Chapter Notes

THERMODYNAMICS

Thermodynamics is the branch of science that studies the transformations of energy from one form to another. While chemical energy can be released as heat (like when burning fuel) or converted into mechanical or electrical energy, thermodynamics helps us understand and quantify these changes.

It focuses on macroscopic systems (those with a large number of molecules) and is concerned with the initial and final states of a system, not the rate or mechanism of the change. The laws of thermodynamics are powerful because they apply to systems in equilibrium or moving between equilibrium states.

Through thermodynamics, we can answer fundamental questions about chemical processes:

• How can we measure the energy changes in a reaction?
• Will a particular reaction happen on its own (is it spontaneous)?
• What is the driving force behind a chemical reaction?
• How far will a reaction proceed before stopping?

THERMODYNAMIC TERMS

To study energy changes, we first need to define a few key terms.

The System and the Surroundings

In thermodynamics, the universe is divided into two parts: the system and the surroundings.

• System: The specific part of the universe we are observing or studying. This could be a chemical reaction in a beaker, a gas in a cylinder, or even a single biological cell.
• Surroundings: Everything else in the universe that is not the system.

Together, they make up the universe: Universe = System + Surroundings

In practice, the surroundings are just the part of the universe that can interact with the system. For a reaction in a beaker, the room it's in is the surroundings. The boundary is the real or imaginary wall that separates the system from its surroundings. This boundary controls how matter and energy are exchanged.

Types of the System

Systems are classified based on how they exchange energy and matter with their surroundings.

1. Open System: Can exchange both energy and matter with the surroundings. [!example] Reactants in an open beaker are an open system. Heat can enter or leave, and matter (like water vapor) can escape into the air.

2. Closed System: Can exchange energy but not matter with the surroundings. [!example] Reactants in a sealed container made of a conducting material like copper is a closed system. Heat can pass through the walls, but the chemicals inside cannot escape.

3. Isolated System: Cannot exchange either energy or matter with the surroundings. [!example] Reactants in a sealed, insulated container like a thermos flask are an example of an isolated system. The boundary prevents any heat or matter from getting in or out.

The State of the System

To describe a system, we need to specify its properties, such as pressure ($p$), volume ($V$), temperature ($T$), and the amount of substance ($n$). These measurable, macroscopic properties define the state of the system.

Variables like $p$, $V$, and $T$ are called state functions or state variables. The key feature of a state function is that its value depends only on the current state of the system, not on the path taken to reach that state.

The Internal Energy as a State Function

Every system has a total energy, which is the sum of all forms of energy it contains (chemical, electrical, mechanical, etc.). In thermodynamics, we call this the internal energy, represented by the symbol U.

Internal energy is a state function. We cannot measure the absolute value of a system's internal energy, but we can measure the change in internal energy, ΔU.

The internal energy of a system can change in three ways:

• Heat passes into or out of the system.
• Work is done on or by the system.
• Matter enters or leaves the system.

Let's look at how heat and work affect internal energy.

Work

Work is a way to transfer energy. An adiabatic process is one where no heat is transferred between the system and surroundings ($q=0$). The system is enclosed by an adiabatic wall (a perfect insulator).

In the 1840s, James Joule showed that for an adiabatic system, a specific amount of work done on the system produces the same change in state (measured by temperature change), regardless of how the work was performed (e.g., mechanical stirring vs. electrical work).

This led to a key conclusion: the change in internal energy in an adiabatic process is equal to the adiabatic work done. $\Delta U = U_2 - U_1 = w_{ad}$ Since $\Delta U$ depends only on the initial and final states, internal energy ($U$) is a state function.

Sign Convention for Work (w)

• w is positive (+): Work is done on the system. This increases the system's internal energy.
• w is negative (-): Work is done by the system. This decreases the system's internal energy.

Heat

Heat is the transfer of energy due to a temperature difference. We use the symbol q for heat. Unlike work and internal energy, heat is not a state function; it depends on the path of the process.

If a system does no work, any change in its internal energy is due to heat transfer. $\Delta U = q \quad (\text{at constant volume, when no work is done})$

Sign Convention for Heat (q)

• q is positive (+): Heat is transferred to the system from the surroundings. This increases the system's internal energy.
• q is negative (-): Heat is transferred from the system to the surroundings. This decreases the system's internal energy.

The General Case and the First Law of Thermodynamics

In most cases, a change in internal energy involves both heat and work. The relationship between them is given by the First Law of Thermodynamics.

