heat capacities are known.