Module 5.2: Enthalpy

5.2.1 Internal Energy and the First Law

I First Law of Thermodynamics: the internal energy of an isolated system is constant.

1. in a closed system, energy is often transferred through both heat and work. Therefore, $\Delta \textrm{U} = \textrm{q} + \textrm{w}$ (Eq. 62)

2. the change of internal energy in an isothermal expansion is zero (Ref. Eq. 61 Derivation)

   1. the internal energy of an ideal gas is independent of volume (no IMFs)

II State function: A function that solely depends on the current state of the system

1. internal energy, temperature, volume, pressure, and density are all state functions

2. its path, or how it got to that specific state does not matter; it is independent of its history of how it got there

3. the change of state functions ($\Delta \textrm{X}$) solely depends on their initial and final states

   1. Temperature is a state function; it does not matter if the change in temperature occurs in the path of $0 \degree \textrm{C} \rightarrow 100 \degree \textrm{C}$, or $0 \degree \textrm{C} \rightarrow 140 \degree \textrm{C} \rightarrow 100 \degree \textrm{C}$, the resulting $\Delta \textrm{T}$ will always be $100 \degree \textrm{C}$.

4. work and heat are not state functions and are path-dependent

5. cyclic process: a series of changes that brings the system back to its initial state.

   1. for state functions, a cyclic process results in the function remaining unchanged.

III in isochoric conditions, expansion work cannot be done as there is no change in volume; assuming there is no non-expansion work,

$\Delta \textrm{U} = \textrm{q}_{v}$ (Eq.63)

1. bomb calorimeters are constant-volume calorimeters that prevent energy loss as heat

   1. because the temperature of the surroundings constantly match the temperature of the system, no energy is transferred as heat to the surroundings.

2. With the relationship defined in Eq. 63, constant-volume heat capacity is defined as:

   $\textrm{C}_{\textrm{V}} = \frac{\textrm{U}}{\textrm{T}}$ (Eq. 64)

   1. in other words, CV expresses how the internal energy of an isochoric system varies with temperature (the heat capacity is the slope of T-U graph)

   2. The change in internal energy of a system can therefore be calculated by: $\Delta \textrm{U} = \textrm{C}_{V} \Delta \textrm{T}$ (Eq. 65)

5.2.2 A Molecular Interlude

IV Although on the macroscopic level heat and work are both ways of transferring energy, they differ on the microscopic level. If energy is transferred from the system to the surroundings as

1. work, the system stimulates molecules of the surroundings to move in a definite direction

2. heat, the system stimulates molecules of the surroundings to move in random directions

V Statistical thermodynamics: a study of thermodynamics that connects macroscopic observations with microscopic parameters

1. often used to explain macroscopic thermodynamic behavior

2. In Section 2.3.2, we have learned that energy levels that an electron can possess is quantized.

   1. the particle-in-a-box illustrates a type of molecular motion, known as electronic motion.

3. the other three types of molecular motion, which are also quantized are:

   1. translation: the motion through space of a molecule in a large space

   2. rotation: the motion around an axis through the molecule

   3. vibration: periodic distortion of the molecule (bending bonds, stretching, etc.)



4. the separation between energy levels are determined by the types of molecular motion, and molecular properties (such as mass)

5. Boltzmann distribution: a probability distribution that determines the probability of a molecule to exist in certain states, the distribution changes with different temperature $\textrm{N}+{i} \propto \textrm{e}^{\frac{-\epsilon_{i}}{\textrm{k} \textrm{T}}}$ (Eq. 66)

   1. you do not need to know this equation or interpret it

   2. at lower temperatures, molecules tend to populate states of lower energy.

   3. As temperature increases, molecules start to occupy higher energy levels.

VI equipartition theorem: the average value of each quadratic contribution to the energy is $\frac{1}{2} \textrm{k} \textrm{T}$.

1. “a quadratic contribution” to the energy is one that depends on the square of velocity or a displacement (such as kinetic energy)

2. the equipartition theorem can be used to estimate the internal energy of an ideal gas if:

   1. the temperature of the system is high enough that many energy levels (states) are occupied (for translational and rotational, room temperature is high enough)

   2. the motion of molecules is nearly classical (quantum mechanics ignored)

3. the internal energy of molecules is shared equally over all the available modes (types of motion)

4. as there are three types of translational motion for each direction (x, y, z), the equipartition theorem states that the overall contribution of translational motion to molar internal energy is:

   $U_{m} = \frac{3}{2} \textrm{N}_{\textrm{A}} \textrm{k} \textrm{t} = \frac{3}{2} \textrm{R} \textrm{T}$ (Eq. 67)

   1. The gas constant R is defined as $\textrm{N}_{A}$ times the Boltzmann constant ($\textrm{k}$).

   2. translational motion is a type of kinetic energy.