The mathematical statement of the First Law is: $\Delta U = q + w$ This equation means that the change in a system's internal energy is the sum of the heat added to it and the work done on it.

While $q$ and $w$ are path-dependent, their sum, $\Delta U$, is path-independent because it is a state function.

For an isolated system, there is no exchange of heat or work with the surroundings, so $q=0$ and $w=0$. Therefore, for an isolated system, $\Delta U = 0$.

This leads to the formal statement of the First Law of Thermodynamics: The energy of an isolated system is constant.

This is also known as the law of conservation of energy: energy can neither be created nor destroyed, only transformed from one form to another.

Solution

(i) Since no heat is absorbed, $q=0$. Work is done on the system, so $w$ is positive. The change in internal energy is $\Delta U = w_{ad}$. The wall must be adiabatic.

(ii) Since no work is done, $w=0$. Heat is taken out of the system, so $q$ is negative. The change in internal energy is $\Delta U = -q$. The wall must be thermally conducting.

(iii) Work is done by the system, so $w$ is negative. Heat is supplied to the system, so $q$ is positive. The change in internal energy is $\Delta U = q - w$. Since energy is exchanged but not matter, this would be a closed system.

APPLICATIONS

Work

In chemistry, we often deal with pressure-volume work, which is the work done by or on a gas during expansion or compression.

Consider a gas in a cylinder with a frictionless piston.

• Initial volume = $V_i$
• Final volume = $V_f$
• Constant external pressure = $p_{ex}$

If the gas is compressed, the piston moves inward. The work done on the system is given by the formula: $w = -p_{ex} \Delta V = -p_{ex} (V_f - V_i)$

• For compression, $V_f < V_i$, so $\Delta V$ is negative. This makes $w$ positive, which matches our sign convention (work done on the system is positive).
• For expansion, $V_f > V_i$, so $\Delta V$ is positive. This makes $w$ negative, which also matches our sign convention (work done by the system is negative).

Reversible and Irreversible Processes

• Irreversible Process: A process that cannot be reversed by an infinitesimal change. Most real-world processes, like a gas expanding against a constant external pressure, are irreversible.
• Reversible Process: A process that is carried out infinitely slowly through a series of equilibrium states, such that it can be reversed at any moment by an infinitesimal change. In this case, the external pressure ($p_{ex}$) is always just infinitesimally different from the internal pressure of the gas ($p_{in}$).

For a reversible, isothermal (constant temperature) process involving an ideal gas, the work done is given by: $w_{rev} = -\int_{V_i}^{V_f} p_{in} dV$ Using the ideal gas law, $p = \frac{nRT}{V}$, this becomes: $w_{rev} = -nRT \ln \frac{V_f}{V_i} = -2.303 nRT \log \frac{V_f}{V_i}$

Free Expansion is the expansion of a gas in a vacuum, where the external pressure ($p_{ex}$) is zero. In this case, no work is done. $w = -p_{ex} \Delta V = -0 \times \Delta V = 0$

Summary of First Law Equations for Different Processes

• General Process: $\Delta U = q + w$
• Process at Constant Volume: Since $\Delta V=0$, $w=0$. So, $\Delta U = q_V$. The subscript $V$ means heat is transferred at constant volume.
• Isothermal Irreversible Change: $\Delta U=0$ for an ideal gas. Thus, $q = -w = p_{ex}(V_f - V_i)$.
• Isothermal Reversible Change: $\Delta U=0$ for an ideal gas. Thus, $q = -w = nRT \ln \frac{V_f}{V_i}$.
• Adiabatic Change: $q=0$. Thus, $\Delta U = w_{ad}$.

Given

• Expansion into a vacuum, so $p_{ex} = 0$
• Initial volume $V_i = 2$ L
• Final volume $V_f = 10$ L

To Find

• Heat absorbed, $q$
• Work done, $w$

Formula

$w = -p_{ex}(V_f - V_i)$ $\Delta U = q + w$ For an ideal gas expanding isothermally into a vacuum, $\Delta U = 0$.

Solution

First, calculate the work done. $w = -p_{ex}(V_f - V_i) = -0 \times (10 - 2) \text{ L} = 0$ Since the process is isothermal and it's an ideal gas, $\Delta U = 0$. Using the first law: $0 = q + w$ $q = -w = 0$

Final Answer No work is done, and no heat is absorbed.

