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# Thermodynamics

Chapter 05

Audio created by Google NotebookLM. *Note:* I have not verified the accuracy of the audio or transcript. Download the transcript here (created by Restream).

**Thermodynamics** is a area of study that explores heat, work, and temperature, and their relation to energy, entropy, and the physical properties of matter and radiation.

**Heat**is an amount of thermal energy in transit between a system and its surroundings due to a temperature difference.**Thermodynamic work**is a process by which a system and its surroundings can exchange energy.**Temperature**is a physical quantity that is a measure of “hotness” and “coldness” and reflects the*average kinetic energy*of atoms in a substance.**Energy**is a physical quantity that measures how systems change over time. Energy can be classified in various ways.**Entropy**is typically stated to be a measure of disorder, randomness, or uncertainty. It is also a measure of how “spread out” or “distributed” energy and matter is.**Matter**is any substance that has mass and volume.**Radiation**is the emission or transmission of energy in the form of waves or particles.

Thermodynamics studies these concepts by examining the interactions between large ensembles of objects. Boundaries between objects are defined to be either the **system** (Table 2), the objects of interest, and the **surroundings**, *everything* that is not the system. The system and surroundings combined is called the **universe**.

## Energy

The **Law of Conservation of Energy** states that, *locally*, energy of an isolated system is conserved, energy cannot be created or destroyed, energy can only be transformed, and the energy of the universe is constant. *Note:* On a cosmological scale, energy is not constant. The **First Law of Thermodynamics** is a version of the *Law of Conservation of Energy* and states that the total energy of an **isolated system** is constant.

There are two principle forms of energy. **Kinetic energy ( E_{k}, T, or K )** is a form of energy that an object possesses due to its motion.

\[T = \dfrac{1}{2}~mv^2\]

Kinetic energy is expressed in SI units as the **joule (J)** and can be approximated as being one half the mass ( *m* ) of an object times its velocity ( *v* ) squared (where velocity is much slower than the **speed of light in a vacuum)**. **Velocity** is a combination of **speed** *and* direction of an object and has SI units of meters per second (m s^{–1}).

##
**Derivation of the joule**

We can derive these units from the kinetic energy equation by visualizing the base SI units for each term. Dimensionless numbers are omitted after the first step for clarity.

\[\begin{align*} T(\mathrm{J}) &= \dfrac{1}{2}~mv^2 \\[1.5ex] &= (\mathrm{kg}) \left ( \dfrac{\mathrm{m}}{\mathrm{s}} \right )^2 \\[1.5ex] &= (\mathrm{kg}) \left ( \dfrac{\mathrm{m}^2}{\mathrm{s}^2}\right ) \\[1.5ex] &= \mathrm{kg~m^2~s^{-2}} \end{align*}\]

Therefore, the joule is

\[\mathrm{J} \equiv \mathrm{kg~m^2~s^{-2}}\]

Energy, work, and heat are typically all expressed in joules.

**Practice**

What is the kinetic energy (in J) of a baseball, with a mass of 145 g, moving at a velocity of 37.6 m s^{–1}?

##
**Solution**

Using the kinetic energy equation

\[\begin{align*} T &= \dfrac{1}{2}mv^2 \\[1.5ex] &= \dfrac{1}{2} \left ( 145~\mathrm{g} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \left ( \dfrac{37.6~\mathrm{m}}{\mathrm{s}} \right )^2 \\[1.5ex] &= 10\bar{2}.4~\mathrm{kg~m^{2}~s^{-2}} \\[1.5ex] &= 10\bar{2}.4~\mathrm{J} \end{align*}\]

**Potential energy ( V )** is a form of energy that an object possesses due to its position or configuration. It can be defined as the capacity for doing work which arises from the object’s position or configuration. For example, a ball at the top of a hill (position 1) has the potential to perform more work than an identical ball at the bottom of a hill (position 2). A system containing a mixture of hydrocarbons and oxygen (configuration 1) can readily burn to produce carbon dioxide and water (configuration 2).

##
**Types of potential energy**

**Chemical potential energy**is related to the structural rearrangement of atoms or molecules. A chemical reaction involving a rearrangement of particles has chemical potential energy that can be transformed into**heat**and**light**, such as the burning of fossil fuels.**Electric potential energy**arises from the interaction of charges. A repulsive force occurs between two positive or two negative charges whereas an attractive force occurs between two opposite charges.**Gravitational potential energy**is the energy an object possesses due to its position in a gravitational field such as what an object experiences by being near the surface of the earth.**Elastic potential energy**is mechanical potential energy stored in the configuration of a material or physical system. A compressed spring (configuration 1) has more elastic potential energy than a relaxed spring (configuration 2).**Magnetic potential energy**is the potential energy of a magnetic moment in an externally produced magnetic field.

Below is a table of various forms of energy.

### System Models

A system can be defined in various ways based their interactions with the surroundings (i.e. the transfer of mass, work, and heat).

## Heat

**Heat ( q )** is thermal energy in transit between a system and surroundings (i.e. it is the transfer of thermal energy).

**Thermal energy**can be defined as the energy that arises from the kinetic energy of particles.

Objects that lose thermal energy give off heat while objects that gain thermal energy absorb heat. The **Second Law of Thermodynamics** states that energy and matter naturally tend to disperse uniformly. One example is the **Clausius theorem** which states heat does not spontaneously pass from a colder object to a hotter body

Below is an example of two objects of different temperatures in contact with each other in an *isolated system*.

Thermal energy would spontaneously transfer from the hot object to the cold object. With time, both objects would reach the same temperature and enter a state of **thermal equilibrium** where the *and* the temperature of both objects are equal *temperature change* of both objects is zero (i.e. there is no net flow of thermal energy between the objects).

When a system experiences an *increase* in thermal energy (*q*_{sys} > 0) due to some process, the process is considered to be **endothermic**. If the system experiences a *decrease* in thermal energy (*q*_{sys} < 0), due to some process, the process is considered to be **exothermic**.

**Practice**

Consider the following processes and determine if the process is endothermic or exothermic.

- Melting of ice
- Freezing of water

##
**Solution**

- The melting of ice is an endothermic process where the ice absorbs thermal energy from its surroundings.
- The freezing of water is an exothermic process where the ice loses thermal energy to its surroundings.

### Entropy

The spontaneous nature of heat transfer from “hot-to-cold” can be illustrated by analyzing the effect of this process on the *entropy of the universe*. **Entropy ( S; J K^{–1})** is related to heat and measured as joule per kelvin and is an extensive property. It is an amount of energy in a system that is unavailable to do work. Entropy is also a measure of the number of ways (number of configurations) a system can be arranged under constraints such as temperature and pressure.

Take two 1.0 L bottles of water and combine them into a single 2.0 L bottle. The entropy of the 2.0 L water is greater than the two individual 1.0 L samples because there are more ways to arrange the water particles in the 2.0 L sample. Entropy is an

extensiveproperty.

All objects have entropy ( *S*_{system} ), all surroundings has entropy ( *S*_{surroundings} ), and the universe has entropy. ( *S*_{universe} ). Therefore,

\[S_{\mathrm{universe}} = S_{\mathrm{surroundings}} + S_{\mathrm{system}}\]

A change in entropy ( Δ*S* ) of the surroundings or system will cause the entropy of the universe to change.

\[\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{surroundings}} + \Delta S_{\mathrm{system}}\]

where, for a reversible process,

\[\Delta S = \dfrac{q_{\mathrm{rev}}}{T}\]

According to the second law of thermodynamics, any process is spontaneous if the process *increases* the entropy of the universe while any process that *decreases* the entropy of the universe is nonspontaneous.

Let us demonstrate entropy and spontaneity. Consider the following setup where two objects of different temperature are placed in an 25 °C environment.

We know from experience that the hot object will not spontaneously become hotter given the cooler temperature of the surroundings. Conversely, we know that the cold object will spontaneously become warmer. Can we show this mathematically by calculating the entropy change of the universe?

Consider a transfer of 25 J of energy from the surroundings to the hot object.

\[\begin{align*} \Delta S_{\mathrm{universe}} &= \Delta S_{\mathrm{surroundings}} + \Delta S_{\mathrm{system~hot}} \\[1.5ex] &= \dfrac{q_{\mathrm{surr}}}{T_{\mathrm{surr}}} + \dfrac{q_{\mathrm{sys}}}{T_{\mathrm{sys}}} \\[1.5ex] &= \dfrac{-25~\mathrm{J}}{298.15~\mathrm{K}} + \dfrac{25~\mathrm{J}}{573.15~\mathrm{K}} \\[1.5ex] &= -0.040~\mathrm{J~K^{-1}} \end{align*}\]

The transfer of heat from the 25 °C surroundings to the 300 °C object would cause a decrease in the entropy of the universe, indicating a nonspontaneous process.

Now consider a transfer of 25 J of energy from the surroundings to the cold object.

\[\begin{align*} \Delta S_{\mathrm{universe}} &= \Delta S_{\mathrm{surroundings}} + \Delta S_{\mathrm{system~cold}} \\[1.5ex] &= \dfrac{q_{\mathrm{surr}}}{T_{\mathrm{surr}}} + \dfrac{q_{\mathrm{sys}}}{T_{\mathrm{sys}}} \\[1.5ex] &= \dfrac{-25~\mathrm{J}}{298.15~\mathrm{K}} + \dfrac{25~\mathrm{J}}{278.15~\mathrm{K}} \\[1.5ex] &= 0.0060~\mathrm{J~K^{-1}} \end{align*}\]

The transfer of heat from the 25 °C surroundings to the 5 °C object would cause an increase in the entropy of the universe, indicating a spontaneous process.