   3. the classical expression (in physics) of translational motion is: $\frac{1}{2} \textrm{m} \textrm{v}^{2}$

   4. the internal energy of an ideal gas is solely dependent on translational motion.

5. A linear molecule has two rotational modes of motion, and a nonlinear molecule has three rotational modes of motion. The equipartition theorem states that:

   $\textrm{U}_{m} (\textrm{rotation, linear}) = \textrm{N}_{\textrm{A}} \textrm{k} \textrm{T} = \textrm{R} \textrm{T}$ (Eq. 68)

   $\textrm{U}_{m} (\textrm{rotation, nonlinear}) = \frac{3}{2} \textrm{N}_{\textrm{A}} \textrm{k} \textrm{t} = \frac{3}{2} \textrm{R} \textrm{T}$ (Eq. 69)

   1. rotational motion is also a type of kinetic energy (rotational kinetic energy).

   2. the classical expression (in physics) of rotational motion is: $\frac{1}{2} \textrm{I} \omega$

6. the separation of energy levels is much larger for vibration modes compared to translation or rotation modes.

   1. At room temperature, it can only occupy the low energy levels compared to other modes due to larger separation, and does not meet the requirements of using the equipartition theorem as an approximation.

5.2.3 Enthalpy

VII Chemical reactions often involve energy transfers through both work and heat

1. most reactions occur in the atmosphere that is in constant-pressure

   1. when gaseous reactants or products are involved, this leads to work done on the atmosphere or done by the atmosphere.

   2. energy transfer also occurs in the form of heat depending on whether the reaction is endothermic or exothermic

VIII enthalpy (H): a state function that considers the expansion work while heat is transferred

1. the definition of enthalpy is: $\textrm{H} = \textrm{U} + \textrm{P} \textrm{' '} {V}$

   1. as enthalpy is defined by state functions (U, P, V) it is also a state function.

2. As enthalpy is an extensive property, it can also be expressed in terms of molar enthalpy

   1. for a perfect gas, the molar enthalpy can be defined as: $\textrm{H}_{m} = \textrm{U}_{m} + \textrm{R} \textrm{T} (\textrm{R} \textrm{T} = \textrm{P} \textrm{V}_{m})$ (Eq. 71)

   2. for solids and liquids, its molar volume is typically 1000 times smaller than the molar volume of gases. Therefore, $\textrm{H}_{m} = \textrm{U}_ {m}$ (Eq. 72)

3. in practice, the change in enthalpy ($\Delta \textrm{H}$) is measured; at constant pressure, $\Delta \textrm{H} = \Delta \textrm{U} + \textrm{P}\Delta \textrm{V}$ (Eq. 73)

   1. at constant pressure with no non-expansion work, the PV term and work cancels out, only leaving: $\Delta \textrm{H} = \textrm{q}_{p}$ (Eq. 74)

4. the signs of enthalpy determine the direction of energy transfer in a reaction:

   1. $\Delta \textrm{H} \gt 0$ is equivalent to an endothermic process.

   2. $\Delta \textrm{H} \lt 0$ is equivalent to an exothermic process.

5.2.4 Heat Capacities in Detail

IX constant pressure heat capacity ($\textrm{C}_{p}$): heat is equivalent to the enthalpy at constant pressure.

$\textrm{C}\_{p} = \frac{\Delta \textrm{H}}{\Delta \textrm{T}}$ (Eq. 75)

X constant volume heat capacity ($\textrm{C}_{v}$): heat is equivalent to the internal energy at constant volume.

$\textrm{C}\_{v} = \frac{\Delta \textrm{U}}{\Delta \textrm{T}}$

1. the relation between the two heat capacities is revealed by substituting $\Delta \textrm{H}$ for its definition:

   $\textrm{C}_{p} = \frac{\Delta \textrm{H}}{\Delta \textrm{T}} = \frac{\Delta \textrm{U} + \textrm{n} \textrm{R} \Delta \textrm{T}}{\Delta \textrm{T}} = \textrm{C}_{v} + \textrm{n} \textrm{R} (\textrm{C}_{v} = \frac{\Delta \textrm{U}}{\Delta \textrm{T}})$ (Eq. 76)

   1. in other words, $\textrm{C}_{P,m} = \textrm{C}_{V,m} + \textrm{R}$

2. heat capacities are dependent on molecular properties such as the equipartition theorem

XI standard state ($\Theta$): the pure substance at exactly one bar (= 105 Pa)

1. the standard state defined for 1 atm (=101.325 kPa) is the old standard but the difference is negligible

2. temperature is not defined by the standard state; it must be defined separately.