Given

• $p_{ex} = 1$ atm
• Initial volume $V_i = 2$ L
• Final volume $V_f = 10$ L
• Volume change $\Delta V = (10 - 2) = 8$ L

To Find

• Heat absorbed, $q$
• Work done, $w$

Formula

$w = -p_{ex}(V_f - V_i)$ For an isothermal expansion of an ideal gas, $\Delta U = 0$, so $q = -w$.

Solution

Calculate the work done by the system. $w = -p_{ex}(V_f - V_i) = -(1 \text{ atm})(8 \text{ L}) = -8 \text{ litre-atm}$ Since the process is isothermal, $\Delta U = 0$. $q = -w = -(-8 \text{ litre-atm}) = 8 \text{ litre-atm}$

Final Answer The work done by the system is 8 litre-atm, and the heat absorbed is 8 litre-atm.

Given

• $n = 1$ mol
• $V_i = 2$ L
• $V_f = 10$ L
• $T = 25^\circ\text{C} = 298$ K
• $R = 0.08206 \text{ L atm mol}^{-1} \text{ K}^{-1}$ (using L-atm units from previous problem)

To Find

• Heat absorbed, $q$
• Work done, $w$

Formula

$w_{rev} = -2.303 nRT \log \frac{V_f}{V_i}$ For an isothermal reversible expansion, $q = -w$.

Solution

Calculate the reversible work done. $w_{rev} = -2.303 \times (1) \times (0.08206) \times (298) \times \log \frac{10}{2}$ $w_{rev} = -2.303 \times 0.08206 \times 298 \times \log(5)$ $w_{rev} = -2.303 \times 0.08206 \times 298 \times 0.6990$ $w_{rev} = -39.366 \text{ L atm}$ Now, find the heat absorbed. $q = -w_{rev} = -(-39.366 \text{ L atm}) = 39.366 \text{ L atm}$ Note: The source text has a calculation error. It uses R=0.8206 instead of 0.08206, leading to an answer of 393.66. The correct value with R=0.08206 is 39.366 L atm.

Final Answer The work done by the system is 39.366 L atm, and the heat absorbed is 39.366 L atm.

Enthalpy, H

A Useful New State Function

Most chemical reactions are carried out at constant atmospheric pressure, not constant volume. At constant pressure, if a gas is produced, the system does work on the surroundings ($w = -p\Delta V$). The heat transferred is denoted as $q_p$.

The first law is: $\Delta U = q_p + w = q_p - p\Delta V$. Rearranging this gives: $q_p = \Delta U + p\Delta V$.

To make this simpler, we define a new state function called enthalpy (H). $H = U + pV$ The change in enthalpy, $\Delta H$, is: $\Delta H = \Delta U + \Delta(pV)$ At constant pressure, this becomes: $\Delta H = \Delta U + p\Delta V$ Comparing this with our rearranged first law equation, we see that: $\Delta H = q_p$ This is a very important result: The change in enthalpy is equal to the heat absorbed or released by the system at constant pressure.

• For an exothermic reaction (releases heat), $q_p$ is negative, so ΔH is negative.
• For an endothermic reaction (absorbs heat), $q_p$ is positive, so ΔH is positive.

Relationship between ΔH and ΔU for Reactions with Gases

For reactions involving only solids and liquids, volume changes are tiny, so $\Delta H \approx \Delta U$. However, for reactions involving gases, the difference can be significant.

Using the ideal gas law ($pV = nRT$), the term $p\Delta V$ can be written as: $p\Delta V = \Delta n_g RT$ where Δn_g is the change in the number of moles of gas in the reaction. $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ Substituting this into the enthalpy equation gives the useful relationship: $\Delta H = \Delta U + \Delta n_g RT$

Given

• Reaction: $H_2O(l) \rightarrow H_2O(g)$
• $\Delta H = 41.00 \text{ kJ mol}^{-1}$
• $n = 1$ mol
• $T = 100^\circ\text{C} = 373$ K
• $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$

To Find

• Internal energy change, $\Delta U$

Formula

$\Delta H = \Delta U + \Delta n_g RT$ $\Delta U = \Delta H - \Delta n_g RT$

Solution

First, find $\Delta n_g$. $\Delta n_g = (\text{moles of gas product}) - (\text{moles of gas reactant})$ $\Delta n_g = 1 - 0 = 1$ Now, substitute the values into the formula. Remember to convert R to kJ. $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} = 8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}$ $\Delta U = 41.00 \text{ kJ mol}^{-1} - (1) \times (8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}) \times (373 \text{ K})$ $\Delta U = 41.00 \text{ kJ mol}^{-1} - 3.099 \text{ kJ mol}^{-1}$ $\Delta U = 37.901 \text{ kJ mol}^{-1}$ Note: The source text uses R=8.3 J, leading to a slightly different result. Using the more precise R=8.314 J is standard practice.