### Heat Capacity

Objects that are heated or cooled *without a change in phase* (e.g. solid, liquid, or gas) experience a change in temperature ( Δ*T* = *T*_{final} - *T*_{initial} ) that is dependent on the object’s **heat capacity ( C ; J K^{–1})**, a physical and extensive property of matter.

\[q = C\Delta T\]

**Molar heat capacity** is given in J mol^{–1} K^{–1}. If the amount of substance is known, one can use the molar heat capacity to determine thermal energy gained or lost based on the change in temperature.

\[q = nC_m\Delta T\]

Similar to heat capacity is **specific heat capacity** which is given in SI units as J kg^{–1} K^{–1}. If the mass of substance is known, one can use the specific heat capacity to determine thermal energy gained or lost based on the change in temperature.

\[q = mc\Delta T\]

Heat capacity changes with the *phase* (c for crystalline/solid, l for liquid, and g for gas) that the object or substance is in to give *C*(c), *C*(l), and *C*(g).

##
**Heat capacity symbols at constant pressure or volume**

Heat capacity quantities can also be denoted with a subscript p for processes occurring under conditions of constant pressure (e.g. *C*_{p}) and a subscript V for processes occurring under conditions of constant volume (e.g. *C*_{V}).

Recall that energy can be transferred in the form of heat and work. When a system is under conditions of constant volume, any energy change is associated only with heat as *pressure-volume* work (or *pV*-work) is not performed.

Under conditions of constant pressure, an energy change is associated with both heat *and* work since the system is allowed to undergo a change in volume. If the volume increases, work is performed by the system (loses energy) whereas if volume decreases, work is done on the system (gains energy).

The first law of thermodynamics states

\[\delta Q = \delta U + p\delta V\]

where δ is just a “really small change”, much smaller than Δ.

**At constant pressure**

For an ideal gas,

\[p\delta V = nR\delta T\]

since *pV=nRT*.

Therefore,

\[\begin{align*} \delta Q &= \delta U + nR\delta T \longrightarrow \\[1.5ex] \dfrac{\delta Q}{\delta T} &= \dfrac{\delta U}{\delta T} + nR \end{align*}\]

Since molar heat capacity, *C*, is δ*Q*/*n*δT, then, at constant pressure

\[C_{\mathrm{p}} = \dfrac{\delta U}{n\delta T} + R\]

**At constant volume**

For an ideal gas,

\[\begin{align*} p\delta V &= nR\delta T \\[1.5ex] R &= \dfrac{p\delta V}{n\delta T} \end{align*}\]

Since no change in volume occurs at constant pressure,

\[\begin{align*} \delta Q &= \delta U + p\delta V \\[1.5ex] \delta Q &= \delta U + nR\delta T\\[1.5ex] \dfrac{\delta Q}{n\delta T} &= \dfrac{\delta U}{n\delta T} + R \\[1.5ex] &= \dfrac{\delta U}{n\delta T} + \dfrac{p\delta V}{n\delta T} \\[1.5ex] &= \dfrac{\delta U}{n\delta T} + \dfrac{p (0)}{n\delta T} \\[1.5ex] &= \dfrac{\delta U}{n\delta T} \end{align*}\]

Since molar heat capacity, *C*, is δ*Q*/*n*δT, then, at constant volume

\[C_{\mathrm{V}} = \dfrac{\delta U}{n\delta T}\]

Comparing both heat capacities (*C*_{p} vs *C*_{V}) gives

\[\begin{align*} C_{\mathrm{p}} &= C_{\mathrm{V}} + R \end{align*}\]

*C*_{P} is always greater than *C*_{P}.

*Note:* For this class, you will always operate under *constant pressure* conditions when using heat capacity.

Notice the small variance in molar heat capacities for pure metals.

One can readily convert between specific heat capacity (*c* in J g^{–1} K^{–1}) and molar heat capacity (*C* in J mol^{–1} K^{–1}) by using the molar mass (*M* in g mol^{–1}) of the substance, *X*.

**Molar Heat Capacity ( C) → Specific Heat Capacity (c) **

\[c(X) = C(X) ~ M(X)^{-1}\]

**Specific Heat Capacity ( c) → Molar Heat Capacity (C) **

\[C(X) = c(X) ~ M(X)\]

**Practice**

An object absorbs 5.0 kJ of heat and experiences a temperature change of 28 °C (28 K). What is the heat capacity (in J K^{–1}) of the object?

##
**Solution**

\[\begin{align*} q &= C\Delta T \longrightarrow \\[1.5ex] C &= \dfrac{q}{\Delta T} \\[1.5ex] &= \dfrac{5.0~\mathrm{kJ} \left ( \dfrac{10^3~\mathrm{J}}{\mathrm{kJ}} \right )} {28~\mathrm{K}} \\[1.5ex] &= 1\bar{7}8~\mathrm{J~K^{-1}} \\[1.5ex] &= 180~\mathrm{J~K^{-1}} \end{align*}\]

**Practice**

Solid aluminum has a molar heat capacity of 24.2 J mol^{–1} K^{–1}. What is the specific heat capacity (in J g^{–1} K^{–1}) of solid aluminum?

##
**Solution**

\[\begin{align*} c_{\mathrm{p}}(\mathrm{Al(s)}) &= C_{\mathrm{p}}(\mathrm{Al(s)}) ~ M(\mathrm{Al})^{-1} \\[1.5ex] &= \dfrac{24.2~\mathrm{J}}{\mathrm{mol~K}} \left ( \dfrac{\mathrm{mol~Al}}{26.98~\mathrm{g}} \right ) \\[1.5ex] &= 0.89\bar{6}9~\mathrm{J~g^{-1}~K^{-1}} \\[1.5ex] &= 0.897~\mathrm{J~g^{-1}~K^{-1}} \\[1.5ex] \end{align*}\]

**Practice**

A 250.0 g solid aluminum object is heated from 25.0 °C to 60.0 °C. How much heat (in kJ) was absorbed? *c*(Al, s) = 0.897 J g^{–1} °C^{–1}

##
**Solution**

The change in temperature was 35.0 °C which is equivalent to a change in temperature of 35.0 K.

\[\begin{align*} q &= mc(\mathrm{Al, (s)}) \Delta T \\[1.5ex] &= 250.0~\mathrm{g} \left ( \dfrac{0.897~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( 60.0~\mathrm{^{\circ}C} - 25.0~\mathrm{^{\circ}C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\[1.5ex] &= 7.8\bar{4}8~\mathrm{kJ} \\[1.5ex] &= 7.85~\mathrm{kJ} \end{align*}\]

### Energy Transfer

Suppose you have an isolated system (Table 2) that consists of a 95.0 °C 35.0 g aluminum sphere that is dropped into a 500. g sample of 5.00 °C water. What will the temperature (in °C) of this system be at thermal equilibrium?

Given that the system is isolated, thermal energy will be conserved between both objects. Recall that, given enough time, both objects will enter into a state of *thermal equilibrium* where both objects will have the same temperature. The aluminum sphere will lose thermal energy (exothermic) and experience a decrease in temperature while the water will gain an equal amount of thermal energy (endothermic) and experience an increase in temperature.

\[\begin{align*} q_{\mathrm{water}} &= -q_{\mathrm{metal}} \\[1.5ex] \left [mc\Delta T \right ]_{\mathrm{water}} &= -\left [mc\Delta T \right ]_{\mathrm{metal}} \\[1.5ex] \left ( 500.~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~K^{-1}} \right ) \left ( T_{\mathrm{final}} - 278~\mathrm{K} \right ) &= -\left ( 35.0~\mathrm{g} \right ) \left ( 0.897~\mathrm{J~g^{-1}~K^{-1}} \right ) \left ( T_{\mathrm{final}} - 368~\mathrm{K} \right ) \\[1.5ex] 2~0\bar{9}2~\mathrm{J~K^{-1}} \left ( T_{\mathrm{final}} - 278~\mathrm{K} \right ) &= -31.\bar{3}9~\mathrm{J~K^{-1}} \left ( T_{\mathrm{final}} - 368~\mathrm{K} \right ) \\[1.5ex] 2~0\bar{9}2~\mathrm{J~K^{-1}} ~ T_{\mathrm{final}} - 58\bar{1}~576~\mathrm{J} &= -31.\bar{3}9~\mathrm{J~K^{-1}}~T_{\mathrm{final}} + 11~\bar{5}51~\mathrm{J} \\[1.5ex] 2~1\bar{2}3~\mathrm{J~K^{-1}} ~ T_{\mathrm{final}} &= 59~\bar{3}~127~\mathrm{J} \\[1.5ex] T_{\mathrm{final}} &= 27\bar{9}.3~\mathrm{K} - 273.15 \\[1.5ex] &= 6~\mathrm{^{\circ}C} \end{align*}\]

### Work

So far we have examined how energy can be transferred as heat ( *q* ); however, energy can also be transferred as **work ( w )**, or as a combination as heat

*and*work.