   1. many thermodynamic constants are defined at 298.15 K
3. a standard state does not have to be stable.

5.2.5 Physical change

XII phase transition: A transition from one state to another

1. heat (enthalpy) is released during phase transitions due to formation and reduction of intermolecular forces

2. if intermolecular forces increase from a phase transition, the overall reaction would be exothermic (heat is released)

3. if intermolecular forces decrease from a phase transition, the overall reaction would be endothermic (heat is absorbed)

XIII standard enthalpy of vaporization ($\Delta \textrm{H}_{vap}^{\Theta}$): the energy that must be supplied as heat in order for one mole of substance to vaporize (liquid $\rightarrow$ vapor)

1. Stronger intermolecular forces results in a larger positive value of $\Delta \textrm{H}_{vap}^{\Theta}$

2. $\Delta \textrm{H}_{vap}^{\Theta}$ is the difference in molar enthalpy between the two phases $\Delta \textrm{H}_{vap} = \textrm{H}_{m}(\textrm{vapor}) − \textrm{H}_{m}(\textrm{liquid})$ (Eq. 77)

XIV standard enthalpy of fusion ($\Delta \textrm{H}_{fus}^{\Theta}$): the energy that must be supplied as heat in order for one mole of substance to melt (solid $\rightarrow$ liquid)

1. $\Delta \textrm{H}_{fus}^{\Theta}$ is often smaller than $\Delta \textrm{H}_{vap}^{\Theta}$ as molecules do not have to be completely separated from each other.

2. similar to vaporization, $\Delta \textrm{H}_{fus}^{\Theta}$ is the difference in molar enthalpy between the two phases $\Delta \textrm{H}_{fus} = \textrm{H}_{m}(\textrm{liquid}) − \textrm{H}_{m}(\textrm{solid})$ (Eq. 78)

XV as enthalpy is a state function, the reverse transition is always equal to the negative enthalpy of the forward transition:

$\Delta \textrm{H}_{forward} = −\Delta \textrm{H}_{reverse}$ (Eq. 79)

1. condensation is the reverse of vaporization, and freezing is the reverse of fusion.

XVI standard enthalpy of sublimation ($\Delta \textrm{H}_{sub}^{\Theta}$): the energy that must be supplied as heat in order for one mole of substance to sublime (solid $\rightarrow$ vapor)

1. as enthalpy is a state function, the sum of $\Delta \textrm{H}_{fus}^{\Theta}$ and $\Delta \textrm{H}_{vap}^{\Theta}$ equals $\Delta \textrm{H}_{sub}^{\Theta}$.

   1. other enthalpy changes for different processes (other than phase changes) can also be calculated indirectly; only the initial and final states must be equivalent. $\Delta \textrm{H}_{sub}^{\Theta} = \Delta \textrm{H}_{fus}^{\Theta} + \Delta \textrm{H}_{vap}^{\Theta}$ (Eq. 80)

XVII standard enthalpy of ionization ($\Delta \textrm{H}_{ion}^{\Theta}$): the change in standard enthalpy per mole of atoms for a loss of an electron for each atom

1. this is different by definition to ionization energy

   1. ionization energy is the change in internal energy at absolute zero

   2. according to the Boltzmann distribution, all of the energies are in ground state at absolute zero.

   3. For the standard enthalpy of ionization, this is not the case; $\textrm{T} \gt 0 \textrm{K}$.

   4. this difference can often be ignored, as most of the population is in the ground state at normal temperatures $\textrm{X}(\textrm{g}) \rightarrow \textrm{X}^{+} (\textrm{g}) + \textrm{e}^{-} (\textrm{g}), \Delta \textrm{H}_{ion}^{\Theta}$ (Eq. 81)

XVIII standard enthalpy of electron gain ($\Delta \textrm{H}_{eg}^{\Theta}$): the change in standard enthalpy per mole of atoms for a gain of an electron for each atom

1. unlike phase transitions, electron gain and ionization can be endothermic or exothermic, and sign representation is necessary

2. electron gain is the reverse process of ionization

5.2.6 Chemical Change

XIX Hess’s Law: The standard enthalpy of a certain reaction is equivalent to the sum of the standard enthalpies of each step of its component reactions

1. in other words, as long as the initial reactant and final product are the same, any path can work

2. many thermochemistry questions utilize Hess’s Law and cycles like the one shown in Figure 68

XX bond enthalpy: the energy released as heat from the dissociation of one mole of bonds

1. breaking bonds requires energy to overcome attractive forces
2. bond enthalpies are correlated to bond strengths
3. bond enthalpies are measured from gas phase molecules