Final Answer The internal energy change is $37.901 \text{ kJ mol}^{-1}$.

Extensive and Intensive Properties

• An extensive property depends on the amount of matter in the system. Examples include mass, volume, internal energy (U), and enthalpy (H). If you double the system, these properties double.
• An intensive property does not depend on the amount of matter. Examples include temperature, pressure, and density. If you divide a system in half, these properties remain the same in each part.

Heat Capacity

Heat capacity (C) is the amount of heat required to raise the temperature of a substance by one degree Celsius (or one Kelvin). $q = C \Delta T$ Heat capacity is an extensive property. We often use two related intensive properties:

• Molar heat capacity ($C_m$): Heat capacity per mole of a substance.
• Specific heat capacity (c): Heat capacity per unit mass of a substance. ($q = c \times m \times \Delta T$)

For gases, the heat capacity depends on the conditions:

• $C_V$: Heat capacity at constant volume.
• $C_p$: Heat capacity at constant pressure.

At constant volume, $q_V = \Delta U = C_V \Delta T$. At constant pressure, $q_p = \Delta H = C_p \Delta T$.

For one mole of an ideal gas, the relationship between the two is: $C_p - C_V = R$ where R is the ideal gas constant.

MEASUREMENT OF ΔU AND ΔH: CALORIMETRY

Calorimetry is the experimental technique used to measure energy changes as heat in chemical or physical processes. The apparatus used is called a calorimeter.

ΔU Measurements

To measure the heat change at constant volume ($q_V$), which equals $\Delta U$, we use a bomb calorimeter.

• A strong steel vessel (the "bomb") contains the sample.
• The bomb is filled with pure oxygen to ensure complete combustion.
• It is then submerged in a known volume of water in an insulated container.
• The reaction is initiated (e.g., by an electrical spark).
• The heat released by the reaction is absorbed by the water and the calorimeter, causing the temperature to rise.
• Since the volume of the sealed bomb is constant, no p-V work is done ($\Delta V=0$).
• The heat evolved is calculated using the known heat capacity of the entire calorimeter system ($C_{cal}$) and the measured temperature change ($\Delta T$). $q_V = -C_{cal} \times \Delta T = \Delta U$ The negative sign indicates that the heat lost by the reaction is gained by the calorimeter.

ΔH Measurements

To measure the heat change at constant pressure ($q_p$), which equals $\Delta H$, a simpler calorimeter can be used, often open to the atmosphere. A styrofoam coffee cup is a common example for introductory labs.

• The reactants (often in solution) are mixed inside the insulated cup.
• The temperature change of the solution is measured.
• Since the process occurs at constant atmospheric pressure, the heat absorbed or evolved is $q_p$. $q_p = \Delta H$ The heat change is often called the enthalpy of reaction or heat of reaction, denoted Δ_r H.

Given

• Mass of graphite = 1 g
• Initial temperature = 298 K
• Final temperature = 299 K
• Heat capacity of calorimeter, $C_V = 20.7 \text{ kJ/K}$

To Find

• Enthalpy change, $\Delta H$ for the reaction per mole of graphite.

Formula

$q_{calorimeter} = C_V \times \Delta T$ $q_{reaction} = -q_{calorimeter}$ At constant volume, $\Delta U = q_{reaction}$ $\Delta H = \Delta U + \Delta n_g RT$

Solution

1. Calculate heat absorbed by the calorimeter: $\Delta T = 299 \text{ K} - 298 \text{ K} = 1 \text{ K}$ $q_{calorimeter} = 20.7 \text{ kJ/K} \times 1 \text{ K} = 20.7 \text{ kJ}$

2. Calculate heat released by the reaction (for 1 g of graphite): $q_{reaction} = -20.7 \text{ kJ}$ Since this is at constant volume, $\Delta U$ for the combustion of 1 g of graphite is $-20.7$ kJ.