##
**Types of work**

- Electrical field work - non-mechanical work
- Gravitational work
- Surface tension work
- Magnetic work - non-mechanical work
- Chemical work - occurs when chemical reactions take place, such as in electrochemical cells, where chemical bonds are broken and formed. The energy involved in these reactions (e.g. Gibbs energy) can be considered as work contributing to the system’s internal energy
- Work due to friction
- Rotational work
- Electrical polarization work - non-mechanical work

**Internal energy ( U )** is the energy of a system necessary to bring the system from its standard (internal) state to its present internal state of interest. The

**change in internal energy ( Δ**is a function of

*U*)*q*and

*w*.

\[\Delta U = q + w\]

##
**Internal vs. Total Energy**

Internal energy ( *U* ) does not account for kinetic and potential energy. More formally, changes in internal energy takes into account temperature ( *T* ), pressure ( *p* ), and changes in entropy ( *S* ), volume ( *V* ) and chemical potentials ( *μ* ) where chemical potentials depend on the activities of the involved substances. Note that *d* represents a “very small change”.

\[dU = T~dS - p~dV + \mu~dn\]

Total energy ( *E* ) of a system accounts for the internal, kinetic, and potential energy of a system.

\[\begin{align*} E &= U + T + V \end{align*}\]

The change in internal energy of a system can be determined by measuring the heat absorbed or released by the system as well as the work performed on or by the system. The sign conventions for heat and work are provided in Table 6.

Internal energy is not a property that can be measured absolutely; however, *changes* in internal energy can.

Imagine an ocean planet where the entire surface was covered in water. It may be “impossible” to accurately quantify how much liquid water is on the planet. However, if one were to pour a cup of water into the ocean, the *change in* the amount of water on the planet is easily quantifiable.

**Practice**

An object absorbs 10.0 kJ worth of heat and performs 5.0 kJ worth of work. What is the change in internal energy (in kJ) of the object?

##
**Solution**

\[\begin{align*} \Delta U &= q + w \\[1.5ex] &= 10.0~\mathrm{kJ} + (-5.0~\mathrm{kJ}) \\[1.5ex] &= 5.0~\mathrm{kJ} \end{align*}\]

#### Pressure-Volume Work

Work can be performed on or by any system experiencing a change in pressure and/or volume.

\[w = -\Delta (pV)\]

**Pressure-volume work** (*p*–*V* work or *pV* work) is work that occurs when the volume of a system changes against a constant external pressure (i.e. an **isobaric process)**.

One can visualize this when considering a system composed of a massless and frictionless piston in a cylinder filled with a gas. At equilibrium, the piston is not moving as the external pressure (*p*_{ext}) is equal to the pressure exerted by the gas inside the cylinder ( *p* ).

Now consider a slow heating process where the gas inside the cylinder is very slowly heated. The gas will expand and the piston moves upward. In this process, the energy of the system increases as it absorbs heat. However, the system *performs work* (i.e. expansion work) on the surroundings and loses some energy at the same time. Therefore, the overall internal energy change of the system must account for the energy gained via heat as well as the energy lost via work.

The work performed by this process is directly related to the volume increase (expansion) that occurred against a constant external pressure.

\[w = -p_{\mathrm{ext}}\Delta V\]

If the system was compressed, Δ*V* is negative, making the work energy quantity positive. We interpret this as the amount of energy a system gained by the work being done on the it.

Solids and liquids are usually not very compressible (Δ*V* ≈ 0) whereas gases are very compressible. Therefore, we can predict work for a reaction by identifying a change in moles of gas between products and reactants. If the change in moles of gas, Δ*n*_{gas}, is positive, volume increases. If Δ*n*_{gas} is negative, volume decreases. If the change in moles of gas is 0, volume does not change.

##
*pV*-work equation derivation

*pV*-work equation derivation

In physics, the work performed is a function of the upward force ( ** F** ) on the piston times the distance (

*d*) that the piston moves (

*w*=

**×**

*F**d*). Force is a function of the mass of the piston, the external pressure, and Earth’s gravitational pull.

\[w = \boldsymbol{F}d\]

Pressure is the force per area.

\[p = \dfrac{\boldsymbol{F}}{A}\]

Force is therefore pressure times area.

\[\boldsymbol{F} = pA\]

Substitute ** F** in terms of pressure and area to give

\[w = pAd\]

Since volume is an area times distance (or length), a *change in volume* ( Δ*V* ), can be expressed.

\[w = p\Delta V \]

Finally, a negative symbol is given to keep with the convention (Table 6) that a system *that performs work* loses energy. *If* the system *expands*, Δ*V* is positive, the system performs work, and therefore loses energy. *If* the system *compresses*, Δ*V* is negative, work is performed *on* the system, and the system gains energy.

\[w = -p\Delta V \] Finally, we label the pressure to clarify that we are referring to the external pressure in the context of a reversible process.

\[w = -p_{\mathrm{ext}}\Delta V \]

**Practice**

A 0.50 L sample of gas in a cylinder subjected to a constant pressure of 101 325 Pa expands in an infinitesimal way to a volume of 1.00 L upon the transfer of 348 J of energy to the gas. What is the change in internal energy (in J) of the gas?

Note that the pascal (Pa) in SI units is kg m^{–1} s^{–2}.

##
**Solution**

Notice that the system absorbs heat (*q*), an endothermic process. At the same time, the system performs work (loses energy) due to an expansion against a constant pressure.

Find the change in volume of the system (in m^{3}).

\[\begin{align*} \Delta V &= V_{\mathrm{final}} - V_{\mathrm{initial}} \\[1.5ex] &= \left ( 1.00~\mathrm{L} - 0.50~\mathrm{L} \right ) \left ( \dfrac{\mathrm{m^3}}{10^3~\mathrm{L}} \right ) \\[1.5ex] &= 5.00\times 10^{-4}~\mathrm{m^3} \end{align*}\]

Next, find the work performed by the system under conditions of constant pressure.

\[\begin{align*} w &= -p\Delta V \\[1.5ex] &= -\left ( 101~325~\mathrm{Pa} \right ) \left ( 5.00\times 10^{-4}~\mathrm{m^3} \right ) \\[1.5ex] &= -\left ( \dfrac{101~325~\mathrm{kg}}{\mathrm{m~s^2}} \right ) \left ( 5.00\times 10^{-4}~\mathrm{m^3} \right ) \\[1.5ex] &= -50.\bar{6}6~\mathrm{kg~m^2~s^2} \\[1.5ex] &= -50.\bar{6}6~\mathrm{J} \end{align*}\]

Finally, determine the change in internal energy.

\[\begin{align*} \Delta U &= q + w \\[1.5ex] &= 348~\mathrm{J} + (-50.\bar{6}6~\mathrm{J}) \\[1.5ex] &= 29\bar{7}.3~\mathrm{J} \\[1.5ex] &= 297~\mathrm{J} \end{align*}\]

If a process is carried out under *constant-volume* conditions (i.e. an **isochoric process)** Δ*V* is 0 and no *pV*-work is performed (i.e. no expansion work is done).

\[\begin{align*} w &= 0 \end{align*}\]

Therefore, a change in internal energy under constant-volume conditions (indicated with a subscript *V*) is directly related to the quantity of heat that was transferred.

\[\begin{align*} \Delta U &= q_{\mathrm{V}} + w_{\mathrm{V}} \\[1.5ex] &= q_{\mathrm{V}} + 0 \\[1.5ex] &= q_{\mathrm{V}} \end{align*}\]

*Bomb calorimetry* is a technique that is used to measure changes in internal energy of a system under constant-volume conditions.

### Enthalpy

Many processes take place under constant-pressure conditions (an *isobaric process*) such as a chemical reaction being carried out in an open flask or beaker which experiences a constant external pressure from the atmosphere. The reaction is allowed to experience a change in volume.

A change in internal energy under conditions of constant pressure can be written as

\[\Delta U = q_{\mathrm{p}} + w_{\mathrm{p}}\]

If work is *pV*-work *at constant-pressure*, then,

\[\Delta U = q_{\mathrm{p}} - p\Delta V\]

Rearranging provides a description of heat under constant-pressure conditions

\[q_{\mathrm{p}} = \Delta U + p\Delta V\]

and this heat ( *q*_{P} ) is defined using a new thermodynamic function called **enthalpy ( H )** (an extensive property) which is the sum of a system’s internal energy

*and*

*pV*-work. A change in enthalpy ( Δ

*H*) under constant pressure is expressed as

\[\Delta H = q_{\mathrm{p}} = \Delta U + p\Delta V\]

or simply

\[\Delta H = q_{\mathrm{p}}\]

A process is *endothermic* if Δ*H* > 0 whereas a process is *exothermic* if Δ*H* is < 0.

Under constant-volume conditions:

\[\Delta U = q_{\mathrm{V}}\]

Under constant-pressure conditions:

\[\Delta H = q_{\mathrm{p}}\]

Δ*U* and Δ*H* are equal when no work is done. Under constant-pressure conditions, Δ*U* and Δ*H* are similar unless a large change in volume occurs.

*Note:* Δ*H* is *preferred* when working under conditions of constant-pressure. Enthalpy still exists for non-isobar processes.

### State vs Path Function

Quantities that only depend on the thermodynamic *state* of a system are called **state functions** (or state quantities).

Examples of state functions in thermodynamics are

- Mass
- Energy (
*E*)- Enthalpy (
*H*) - Internal energy (
*U*) - Gibbs energy (
*G*)

- Enthalpy (
- Entropy (
*S*) - Pressure (
*p*) - Temperature (
*T*) - Volume (
*V*) - Chemical composition

Quantities that depend on the “path taken” between two thermodynamic states are called **path functions** (or process functions) and include heat and work.

**Example**

Imagine climbing a mountain. The distance traveled depends on the *path* that was taken (a winding hiking trail) and is therefore a “path function”. The *change in altitude* is analogous to a *state function* such that this quantity only depends on the initial and final elevations.