4. as bond enthalpies also depend on the surroundings of the two atoms that are bonded together, the average value, or the mean bond enthalpy is often used. (Also only for gaseous state)

XXI standard enthalpy of formation ($\textrm{H}_{f}^{\Theta}$): the change in standard enthalpy (per mole) from its elements in their most stable forms (known as reference state) to its final product

$\textrm{H}_{2} (\textrm{g}) + \frac{1}{2} \textrm{O}_{2} (\textrm{g}) \rightarrow \textrm{H}_{2} \textrm{O} (\textrm{l})$ $\textrm{H}_{f}^{\Theta} = − 286 \frac{\textrm{kJ}}{\textrm{mol}}$

1. Most of these reactions are purely hypothetical and do not occur

2. for pure elements in their most stable form (like $\textrm{H}_{2}$ or $\textrm{O}_{2}$), $\textrm{H}_{f}^{\Theta} = 0$.

3. allows breakdown of elements into their reference states and then building new molecules from the reference states (Ref. Figure 69)

   1. the breakdown of reactants can be represented as: $\Delta \textrm{H}^{\Theta} = − \sum \textrm{n} \Delta \textrm{H}_{f}^{\Theta}$

   2. the formation of new products can be represented as: $\Delta \textrm{H}^{\Theta} = \sum \textrm{n} \Delta \textrm{H}_{f}^{\Theta}$

   3. overall, the enthalpy of reaction can be calculated by: $\Delta \textrm{H}_{r}^{\Theta} = \sum_{\textrm{products}} \textrm{H}_{f}^{\Theta} - \sum_{\textrm{reactants}} \textrm{H}_{f}^{\Theta}$ (Eq. 82)

XXIII standard lattice enthalpy ($\Delta \textrm{H}_{L}^{\Theta}$): the change in molar enthalpy between the solid and a gas of separated ions

1. lattice energy and lattice enthalpy are slightly different but their difference can be ignored

2. the lattice enthalpies give an estimate of how accurate the ionic model is for a particular compound

   1. if the measured and calculated enthalpies are similar, the ionic model is accurate.

   2. if they are not similar, the ionic model must be improved or discarded for that particular substance.

3. The lattice enthalpy can be calculated through the Born-Haber cycle (Ref. Figure 69, Figure 19)

   1. Born-Haber cycle: a closed path of steps in which the initial and final states are the same; the overall enthalpy change is zero.

   2. The lattice enthalpy can be calculated as all of the other thermodynamic constants are known.

   3. the Born-Haber cycle for the lattice enthalpy of potassium chloride is shown in Figure 68

XXIV standard enthalpy of combustion ($\Delta \textrm{H}_{c}^{\Theta}$): the change in enthalpy resulting from a mole of substance that undergoes a combustion reaction (under standard conditions)

1. unlike $\Delta \textrm{H}_{f}^{\Theta}$, most of these combustion reactions can occur in real life

2. organic compounds tend to have similar combustion products (e.g. $\textrm{C}\textrm{O}_{2}, \textrm{H}_{2}\textrm{O}, \textrm{N}_{2}$ or nitric oxides if the compound includes N)

XXV the relationship between $\Delta \textrm{H}$ and $\Delta \textrm{U}$ can be summarized by:

1. in the solid and liquid phase, $\Delta \textrm{H} \approx \Delta \textrm{U}$

2. in the gas phase, where ∆ngas is the change of gas molecules resulting from the reaction, $\Delta \textrm{H} = \Delta \textrm{U} + \Delta \textrm{n}_{\textrm{gas}} \textrm{R} \textrm{T}$ (Eq. 83)

XXIII Kirchhoff’s Law: the change in $\Delta \textrm{H}^{\Theta}$ resulting from the change in temperature can be predicted using the difference between the constant-pressure heat capacities of the products and reactants ($\Delta \textrm{C}_{P} = \sum \textrm{n} \textrm{C}_{\textrm{P},\textrm{m}}(\textrm{products}) − \sum \textrm{n} \textrm{C}_{\textrm{P},\textrm{m}}(\textrm{reactants})$)

$\Delta \textrm{H}^{\Theta} (\textrm{T}_{2}) = \Delta \textrm{H}^{\Theta} (\textrm{T}_{1}) + (\textrm{T}_{2} - \textrm{T}_{1}) \times \Delta \textrm{C}_{P}$ (Eq. 84)

1. Kirchhoff’s law states that enthalpies of both reactants and products increase with temperature

   1. this increase in temperature can sometimes be negligible; if the difference in temperature is big, you should use Kirchhoff’s law.