3. Calculate ΔU for 1 mole of graphite: Molar mass of Carbon (graphite) = $12.0 \text{ g/mol}$. $\Delta U_{molar} = \frac{-20.7 \text{ kJ}}{1 \text{ g}} \times \frac{12.0 \text{ g}}{1 \text{ mol}} = -248.4 \text{ kJ/mol}$

4. Calculate ΔH from ΔU: The reaction is $C(s) + O_2(g) \rightarrow CO_2(g)$. $\Delta n_g = (\text{moles of gas product}) - (\text{moles of gas reactant})$ $\Delta n_g = 1 - 1 = 0$ Since $\Delta n_g = 0$, the term $\Delta n_g RT$ is zero. $\Delta H = \Delta U + 0 = \Delta U$ Therefore, $\Delta H = -248.4 \text{ kJ/mol}$.

Final Answer The enthalpy change for the reaction is $-248.4 \text{ kJ mol}^{-1}$.

ENTHALPY CHANGE, Δ_r H OF A REACTION - REACTION ENTHALPY

The enthalpy change of a reaction (Δ_r H) is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. $\Delta_r H = \sum (\text{enthalpies of products}) - \sum (\text{enthalpies of reactants})$ For a general reaction: $b_1 R_1 + b_2 R_2 \rightarrow a_1 P_1 + a_2 P_2$ $\Delta_r H = \sum_i a_i H_{products} - \sum_i b_i H_{reactants}$ Here, $a_i$ and $b_i$ are the stoichiometric coefficients from the balanced equation.

Standard Enthalpy of Reactions

Since enthalpy depends on conditions like temperature and pressure, we define a set of standard conditions to compare reactions.

• The standard state of a substance is its pure form at a pressure of 1 bar and a specified temperature (usually 298 K or $25^\circ\text{C}$).
• The standard enthalpy of reaction ($\Delta H^\ominus$) is the enthalpy change when all reactants and products are in their standard states. The superscript $\ominus$ denotes standard conditions.

Enthalpy Changes during Phase Transformations

Changing the phase of a substance involves an enthalpy change at constant temperature and pressure.

• Standard Enthalpy of Fusion ($\Delta_{fus}H^\ominus$): The enthalpy change when one mole of a solid melts into a liquid at its melting point and 1 bar pressure. It is always positive (endothermic).

  • For ice: $H_2O(s) \rightarrow H_2O(l) \quad \Delta_{fus}H^\ominus = 6.00 \text{ kJ mol}^{-1}$

• Standard Enthalpy of Vaporization ($\Delta_{vap}H^\ominus$): The enthalpy change when one mole of a liquid vaporizes into a gas at its boiling point and 1 bar pressure. It is also always positive (endothermic).

  • For water: $H_2O(l) \rightarrow H_2O(g) \quad \Delta_{vap}H^\ominus = 40.79 \text{ kJ mol}^{-1}$

• Standard Enthalpy of Sublimation ($\Delta_{sub}H^\ominus$): The enthalpy change when one mole of a solid turns directly into a gas at a constant temperature and 1 bar pressure.

  • For dry ice: $CO_2(s) \rightarrow CO_2(g) \quad \Delta_{sub}H^\ominus = 25.2 \text{ kJ mol}^{-1}$

Standard Enthalpy of Formation

The standard molar enthalpy of formation ($\Delta_f H^\ominus$) is the enthalpy change when one mole of a compound is formed from its constituent elements, with all substances in their standard states.

• The reference state of an element is its most stable form at 298 K and 1 bar (e.g., $H_2$ gas, $O_2$ gas, C as graphite, $S$ as rhombic sulfur).

By convention, the standard enthalpy of formation of any element in its reference state is defined as zero.

Using standard enthalpies of formation, we can calculate the standard enthalpy of any reaction: $\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants})$

Thermochemical Equations

A thermochemical equation is a balanced chemical equation that includes the enthalpy change ($\Delta_r H$). Example: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \quad \Delta_r H^\ominus = -1367 \text{ kJ mol}^{-1}$

Conventions for Thermochemical Equations:

1. Coefficients refer to the number of moles of substances.
2. The value of $\Delta_r H^\ominus$ corresponds to the reaction as written (per mole of reaction).
3. If you reverse a reaction, you must change the sign of $\Delta_r H^\ominus$.

Hess's Law of Constant Heat Summation

Because enthalpy is a state function, the total enthalpy change for a reaction is the same whether the reaction occurs in one step or in a series of steps.