**Example**

Two swimmers swim exactly 50 yards along a straight line. The first swimmer uses the breaststroke technique while the second swimmer uses the butterfly technique. Both swimmers traveled the same distance (in this case a state function since the paths were the same) but the energy expended was different since two different swimming techniques were used.

## Enthalpy of Reaction

The molar **enthalpy of reaction** or molar **heat of reaction** (Δ_{r}*H* in kJ mol^{–1}), is the enthalpy change associated with a chemical reaction.

Consider the vaporization of water.

\[\mathrm{H_2O(l)} \longrightarrow \mathrm{H_2O(g)}\]

This process (or reaction) is endothermic ( Δ*H* > 0 ) as liquid water needs to absorb heat in order to vaporize.

\[\begin{align*} \Delta_{\mathrm{r}} H &= H_{\mathrm{final}} - H_{\mathrm{initial}} \\[1.5ex] &= H_{\mathrm{products}} - H_{\mathrm{reactants}} \\[1.5ex] &= H(\mathrm{H_2O, g}) - H(\mathrm{H_2O, l}) \end{align*}\]

where

\[H(\mathrm{H_2O, g}) > H(\mathrm{H_2O, l})\]

The enthalpy of this reaction can be referred to as *the enthalpy of vaporization* or *the (latent) heat of vaporization* for water and is a per-mole of reaction quantity ( Δ_{vap}*H* in kJ mol^{–1}). One would need to know the enthalpy of one mole of gaseous water and one mole of liquid water to determine the enthalpy change (per mole) for this process.

##
**Labeling Enthalpy**

Enthalpy is often simply given as Δ*H* ; however, enthalpy can be labeled to denote a particular process it is associated with. See table below for some examples.

### Standard State

To help simplify thermodynamic evaluations, a reference point was adopted called the *standard state*, denoted with a Plimsoll (^{⦵}) or degree symbol (°).

**Standard state** is defined to be for

**Gases**: A hypothetical state a gas would have as a pure substance obeying the*ideal gas equation*(i.e. behaves as an ideal gas) at standard state pressure**Liquids**: The state of a pure substance as a liquid under standard state pressure**Solids**: The state of a pure, crystalline substance under standard state pressure**Solutes**: A hypothetical state a solute would have in an ideal solution with a standard amount concentration,*c*°, of exactly 1 mol L^{–1}

Standard state pressure, *p*°, is defined to be a pressure of 1 bar or 10^{5} Pa.

Note that temperature is not included in the definition of standard state; however, most thermodynamic quantities are compiled at specific temperatures, *usually* at 25 °C (298.15 K).

### Standard Enthalpy of Formation

A **standard enthalpy of formation (Δ _{f}H° in kJ mol^{–1})**, also called the

*standard heat of formation*, of a compound is the change in enthalpy during the formation of exactly 1 mole of a substance from its constituent elements in their reference state, with all substances in the standard state.

The standard heat of formation for substances in their pure, elemental, and most stable form is zero.

Consider the formation of carbon dioxide from its constituent elements in their reference states. This reaction is exothermic with a standard heat of reaction –393.5 kJ mol^{–1}, and can be written as

\[\mathrm{C(s,graphite)} + \mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \qquad \Delta_{\mathrm{r}}H^{\circ} = -393.5~\mathrm{kJ~mol^{-1}}\]

Notice the reactants. Elemental carbon is given in its standard state (a crystalline solid in the form of graphite, an allotrope of solid carbon; solid carbon also exists in the form as diamond, another allotrope, but graphite is slightly lower in energy). Elemental oxygen exists naturally as oxygen gas at its standard state. Carbon dioxide exists as a gas at its standard state.

The standard heat of formation for carbon dioxide, Δ_{f}*H*°(CO_{2}, g), can be determined from the heats of formation of the reactants (where ν is the stoichiometric coefficient from the balanced chemical equation) using values tabulated here are compiled at specific temperatures, *usually* at 25 °C (298.15 K).

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ} \\[1.5ex] \Delta_{\mathrm{r}} H^{\circ} &= \biggl [ \Delta_{\mathrm{f}} H^{\circ}(\mathrm{CO_2,g}) \biggl ] - \biggl [ \Delta_{\mathrm{f}} H^{\circ}(\mathrm{C, s, graphite}) + \Delta_{\mathrm{f}} H^{\circ}(\mathrm{O_2, g}) \biggl ] \\[3ex] -393.5~\mathrm{kJ~mol^{-1}} &= \biggl [ \Delta_{\mathrm{f}} H^{\circ}(\mathrm{CO_2, g}) \biggl ] - \biggl [ 0~\mathrm{kJ~mol^{-1}} + 0~\mathrm{kJ~mol^{-1}} \biggl ] \\[1.5ex] -393.5~\mathrm{kJ~mol^{-1}} &= \Delta_{\mathrm{f}} H^{\circ}(\mathrm{CO_2, g}) \end{align*}\]

Therefore, to create CO_{2} from pure elements in the standard state is an exothermic process.

##
**Energy level diagram for formation of CO**_{2}

_{2}

See tabulated heats of formation for around 2500 substances.

**Practice**

Write out a balanced chemical reaction that is associated with the standard enthalpy of formation of H_{2}O(l).

##
**Solution**

\[\mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{H_2O(l)}\]

Note that we are asked to find the standard enthalpy of formation which is defined as being the change in heat for the formation of *exactly 1 mole* of a substance. Here, exactly 1 mole of product is represented which requires a fraction of 1/2 to appear in front of O_{2}.

**Practice**

How much heat (in kJ) is required to melt 250. g of water at exactly water’s melting point of 0 °C? Note that “melting” is also referred to as “fusion”. The standard heat of fusion for water is 6.89 kJ mol^{–1}

##
**Solution**

**Balanced Reaction**

\[\mathrm{H_2O(s)} \longrightarrow \mathrm{H_2O(l)}\] **Heat of Reaction**

\[\begin{align*} q &= n(\mathrm{H_2O}) ~ \Delta_{\mathrm{fus}} H^{\circ} \\[1.5ex] &= m(\mathrm{H_2O}) ~ M(\mathrm{H_2O})^{-1} ~ \Delta_{\mathrm{fus}} H^{\circ} \\[1.5ex] &= 250.~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \left ( \dfrac{6.89~\mathrm{kJ}}{\mathrm{mol}} \right ) \\[1.5ex] &= 95.\bar{5}8~\mathrm{kJ} \\[1.5ex] &= 95.6~\mathrm{kJ} \end{align*}\]

**Practice**

A 5.50 mol sample of aqueous HCl is reacted with a 7.50 mol sample of NaOH. The 1.000 L solution is initially 22.0 °C. What is the final temperature of the solution if the enthalpy of reaction is –57.3 kJ mol^{–1}? Assume the density of the solution to be approximately 1.000 g mL^{–1} and the specific heat capacity of the solution to be the same as water.

##
**Solution**

**Balanced Reaction**

\[\mathrm{HCl(aq)}+\mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(s)}\]

**Mass of Solution**

\[\begin{align*} d &= \dfrac{m}{V} \longrightarrow \\[1.5ex] m(\mathrm{solution}) &= dV \\[1.5ex] &= \left ( \dfrac{1.000~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) ~ 1.000~\mathrm{L} \\[1.5ex] &= 1000.~\mathrm{g} \end{align*}\]

**Reaction Heat**

Determine the moles of product that is produced from the limiting reactant. HCl is limiting.

\[\begin{align*} n(\mathrm{NaCl}) &= n(\mathrm{HCl}) ~ r(\mathrm{NaCl, HCl}) \\[1.5ex] &= 5.50~\mathrm{mol~HCl} \left ( \dfrac{1~\mathrm{mol~NaCl}}{1~\mathrm{mol~HCl}} \right ) \\[1.5ex] &= 5.50~\mathrm{mol} \end{align*}\]

Next, find the change in heat of the reaction.

\[\begin{align*} q_{\mathrm{reaction}} &= n \Delta H \\[1.5ex] &= 5.50~\mathrm{mol} \left ( \dfrac{-57.3~\mathrm{kJ}}{\mathrm{mol}} \right ) \\[1.5ex] &= -31\bar{5}.15~\mathrm{kJ} \end{align*}\]

**Find final temperature of solution**

\[\begin{align*} q_{\mathrm{solution}} &= -q_{\mathrm{reaction}} \\[1.5ex] \left [mc\Delta T \right ]_{\mathrm{solution}} &= -q_{\mathrm{reaction}} \\[1.5ex] \left [mc(\mathrm{l}) (T_{\mathrm{final}} - T_{\mathrm{initial}}) \right ]_{\mathrm{solution}} &= -q_{\mathrm{reaction}} \\[1.5ex] \left ( 1~000.~\mathrm{g} \right ) \left ( \dfrac{4.184~\mathrm{J}}{\mathrm{mol^{-1}~K^{-1}}} \right ) \left ( T_{\mathrm{final}} - 295~\mathrm{K} \right ) &= -\left ( -31\bar{5}.15~\mathrm{kJ} \right ) \left ( \dfrac{10^3~\mathrm{J}}{\mathrm{kJ}} \right ) \\[1.5ex] 4~1\bar{8}4~\mathrm{J~K^{-1}} \left ( T_{\mathrm{final}} - 295~\mathrm{K} \right ) &= 31\bar{5}~150~\mathrm{J} \\[1.5ex] T_{\mathrm{final}} &= 37\bar{0}.3~\mathrm{K} - 273.15 \\[1.5ex] &= 9\bar{7}.1~\mathrm{^{\circ}C} \\[1.5ex] &= 97~\mathrm{^{\circ}C} \end{align*}\]

### Standard Enthalpy of Reaction

The **standard enthalpy of reaction (Δ _{r}H° or Δ_{r}H^{⦵} in kJ mol^{–1})** can be determined as the difference in the total standard molar product and total standard molar reactant enthalpies of formation.