Hess's Law: If a reaction can be expressed as the sum of two or more other reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.

This law is incredibly useful for calculating the enthalpy change of reactions that are difficult to measure directly.

We can use two reactions that are easy to measure: (i) $C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \quad \Delta_r H^\ominus = -393.5 \text{ kJ mol}^{-1}$ (ii) $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g) \quad \Delta_r H^\ominus = -283.0 \text{ kJ mol}^{-1}$

To get our target equation, we need $CO(g)$ as a product. So, we reverse equation (ii) and change the sign of its $\Delta H$: (iii) $CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \quad \Delta_r H^\ominus = +283.0 \text{ kJ mol}^{-1}$

Now, we add equation (i) and (iii): $C(\text{graphite}) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2}O_2(g)$

Canceling the species that appear on both sides ($CO_2$ and $\frac{1}{2}O_2$), we get our target equation: $C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

The enthalpy change is the sum of the enthalpies of the steps: $\Delta_r H^\ominus = (-393.5 \text{ kJ mol}^{-1}) + (+283.0 \text{ kJ mol}^{-1}) = -110.5 \text{ kJ mol}^{-1}$

Final Answer The standard enthalpy of formation of CO(g) is $-110.5 \text{ kJ mol}^{-1}$.

ENTHALPIES FOR DIFFERENT TYPES OF REACTIONS

Standard Enthalpy of Combustion ($\Delta_c H^\ominus$)

This is the enthalpy change when one mole of a substance undergoes complete combustion, with all reactants and products in their standard states at a specified temperature. Combustion reactions are always exothermic, so $\Delta_c H^\ominus$ is always negative.

Example for butane ($C_4H_{10}$): $C_4H_{10}(g) + \frac{13}{2}O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l) \quad \Delta_c H^\ominus = -2658.0 \text{ kJ mol}^{-1}$

Enthalpy of Atomization ($\Delta_a H^\ominus$)

This is the enthalpy change that occurs when one mole of bonds in a substance is broken completely to obtain atoms in the gas phase.

• For diatomic molecules, it is the same as the bond dissociation enthalpy. $H_2(g) \rightarrow 2H(g) \quad \Delta_a H^\ominus = 435.0 \text{ kJ mol}^{-1}$
• For elements in their solid state, it is the same as the enthalpy of sublimation. $Na(s) \rightarrow Na(g) \quad \Delta_a H^\ominus = 108.4 \text{ kJ mol}^{-1}$

Bond Enthalpy ($\Delta_{bond} H^\ominus$)

Chemical reactions involve breaking old bonds and forming new ones.

• Bond breaking requires energy (endothermic).
• Bond formation releases energy (exothermic).

1. Bond Dissociation Enthalpy: The enthalpy change required to break one mole of a specific type of covalent bond in a gaseous compound. This is precise for diatomic molecules like $H_2$ or $Cl_2$.
2. Mean Bond Enthalpy: For polyatomic molecules (like $CH_4$), the energy required to break each successive bond is different. We therefore use an average value, called the mean bond enthalpy. For example, the mean C-H bond enthalpy is about $416 \text{ kJ mol}^{-1}$.

We can estimate the enthalpy of a gas-phase reaction using mean bond enthalpies: $\Delta_r H^\ominus = \sum (\text{bond enthalpies of reactants}) - \sum (\text{bond enthalpies of products})$ [!note] This formula is reactants minus products because we are calculating the energy input to break reactant bonds minus the energy released when product bonds form.

Lattice Enthalpy ($\Delta_{lattice} H^\ominus$)

Lattice enthalpy is the enthalpy change when one mole of an ionic compound dissociates into its constituent ions in the gaseous state. It is a measure of the strength of the ionic bond. $Na^+Cl^-(s) \rightarrow Na^+(g) + Cl^-(g) \quad \Delta_{lattice} H^\ominus = +788 \text{ kJ mol}^{-1}$ Lattice enthalpy is always positive and cannot be measured directly. It is calculated indirectly using a Born-Haber Cycle, which is an application of Hess's Law. The cycle relates the lattice enthalpy to other measurable quantities like the enthalpy of formation, ionization enthalpy, electron gain enthalpy, and sublimation/atomization enthalpies.