Standard enthalpy of reaction represents reactants and products in the standard state where the reactants are *unmixed* (i.e. in pure, isolated form) and the products are *unmixed*.

Therefore, the a standard enthalpy of a reaction compares the state of separated/isolated reactants to the state of separated/isolated products.

Let us generalize this enthalpy evaluation for any reaction. Consider the following general reaction equation.

\[\nu_{\mathrm{A}}~\mathrm{A} + \nu_{\mathrm{B}}~\mathrm{B} \longrightarrow \nu_{\mathrm{C}}~\mathrm{C} + \nu_{\mathrm{D}}~\mathrm{D}\]

The standard enthalpy of reaction would be

\[\Delta_{\mathrm{r}} H^{\circ} = \bigg[ \nu_{\mathrm{C}}~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{C}) + \nu_{\mathrm{D}}~ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{D}) \bigg] - \bigg [ \nu_{\mathrm{A}}~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{A}) + \nu_{\mathrm{B}}~ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{B}) \bigg ] \]

or, more compactly,

\[\Delta_{\mathrm{r}} H^{\circ} = \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ}\]

This is the *difference of sums* of product and reactant enthalpies.

**Example: Standard heat of combustion of methane**

Let us evaluate the standard enthalpy of reaction for the combustion of methane.

\[\mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)}\]

Grabbing the appropriate heats of formation gives

Now, calculate the standard enthalpy of reaction.

\[\begin{align*} \Delta_{\mathrm{c}} H^{\circ}(\mathrm{CH_4,g}) &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ} \\[1.5ex] &= \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{CO_2, g}) + 2~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2O, g}) \bigg ] - \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{CH_4, g}) + 2~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{O_2, g}) \bigg ] \\[1.5ex] &= \bigg [ (-393.5) + (2 \times -241.826) \bigg ]~\mathrm{kJ~mol^{-1}} - \bigg [ (-74.6) + (2\times 0) \bigg ]~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -802.\bar{5}5~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -802.6~\mathrm{kJ~mol^{-1}} \end{align*}\]

This reaction is very exothermic. A lot of heat is given off when 1 mole of methane is combusted with oxygen.

The heat of combustion, calculated above, is a per mole-reaction quantity. What does this mean? Notice how the stoichiometric coefficients in the balanced chemical equation are not all 1.

We interpret molar thermodynamic quantities to be per mole-reaction meaning, per mole as indicated in the balanced chemical reaction.

Therefore, the heat of combustion is –802.6 kJ…

- …per 1 mole of CH
_{4}(g) consumed - …per 2 moles of O
_{2}(g) consumed - …per 1 mole of CO
_{2}(g) produced - …per 2 moles of H
_{2}O(g) produced

**Practice**

Determine the standard enthalpy of combustion of methane if exactly

- 2 moles of CH
_{4}(g) was combusted - 5 moles of O
_{2}(g) was combusted - 0.25 moles of H
_{2}(g) was produced

\[\mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)}\]

Use the standard enthalpy of combustion for methane that was calculated in the previous example, given as Δ_{c}*H*° = –802.5 kJ mol^{–1}.

##
**Solution**

**2 moles of CH _{4}(g) combusted**

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= n(\mathrm{CH_4, g}) ~ \Delta_{\mathrm{c}} H^{\circ} \\[1.5ex] &= 2~\mathrm{mol~CH_4(g)} ~ \left ( \dfrac{-802.5~\mathrm{kJ}}{\mathrm{1~mol~CH_4(g)}} \right ) \\[1.5ex] &= -1~605~\mathrm{kJ} \end{align*}\]

**5 moles of O _{2}(g) combusted**

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= n(\mathrm{O_2, g}) ~ \Delta_{\mathrm{c}} H^{\circ} \\[1.5ex] &= 5~\mathrm{mol~O_2(g)} \left ( \dfrac{-802.5~\mathrm{kJ}}{\mathrm{2~mol~O_2(g)}} \right ) \\[1.5ex] &= -2~006~\mathrm{kJ} \end{align*}\]

**0.25 moles of H _{2}O(g) produced**

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= n(\mathrm{H_2O, g}) ~ \Delta_{\mathrm{c}} H^{\circ} \\[1.5ex] &= 0.25~\mathrm{mol~H_2O(g)} \left ( \dfrac{-802.5~\mathrm{kJ}}{\mathrm{2~mol~H_2O(g)}} \right ) \\[1.5ex] &= -100.3~\mathrm{kJ} \end{align*}\]

**Practice**

Determine the standard molar enthalpy of reaction (in kJ mol^{–1}) for a reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas. Is the reaction endothermic or exothermic? What is the enthalpy of reaction (in kJ mol^{–1}) if 10 moles of HCl were produced?

Look up thermodynamic values reported here.

##
**Solution**

Write a balanced chemical equation.

\[\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{HCl(g)}\]

Collect the appropriate thermodynamic values.

Calculate the standard molar enthalpy of reaction.

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ} \\[1.5ex] &= \bigg [ 2~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{HCl, g}) \bigg ] - \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2, g}) + \Delta_{\mathrm{f}}H^{\circ}(\mathrm{O_2, g}) \bigg ] \\[1.5ex] &= \bigg [ (2 \times -92.3) \bigg ]~\mathrm{kJ~mol^{-1}} - \bigg [ (2 \times 0) + (2 \times 0) \bigg ]~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -184.6~\mathrm{kJ~mol^{-1}} \end{align*}\]

Calculate the enthalpy of reaction if exactly 10 moles of hydrogen chloride gas was produced.

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= n(\mathrm{HCl, g}) ~ \Delta_{\mathrm{r}} H^{\circ} \\[1.5ex] &= 10~\mathrm{mol~HCl(g)} \left ( \dfrac{-184.6~\mathrm{kJ}}{\mathrm{2~mol~HCl(g)}} \right ) \\[1.5ex] &= -923.0~\mathrm{kJ} \end{align*}\]

**Example: Standard heat of vaporization of water**

Now, let us consider *phase change* of a substance by revisiting the vaporization of water.

\[\mathrm{H_2O(l)} \longrightarrow \mathrm{H_2O(g)}\]

This reaction is endothermic. Using the appropriate heats of formation gives

\[\begin{align*} \Delta_{\mathrm{vap}} H^{\circ} &= \Delta_{\mathrm{f}} H^{\circ}(\mathrm{H_2O, g}) - \Delta_{\mathrm{f}} H^{\circ}(\mathrm{H_2O, l}) \\[1.5ex] &= (-241.83~\mathrm{kJ~mol^{-1}}) - (-285.83~\mathrm{kJ~mol^{-1}}) \\[1.5ex] &= 44.00~\mathrm{kJ~mol^{-1}} \end{align*}\]

Therefore, the standard molar heat of vaporization of water is 44.00 kJ mol^{–1}, an endothermic process.

The opposite of vaporization (liquid to gas) of condensation (gas to liquid). What would the heat of condensation be?

Reversing the chemical equation

\[\mathrm{H_2O(g)} \longrightarrow \mathrm{H_2O(l)}\]

gives

\[\begin{align*} \Delta_{\mathrm{con}} H^{\circ} &= \Delta_{\mathrm{f}} H^{\circ}(\mathrm{H_2O, l}) - \Delta_{\mathrm{f}} H^{\circ}(\mathrm{H_2O, g}) \\[1.5ex] &= (-285.83~\mathrm{kJ~mol^{-1}}) - (-241.83~\mathrm{kJ~mol^{-1}}) \\[1.5ex] &= -44.00~\mathrm{kJ~mol^{-1}} \end{align*}\]

The standard molar heat of condensation of water is –44.00 kJ mol^{–1}.

Thermodynamic quantities that are state functions *change in sign* for any process that is reversed.

Δ_{vap}*H*°(H_{2}O) = 44.00 kJ mol^{–1}

Δ_{con}*H*°(H_{2}O) = –44.00 kJ mol^{–1}

##
**Energy level diagram for formation of H**_{2}O

_{2}O

### Enthalpy at Other Temperatures (*T* ≠ 298.15 K)

So far, we have examined how to determine an enthalpy for a reaction using standard thermodynamic values all set to one temperature (298.15 K). However, we know that many reactions take place at other temperatures, such as the boiling of water which takes place at 373.15 K (100 °C).

NIST-JANAF provides many thermodynamic values for substances across a range of temperatures including enthalpy, entropy, free energy, and heat capacity.

##
**Example: Thermodynamic properties for H**_{2}O(l) from 280 K to 500 K

_{2}O(l) from 280 K to 500 K

\[\mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \longrightarrow \mathrm{H_2O}(l)\]

##
**Tabulated Thermodynamic Data**

Enthalpy reference temperature (*T*_{r} = 298.15 K); Standard state pressure = *p*° = 0.1 MPa

*C*_{p}°, S°, and -[*G*°-*H*°(*T*_{r})] /*T*in J mol^{–1}K^{–1}- Δ
_{f}*H*° and Δ_{f}*G*° in kJ mol^{–1}

T (K) |
C°_{p} |
S° |
–[G°-H°(T_{r})]/T |
H° -H°(T_{r}) |
Δ_{f}H° |
Δ_{f}G° |
log K_{f} |
---|---|---|---|---|---|---|---|

0 |
|||||||

100 |
|||||||

200 |
|||||||

280 |
75.563 |
65.215 |
70.102 |
-1.368 |
-286.41 |
-240.123 |
44.796 |

298.15 |
75.351 |
69.95 |
69.95 |
0 |
-285.83 |
-237.141 |
41.546 |

300 |
75.349 |
70.416 |
69.952 |
0.139 |
-285.771 |
-236.839 |
41.237 |

320 |
75.344 |
75.279 |
70.134 |
1.646 |
-285.137 |
-233.598 |
38.131 |

340 |
75.388 |
79.847 |
70.573 |
3.153 |
-284.506 |
-230.396 |
35.396 |

360 |
75.679 |
84.164 |
71.209 |
4.664 |
-283.874 |
-227.231 |
32.97 |

372.78 |
75.962 |
86.808 |
71.699 |
5.633 |
LIQUID |
<–> |
REAL GAS |

380 |
76.154 |
88.267 |
72 |
6.182 |
283.237 |
224.102 |
30.805 |

400 |
76.77 |
92.189 |
72.912 |
7.711 |
-282.591 |
-221.006 |
28.86 |

420 |
77.547 |
95.952 |
73.92 |
9.254 |
-281.934 |
-217.943 |
27.105 |

440 |
78.543 |
99.582 |
75.004 |
10.814 |
-281.262 |
-214.912 |
25.513 |

460 |
79.793 |
103.1 |
76.15 |
12.397 |
-280.569 |
-211.911 |
24.063 |

480 |
81.463 |
106.53 |
77.344 |
14.009 |
-279.85 |
-208.941 |
22.737 |

500 |
83.694 |
109.898 |
78.579 |
15.659 |
-279.095 |
-206.002 |
21.521 |

One can transform an enthalpy determined at 298.15 K to some other temperature using

\[\Delta_{\mathrm{r}} H_{T} = \Delta_{\mathrm{r}} H_{298.15~\mathrm{K}} + \int_{298}^{T} \Delta C_{\mathrm{p}}~dT\]

or

\[\Delta_{\mathrm{r}} H_{T} = \Delta_{\mathrm{r}} H_{298.15~\mathrm{K}} + \int_{298}^{T} \left [ C_{\mathrm{p}}^{\mathrm{products}} - C_{\mathrm{p}}^{\mathrm{reactants}}\right ]~dT\]

where

\[\Delta C_{\mathrm{p}} = \!\!\!\!\sum_{products,~p} \!\!\! \nu_p ~ C_{\mathrm{p}}- \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~ C_{\mathrm{p}}\]

and

- Δ
_{r}*H*_{T}is the enthalpy of reaction at temperature*T* - Δ
_{r}*H*_{298.15 K}is the enthalpy of reaction at 298.15 K - Δ
*C*_{p}is the difference in heat capacities (at constant pressure) between the products and the reactants where the stoichiometric coefficients must be considered *T*is new temperature of interest*dT*is the change in temperature from 298.15 K to*T*

The equation can be simplified if the heat capacities are assumed to constant over the temperature range of interest (eliminating the integral) to give

\[\Delta_{\mathrm{r}} H_{T} = \Delta_{\mathrm{r}} H_{298.15~\mathrm{K}} + \Delta C_{\mathrm{p}}~(T-298.15~\mathrm{K})\]

**Example: Heat of vaporization of water at 100 °C**

Let us now calculate the standard molar heat of vaporization of water (in kJ mol^{–1}) at 100 °C (373.15 K). The molar heat capacity for liquid water is 75.35 J mol^{–1} K^{–1} and the molar heat capacity for gaseous water is 33.60 J mol^{–1} K^{–1}. Use the standard molar heat of vaporization for water at 298.15 K that we calculated above (Δ_{vap}*H*° = 44.00 kJ mol^{–1}).

\[\begin{align*} \Delta_{\mathrm{vap}} H^{\circ}_{373.15~\mathrm{K}} &= \Delta_{\mathrm{vap}} H^{\circ}_{298.15~\mathrm{K}} + \Delta C_{\mathrm{p}}~(T-298.15~\mathrm{K}) \\[1.5ex] &= \dfrac{44.00~\mathrm{kJ}}{\mathrm{mol}} + \left [ \dfrac{\left ( 33.60 - 75.35 \right )~\mathrm{J}}{\mathrm{mol~K}} \right ] \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 373.15~\mathrm{K} - 298.15~\mathrm{K} \right ) \\[1.5ex] &= 40.8\bar{6}8~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= 40.87~\mathrm{kJ~mol^{-1}} \end{align*}\]

The experimental standard molar heat of vaporization of water at 373.15 K is 40.65 kJ mol^{–1} so our calculated (or approximated) value is very close.

Error in our calculated value includes assuming that heat capacity remains constant with temperature change. However, given the close agreement of the experimental value and the computed value, this approximation seems reasonable.

Thermodynamic quantities such as enthalpy usually do not change very much across broad temperature ranges. Therefore, approximations using values at 298.15 K are generally reasonable.

**Practice**

What is the molar heat of combustion of methane if the flame temperature with pure oxygen is 2 810°C (3 083.15 K)?

\[\mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)}\]

Use the standard molar enthlapy of combustion for methane calculated earlier (Δ_{c}*H*°_{298 K} = –877.2 kJ mol^{–1}).

The heat capacities (in J mol^{–1} K^{–1}) are:

*C*_{p}(CH_{4}, g) = 35.7*C*_{p}(O_{2}, g) = 29.4*C*_{p}(CO_{2}, g) = 37.1*C*_{p}(H_{2}O, g) = 33.6

##
**Solution**

For simplicity, calculate Δ*C*_{p} first.

\[\begin{align*} \Delta C_{\mathrm{p}} &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p ~ C_{\mathrm{p}} - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~ C_{\mathrm{p}} \\[1.5ex] &= \biggl\{ \Bigl [ C_{\mathrm{p}}(\mathrm{CO_2, g}) \Bigr ] + \Bigl [ 2~C_{\mathrm{p}}(\mathrm{H_2O, g}) \Bigr ] \biggr\} - \biggl\{ \Bigl [ C_{\mathrm{p}}(\mathrm{CH_4, g}) \Bigr ] + \Bigl [ 2~C_{\mathrm{p}}(\mathrm{O_2, g}) \Bigr ] \biggr\} \\[1.5ex] &= \biggl\{ \Bigl ( 37.1 \Bigr ) + \Bigl ( 2\times 33.6 \Bigr ) \biggr\} - \biggl\{ \Bigl ( 35.7 \Bigr ) + \Bigl ( 2\times 29.4 \Bigr ) \biggr\}~\mathrm{J~mol^{-1}~K^{-1}} \\[1.5ex] &= 9.8~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}\]

Next, find the molar enthalpy of combustion at 3 083.15 K.

\[\begin{align*} \Delta_{\mathrm{c}} H^{\circ}_{3~083.15~\mathrm{K}} &= \Delta_{\mathrm{c}} H^{\circ}_{298.15~\mathrm{K}} + \Delta C_{\mathrm{p}}~(T-298.15~\mathrm{K}) \\[1.5ex] &= \dfrac{-877.2~\mathrm{kJ}}{\mathrm{mol}} + \left ( \dfrac{9.8~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 3~083.15~\mathrm{K} - 298.15~\mathrm{K} \right ) \\[1.5ex] &= -84\bar{9}.9~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -850~\mathrm{kJ~mol^{-1}} \end{align*}\]

Notice that the heat of combustion changed very little (less than 20 kJ mol^{–1}) even though the temperature changed by nearly 3 000 K.

We find that enthalpy values, in general, do not change very much with temperature, and therefore enthalpies calculated at 298.15 K are very reasonable approximations.

### Relationship Between Enthalpy and Internal Energy

Enthalpy and internal energy are related by

\[\Delta H = \Delta U + \Delta (pV)\]

Constant-pressure conditions gives

\[\Delta H = \Delta U + p\Delta V\]

The ideal gas law (*pV* = *nRT*) gives the following relationship

\[\Delta (pV) = \Delta (nRT) \]

where

- Δ is “change in”
*p*is pressure*V*is volume*n*is moles of gas*R*is the molar gas constant (8.314 J mol^{–1}K^{–1})*T*is temperature

The change in moles of gas (Δ*n*_{gas}) is

\[\Delta n_{\mathrm{gas}} = \!\!\!\!\sum_{products,~p} \!\!\! \nu_p ~ n_{\mathrm{gas}} - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~n_{\mathrm{gas}} \]

Therefore, we can rewrite the Δ*H*/Δ*U* relation as

\[\Delta H = \Delta U + \Delta (n_{\mathrm{gas}}RT)\]

Typically, a temperature of 298.15 K is used when assuming standard conditions. Therefore, Δ*T* → *T* = 298.15 K.

\[\Delta H = \Delta U + \Delta n_{\mathrm{gas}} RT\]

**Example: Decomposition of ammonium nitrate**

The decomposition of ammonium nitrate can be written as

\[\mathrm{NH_4NO_3(s)} \longrightarrow \mathrm{N_2O(g)} + \mathrm{H_2O}(g)\]

We can find the molar internal energy change ( Δ*U* in kJ mol^{–1} ) of this reaction under conditions of constant temperature and pressure by first finding the standard molar enthalpy change for this reaction.