Enthalpy of Solution ($\Delta_{sol} H^\ominus$)

This is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent. When an ionic compound dissolves, two processes occur:

1. The crystal lattice breaks apart (requires energy = $\Delta_{lattice} H^\ominus$).
2. The ions are solvated (hydrated by water), which releases energy ($\Delta_{hyd} H^\ominus$).

The overall enthalpy of solution is the sum of these two terms: $\Delta_{sol} H^\ominus = \Delta_{lattice} H^\ominus + \Delta_{hyd} H^\ominus$ If $\Delta_{sol} H^\ominus$ is positive (endothermic), the solubility of the salt usually increases with temperature.

Enthalpy of Dilution

This is the enthalpy change that occurs when more solvent is added to a solution. It is the heat change associated with diluting a solution from one concentration to another.

SPONTANEITY

The first law tells us about energy conservation, but it doesn't tell us if a process will happen on its own. A spontaneous process is a process that has the potential to occur without the assistance of an external agency.

• Spontaneity does not mean the process is fast. Rusting is spontaneous but slow.
• Spontaneous processes are irreversible and only proceed in one direction on their own. Water flows downhill spontaneously, but never uphill.

Is Decrease in Enthalpy a Criterion for Spontaneity?

Many spontaneous processes are exothermic ($\Delta H < 0$), releasing energy and becoming more stable. This suggests that a decrease in enthalpy might be the driving force. For example, the combustion of fuel is both spontaneous and highly exothermic.

However, this is not always true. Some endothermic processes ($\Delta H > 0$) are also spontaneous. [!example] The melting of ice above $0^\circ\text{C}$ is spontaneous, yet it is an endothermic process that requires heat from the surroundings. The dissolution of some salts in water is also spontaneous and endothermic.

Therefore, a decrease in enthalpy is a contributing factor but not the only criterion for spontaneity.

Entropy and Spontaneity

There must be another driving force. Consider two different gases in a container separated by a partition. When the partition is removed, the gases mix spontaneously. This process involves a negligible change in enthalpy ($\Delta H \approx 0$).

What has changed? The system has become more disordered. The arrangement of gas molecules is now more random and chaotic.

This leads to the concept of entropy (S), a thermodynamic state function that is a measure of the degree of randomness or disorder in a system.

• Higher disorder = Higher entropy.
• The gaseous state has the highest entropy, followed by the liquid state, and the crystalline solid state has the lowest entropy.

The driving force for spontaneous processes appears to be a tendency toward maximum disorder. For a spontaneous process in an isolated system, the entropy always increases.

The change in entropy ($\Delta S$) can be related to the heat transferred in a reversible process: $\Delta S = \frac{q_{rev}}{T}$ where T is the absolute temperature.

The Second Law of Thermodynamics provides the true criterion for spontaneity. For any spontaneous process, the total entropy of the system and its surroundings must increase. $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0$

• If $\Delta S_{total} > 0$, the process is spontaneous.
• If $\Delta S_{total} < 0$, the process is non-spontaneous (the reverse process is spontaneous).
• If $\Delta S_{total} = 0$, the system is at equilibrium.

Given

• $\Delta S_{sys} = -549.4 \text{ J K}^{-1} \text{ mol}^{-1}$
• $\Delta H_{sys} = -1648 \times 10^3 \text{ J mol}^{-1}$
• $T = 298$ K

To Find

• Why the reaction is spontaneous. This means we need to show that $\Delta S_{total} > 0$.

Formula

$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$ $\Delta S_{surr} = \frac{- \Delta H_{sys}}{T}$

Solution

The reaction is exothermic, so it releases heat into the surroundings. The heat gained by the surroundings is $-\Delta H_{sys}$.

1. Calculate the entropy change of the surroundings: $\Delta S_{surr} = \frac{-(-1648 \times 10^3 \text{ J mol}^{-1})}{298 \text{ K}}$ $\Delta S_{surr} = +5530 \text{ J K}^{-1} \text{ mol}^{-1}$

2. Calculate the total entropy change: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$ $\Delta S_{total} = (-549.4 \text{ J K}^{-1} \text{ mol}^{-1}) + (5530 \text{ J K}^{-1} \text{ mol}^{-1})$ $\Delta S_{total} = +4980.6 \text{ J K}^{-1} \text{ mol}^{-1}$

Since $\Delta S_{total}$ is positive, the reaction is spontaneous. The large increase in the entropy of the surroundings (due to the release of heat) outweighs the decrease in the entropy of the system.

Final Answer The reaction is spontaneous because the total entropy change of the system and surroundings is positive.

Gibbs Energy and Spontaneity

Calculating $\Delta S_{surroundings}$ for every reaction can be inconvenient. To focus only on the system, we define a new state function called Gibbs energy (G), also known as free energy.

Gibbs energy combines enthalpy and entropy into a single value: $G = H - TS$ The change in Gibbs energy at constant temperature is given by the Gibbs equation: $\Delta G = \Delta H - T\Delta S$ The sign of $\Delta G$ for the system provides a direct criterion for spontaneity at constant temperature and pressure:

• If ΔG is negative (< 0), the process is spontaneous.
• If ΔG is positive (> 0), the process is non-spontaneous.
• If ΔG is zero (= 0), the system is at equilibrium.

$\Delta G$ represents the net energy available to do useful work. It is the "free energy" because it is the portion of the total energy change ($\Delta H$) that is not lost to increasing the disorder ($T\Delta S$).

The effect of temperature on spontaneity depends on the signs of $\Delta H$ and $\Delta S$:

Absolute Entropy and the Third Law of Thermodynamics

The Third Law of Thermodynamics states: The entropy of any pure, perfectly crystalline substance approaches zero as the temperature approaches absolute zero (0 K).

At 0 K, particles in a perfect crystal are in their lowest possible energy state and are perfectly ordered, meaning there is essentially zero disorder, so $S=0$. This law allows us to determine the absolute entropy of a substance at any temperature.

GIBBS ENERGY CHANGE AND EQUILIBRIUM

For a reversible reaction, a dynamic equilibrium is established where the forward and reverse reactions occur at the same rate. At equilibrium, the Gibbs energy of the system is at its minimum, and the overall change in Gibbs energy is zero. $\Delta_r G = 0 \quad (\text{at equilibrium})$ The standard Gibbs energy change ($\Delta_r G^\ominus$) is related to the equilibrium constant (K) of a reaction by the following important equation: $\Delta_r G^\ominus = -RT \ln K$ or $\Delta_r G^\ominus = -2.303 RT \log K$

This equation links thermodynamics ($\Delta_r G^\ominus$) with chemical equilibrium (K).

• If $\Delta_r G^\ominus$ is large and negative, K will be much larger than 1, meaning the reaction goes nearly to completion (products are favored).
• If $\Delta_r G^\ominus$ is large and positive, K will be much smaller than 1, meaning the reaction hardly proceeds at all (reactants are favored).
• If $\Delta_r G^\ominus$ is close to zero, K will be close to 1, meaning significant amounts of both reactants and products exist at equilibrium.

Given

• $K_p = 2.47 \times 10^{-29}$
• $T = 298$ K
• $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

To Find

• Standard Gibbs energy change, $\Delta_r G^\ominus$

Formula

$\Delta_r G^\ominus = -2.303 RT \log K_p$

Solution

Substitute the given values into the formula. $\Delta_r G^\ominus = -2.303 \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K}) \times (\log(2.47 \times 10^{-29}))$ $\log(2.47 \times 10^{-29}) = \log(2.47) + \log(10^{-29}) = 0.3927 - 29 = -28.6073$ $\Delta_r G^\ominus = -2.303 \times 8.314 \times 298 \times (-28.6073)$ $\Delta_r G^\ominus = +163229 \text{ J mol}^{-1}$ $\Delta_r G^\ominus = +163.2 \text{ kJ mol}^{-1}$

Final Answer The standard Gibbs energy change for the reaction is $+163 \text{ kJ mol}^{-1}$.

Given

• $\Delta_r G^\ominus = -13.6 \text{ kJ mol}^{-1} = -13600 \text{ J mol}^{-1}$
• $T = 298$ K
• $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

To Find

• Equilibrium constant, K

Formula

$\Delta_r G^\ominus = -2.303 RT \log K$ $\log K = \frac{-\Delta_r G^\ominus}{2.303 RT}$

Solution

Substitute the values to find $\log K$. $\log K = \frac{-(-13600 \text{ J mol}^{-1})}{2.303 \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})}$ $\log K = \frac{13600}{5705.8} = 2.383$ Now, find K by taking the antilog. $K = \text{antilog}(2.383) = 10^{2.383} = 10^{0.383} \times 10^2$ $K = 2.415 \times 10^2$

Final Answer The equilibrium constant for the reaction is $2.4 \times 10^2$.