Collect the appropriate thermodynamic values reported here

Next, calculate the standard molar heat of reaction.

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ} \\[1.5ex] &= \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{N_2O, g}) + \Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2O, g}) \bigg ] - \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{NH_4NO_3, s}) \bigg ] \\[1.5ex] &= \left ( 81.6 + -241.826 \right )~\mathrm{kJ~mol^{-1}} - \left ( -365.6 \right )~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= 20\bar{5}.3~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= 205~\mathrm{kJ~mol^{-1}} \end{align*}\]

Determine the change in moles of gas.

\[\begin{align*} \Delta n_{\mathrm{gas}} &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p ~ n_{\mathrm{gas}} - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~n_{\mathrm{gas}} \\[1.5ex] &= \left (1 + 1 \right ) ~\mathrm{mol} - \left ( 0 \right )~\mathrm{mol} \\[1.5ex] &= 2~\mathrm{mol} \end{align*}\]

Next, find the standard molar internal energy change under conditions of constant temperature and pressure. Since we are finding the *standard* molar value, the temperature is 298.15 K.

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \Delta_{\mathrm{r}} U^{\circ} + \Delta n_{\mathrm{gas}} RT \longrightarrow \\[1.5ex] \Delta_{\mathrm{r}}U^{\circ} &= \Delta_{\mathrm{r}}H^{\circ} - \Delta n_{\mathrm{gas}} RT \\[1.5ex] &= \dfrac{20\bar{5}.3~\mathrm{kJ}}{\mathrm{mol}} - \biggl [ \left ( 2\right ) \left ( \dfrac{8.314~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 298.15~\mathrm{K} \right ) \biggl ] \\[1.5ex] &= 20\bar{0}.3~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= 200~\mathrm{kJ~mol^{-1}} \end{align*}\]

**What if Δ n_{gas} = 0?**

If a reaction occurs where the change in moles of gas is zero at constant temperature, then

\[\Delta_{\mathrm{r}} H = \Delta_{\mathrm{r}} U \qquad (\mathrm{when}~\Delta n_{\mathrm{gas}} = 0~\mathrm{at~constant}~T)\]

An example of this would be an isomerization process of butane to isobutane where the reactants and products molecules have the same molecular formula but a different structural arrangement.

\[\mathrm{C_4H_{10}(g)} \longrightarrow \mathrm{isobutane~(g)} \qquad \Delta_{\mathrm{isom}}H^{\circ} = -9.5~\mathrm{kJ~mol^{-1}}\]

Because the change in moles of gas is zero *and* the temperature remains constant, the standard molar internal energy change is also –9.5 kJ mol^{–1}.

\[\Delta_{\mathrm{isom}}H^{\circ} = \Delta_{\mathrm{isom}}U^{\circ} = -9.5~\mathrm{kJ~mol^{-1}}\]

### Transforming Thermodynamic Quantities

Presented are the rules for transforming thermodynamic quantities when manipulating reactions. These rules extend to state functions (state quantities) such as Δ*H*, Δ*U*, Δ*S*, etc. The examples below use Δ*X* to denote any state quantity.

**Reversing Reactions**

Reverse the sign of a state function when reversing a reaction.

*Example:*

If

\[\mathrm{A} \longrightarrow \mathrm{B} \qquad \Delta X\]

then

\[\mathrm{B} \longrightarrow \mathrm{A} \qquad -\!\Delta X\]

**Multiplying Reactions**

If multiplying a reaction by a number, multiply the state function by the same number. This number is *exact*.

*Example:*

If

\[\mathrm{A} \longrightarrow \mathrm{B} \qquad \Delta X\]

then, multiplying by a number, *x*, gives

\[\begin{align*} x~(~\mathrm{A} &\longrightarrow \mathrm{B}~) \qquad x \left ( \Delta X \right ) \\[1.5ex] x~\mathrm{A} &\longrightarrow x~\mathrm{B} \qquad x \left ( \Delta X \right ) \end{align*}\]

**Combining Reactions**

If combining multiple reactions into one reaction, add the state functions.

*Example:*

If

\[\textrm{Step 1:} ~~\mathrm{A} \longrightarrow \mathrm{B} \qquad \Delta X_1\]

and

\[\textrm{Step 2:} ~~ \mathrm{B} \longrightarrow \mathrm{C} \qquad \Delta X_2\]

then, combining these reactions give

\[\begin{align*} &\textrm{Step 1:} \: \quad \mathrm{A} \longrightarrow \mathrm{B} \:\qquad \Delta X_1 \\[1ex] &\underline{\textrm{Step 2:} \quad\:\mathrm{B} \longrightarrow \mathrm{C} \quad\quad\: \Delta X_2 \qquad\qquad\qquad\qquad\phantom{x}} \\[1ex] &\textrm{Overall:} \!\!\quad \mathrm{A} \longrightarrow \mathrm{C} \qquad\: \Delta X_{\mathrm{overall}} = \Delta X_1 + \Delta X_2 \end{align*}\]

**Practice**

What is the heat of the following reaction (in kJ mol^{–1}) if the reaction was reversed and multiplied by 2.5?

\[\mathrm{NH_4NO_3(s)} \longrightarrow \mathrm{N_2O(g)} + \mathrm{H_2O}(g) \quad \Delta_{\mathrm{r}}H^{\circ} = 205~\mathrm{kJ~mol^{-1}}\]

##
**Solution**

\[2.5~\mathrm{N_2O(g)} + 2.5~\mathrm{H_2O}(g) \longrightarrow 2.5~\mathrm{NH_4NO_3(s)} \quad \Delta_{\mathrm{r}}H^{\circ} = ~?~\mathrm{kJ~mol^{-1}}\]

- Reversing the reaction flips the sign of Δ
*H* - Multiplying the reaction by 2.5 means Δ
*H*gets multiplied by 2.5

\[\begin{align*} \Delta_{\mathrm{r}} H^{\circ}_{f} &= -1 \times 2.5 \times \Delta_{\mathrm{r}} H^{\circ}_{i} \\[1.5ex] &= -1 \times 2.5 \times \dfrac{205~\mathrm{kJ}}{\mathrm{mol}} \\[1.5ex] &= -51\bar{2}.5~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -512~\mathrm{kJ~mol^{-1}} \end{align*}\]

## Calorimetry

**Calorimetry** is a technique used to determine the energy evolved from a chemical reaction or physical process by measuring changes in temperature. These techniques are carried out in a calorimeter.

### Constant pressure calorimetry

Constant-pressure calorimetry is a technique used to measure enthalpy changes ( Δ*H* ). These measurements can be performed in something as simple as a covered Styrofoam container such as a coffee cup.

This apparatus acts as an insulator, allowing the system to be treated as a pseudo-closed system where work and heat is not exchanged with the surroundings.

Then enthalpy of a reaction could be measured or the heat transfer between a hot metal object and the surrounding water, for example, by measuring the change in temperature of the system.

**Example**

Imagine the following acid-base neutralization reaction where a hydrochloric acid solution and a sodium hydroxide solution, both at 21.0 °C, being mixed into a coffee cup calorimeter to give a final volume of 175.0 mL.

\[\mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(s)}\]

Upon reaction, the temperature of the solution increases to 26.2 °C. Find the heat of reaction (in kJ mol^{–1}).

The heat evolved by the reaction ( *q*_{reaction} ) is absorbed by the solution ( *q*_{solution} ).

\[-q_{\mathrm{reaction}} = q_{\mathrm{solution}}\]

Recall that

\[q_{\mathrm{solution}} = mc\Delta T\]

Since the solution is mostly water, let us approximate the specific heat of the solution to be the same as water ( *c*_{solution} ≈ *c*_{water} = 4.184 J mol^{–1} °C^{–1} ) *and* approximate the density of the solution to be the same as water ( *d*_{solution} ≈ *d*_{water} ≈ 1.0 g mL^{–1} ).

We can now solve for the heat of reaction (in kJ).

\[\begin{align*} -q_{\mathrm{reaction}} &= q_{\mathrm{solution}} \longrightarrow \\[1.5ex] q_{\mathrm{reaction}} &= -q_{\mathrm{soln}} \\[1.5ex] &= -m_{\mathrm{soln}}~c_{\mathrm{soln}}~\Delta T_{\mathrm{soln}} \\[1.5ex] &= -d_{\mathrm{soln}}~V_{\mathrm{soln}}~c_{\mathrm{soln}}~\Delta T_{\mathrm{soln}} \\[1.5ex] &= - \left ( \dfrac{1.0~\mathrm{g}}{\mathrm{mL}} \right ) \left ( 175~\mathrm{mL} \right ) \left ( \dfrac{4.184~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( 26.2 - 21.0 \right )~^{\circ}\mathrm{C} \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\[1.5ex] &= -3.\bar{8}0~\mathrm{kJ} \\[1.5ex] &= -3.8~\mathrm{kJ} \end{align*}\]

### Constant volume calorimetry

Constant volume calorimetry is a technique used to measure internal energy changes ( Δ*U* ). These measurements are typically performed in a *bomb calorimeter*, an insulated container that contains a steel vessel, called a *bomb*, that is submerged in water. A combustion reaction is carried out inside the bomb that is pressurized with oxygen gas.

This exothermic reaction releases heat that is absorbed by the water *and* the bomb calorimeter apparatus. An increase in temperature is recorded by a thermometer that is submerged in the water. This data can be used to determine the internal energy change ( Δ*U* ) of the reaction.

**Exmaple**

Consider a 0.6475 g sample of napthalene (C_{10}H_{8}) that is combusted in an oxygen-rich environment under constant volume conditions. The surrounding water *and* the entire bomb calorimeter itself is at a temperature of 22.00 °C. Upon combustion, a final temperature of 25.21 °C was recorded. If the heat capacity of the calorimeter (which includes the water) is 8.0931 kJ °C^{–1}, what is Δ_{c}*U* (in kJ mol^{–1}) of the reaction (where c stands for combustion)?

\[\mathrm{C_{10}H_8(s)} + 12~\mathrm{O_2(g)} \longrightarrow \mathrm{10~CO_2(g)} + 4~\mathrm{H_2O(g)}\]

Note that

\[q_{\mathrm{calorimeter}} = C_{\mathrm{calorimeter}}\Delta T\]

First, find the heat of reaction (in kJ).

\[\begin{align*} -q_{\mathrm{reaction}} &= q_{\mathrm{calorimeter}} \longrightarrow \\[1.5ex] q_{\mathrm{reaction}} &= -C_{\mathrm{cal}} \Delta T \\[1.5ex] &= -\left ( \dfrac{8.0931~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( 25.21~^{\circ}\mathrm{C} - 22.00~^{\circ}\mathrm{C} \right )\\[1.5ex] &= -25.9\bar{7}8~\mathrm{kJ} \end{align*}\]

Next, find the specific internal energy of reaction (Δ_{c}*u* in kJ g^{–1}) per gram of napthalene.

\[\begin{align*} \Delta_{\mathrm{c}} u &= \dfrac{q_{\mathrm{reaction}}}{m_{\mathrm{napthalene}}} \\[1.5ex] &= \dfrac{-25.9\bar{7}8~\mathrm{kJ}}{0.6475~\mathrm{g}} \\[1.5ex] &= -40.1\bar{2}2~\mathrm{kJ~g^{-1}} \\[1.5ex] &= -40.12~\mathrm{kJ~g^{-1}} \end{align*}\]

Next, find the molar internal energy of reaction (Δ_{c}*U* in kJ mol^{–1}) per mole of napthalene.

\[\begin{align*} \Delta_{\mathrm{c}} U &= \Delta_{\mathrm{c}} u ~ M(\mathrm{C_{10}H_8}) \\[1.5ex] &= \dfrac{-40.1\bar{2}2~\mathrm{kJ}}{\mathrm{g}} \left ( \dfrac{128.18~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= -5~14\bar{2}.8~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -5~143~\mathrm{kJ~mol^{-1}} \end{align*}\]

**Practice**

The combustion of 6.491 g methanol (CH_{3}OH) is carried out in a bomb calorimeter.

\[\mathrm{CH_3OH(l)} + \dfrac{3}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(l)}\]

The calorimeter has a heat capacity of 14.402 kJ °C^{–1} and a change in temperature of 3.74 °C is recorded. What is the standard molar enthalpy of reaction (in kJ mol^{–1})?

##
**Solution**

**Find Heat of Reaction**

\[\begin{align*} q_{\mathrm{reaction}} &= -q_{\mathrm{calorimeter}} \\[1.5ex] &= -C\Delta T \\[1.5ex] &= -\left ( \dfrac{14.402~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( 3.74~\mathrm{^{\circ}C} \right ) \\[1.5ex] &= -53.\bar{8}6~\mathrm{kJ} \end{align*}\]

Under constant volume conditions,

\[\Delta H = q_{\mathrm{V}} + \Delta n_{\mathrm{gas}}RT\]

and

\[\Delta n_{\mathrm{gas}} = 1 - \dfrac{3}{2} = -\dfrac{1}{2}\]

Since we are finding the standard molar enthalpy of reaction, use 298.15 K for *T*.

\[\begin{align*} \Delta_{\mathrm{c}} H &= q_{\mathrm{reaction}} + \Delta n_{\mathrm{gas}}RT \\[1.5ex] &= q_{\mathrm{V}} + \Delta n_{\mathrm{gas}}RT \\[1.5ex] &= -53.\bar{8}6~\mathrm{kJ} + \left [ \left ( -\dfrac{1}{2} \right ) \left ( \dfrac{8.314~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 298.15~\mathrm{K} \right ) \right ] \\[1.5ex] &= -55.\bar{0}9~\mathrm{kJ} \end{align*}\]

**Find Standard Molar Heat of Reaction**

\[\begin{align*} \Delta_{\mathrm{c}} H^{\circ} &= \dfrac{\Delta_{\mathrm{c}} H}{n_{\mathrm{methanol}}} \\[1.5ex] &= \dfrac{\Delta_{\mathrm{c}} H}{m~M^{-1}} \\[1.5ex] &= \dfrac{-55.\bar{0}9~\mathrm{kJ}}{2.429~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{32.04~\mathrm{g}} \right )} \\[1.5ex] &= -72\bar{6}.6~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -727~\mathrm{kJ~mol^{-1}} \end{align*}\]

## Hess’s Law

Enthalpies can be measured experimentally such as the case with calorimetry. However, sometimes it is difficult or impossible to accurately measure the enthalpy of some reactions. However, according to **Hess’s Law**, a total enthalpy change of a complete set of reactions is independent of the sequence of steps taken. This is a consequence of enthalpy being a *state function*.

Take, for example, the formation of carbon dioxide under standard conditions and 298.15 K.

\[\begin{align*} \mathrm{C(s, graphite)} + \mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \qquad \Delta_{\mathrm{r}}H^{\circ} = -393.5~\mathrm{kJ} \end{align*}\]

We could examine this process as a two-step process where in Step 1, CO is formed from its constituent elemental forms and, in Step 2, CO_{2} is produced from the CO(g) intermediate and oxygen gas. Each individual step has its own heat of formation. Combining these steps gives the heat of formation of CO_{2}(g).

\[\begin{align*} &\textrm{Step 1:} \: \quad \mathrm{C(s, graphite)} + \frac{1}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{CO(g)} \qquad\! \Delta_{\mathrm{r}} H^{\circ} = -110.5~\mathrm{kJ} \\[1ex] &\underline{\textrm{Step 2:} \: \quad \mathrm{CO(g)} + \frac{1}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \qquad\qquad\:\,\,\Delta_{\mathrm{r}} H^{\circ} = -283.0~\mathrm{kJ}} \\[1ex] &\textrm{Overall:} \!\!\quad \mathrm{C(s, graphite)} + \mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \qquad\:\:\, \Delta_{\mathrm{r}} H^{\circ} = -393.5~\mathrm{kJ} \end{align*}\]

Therefore, known enthalpies of reactions (formal or real, meaning, the reaction or step does not have to take place in actuality) can be used to determine the enthalpies of other reactions who may not be known. Reaction transformations may be necessary to carry out this determination.

**Example**

Find the enthalpy of formation for acetylene (C_{2}H_{2}), given the following data.

\[\begin{align*} &\textrm{Reaction 1:} \quad \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2}(g) \longrightarrow \mathrm{H_2O(l)} \qquad\qquad\qquad\qquad\, \Delta_{\mathrm{r}} H^{\circ} = -285.83~\mathrm{kJ} \\[1ex] &\textrm{Reaction 2:} \quad \mathrm{C(s,graphite)} + \mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \qquad\qquad\qquad \!\!\!\!\:\Delta_{\mathrm{r}} H^{\circ} = -393.5~\mathrm{kJ} \\[1ex] &\textrm{Reaction 3:} \quad 2~\mathrm{C_2H_2(g)} + 5~\mathrm{O_2(g)} \longrightarrow 4~\mathrm{CO_2(g)} + 2~\mathrm{H_2O(l)} \quad\!\!\!\, \Delta_{\mathrm{r}} H^{\circ} = -2598~\mathrm{kJ} \end{align*}\]

Recall the definition for a enthalpy of formation reaction. One mole of a product is formed from its substances in their standard states.

Transform the reactions.

- Leave Reaction 1 alone
- Multiply Reaction 2 by 2
- Reverse Reaction 3 and divide by 2

Write out the reaction transformations and state function transformations and then combine.

\[\begin{align*} &\textrm{Reaction 1:} \quad \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2}(g) \longrightarrow \mathrm{H_2O(l)} \qquad\qquad\qquad\, \qquad\,\Delta_{\mathrm{r}} H^{\circ} = -285.8~\mathrm{kJ} \\[1ex] &\textrm{Reaction 2:} \quad 2~\mathrm{C(s,graphite)} + 2~\mathrm{O_2(g)} \longrightarrow 2~\mathrm{CO_2(g)} \qquad\qquad\, \!\!\!\! \Delta_{\mathrm{r}} H^{\circ} = -787.0~\mathrm{kJ} \\[1ex] &\underline{\textrm{Reaction 3:} \quad 2~\mathrm{CO_2(g)} + \mathrm{H_2O(l)} \longrightarrow \mathrm{C_2H_2(g)} + \dfrac{5}{2}~\mathrm{O_2(g)} \quad\,\,\:\:\!\:\Delta_{\mathrm{r}} H^{\circ} = 1299~\mathrm{kJ} \qquad} \\[1ex] &\textrm{Overall:} \quad\,\:\quad 2~\mathrm{C(s, graphite)} + \mathrm{H_2(g)} \longrightarrow \mathrm{C_2H_2(g)} \qquad\quad\, \quad \:\,\Delta_{\mathrm{f}}H^{\circ} = 226.\bar{1}7~\mathrm{kJ} \end{align*}\]