The energy stored within the chemical bonds of molecules is a form of chemical energy. This energy can be released as heat during chemical reactions (e.g., combustion of fuels like methane or coal) or converted into other forms of energy, such as mechanical work (in engines) or electrical energy (in batteries).

Thermodynamics is the branch of science that studies these energy transformations and the relationships between different forms of energy. It focuses on macroscopic systems (dealing with large numbers of molecules) and is concerned with the initial and final states of a system undergoing a change, rather than the specific pathway or rate of the change.

The laws of thermodynamics apply to systems that are in equilibrium or moving between equilibrium states. Key questions addressed by thermodynamics include:

• How can we quantify the energy changes that occur during chemical reactions or physical processes?
• Can we predict whether a specific reaction or process will occur spontaneously?
• To what extent will a reaction proceed before reaching equilibrium?

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Thermodynamic Terms

To study energy changes in chemical systems, we need to define certain thermodynamic terms.

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The System And The Surroundings

• A system is the specific part of the universe that we are observing or studying.
• The surroundings are everything else in the universe outside the system.
• The system and the surroundings together constitute the universe.

The Universe = System + Surroundings

Practically, the surroundings are considered to be the portion of the universe near the system that can interact with it.

The boundary is the real or imaginary surface that separates the system from the surroundings. It allows for tracking the exchange of matter and energy.

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Types Of The System

Systems are classified based on whether they exchange matter and/or energy with their surroundings.

1. Open System: Exchanges both energy and matter with the surroundings. Example: Reactants in an open beaker.
2. Closed System: Exchanges energy but not matter with the surroundings. Example: Reactants in a sealed vessel made of a conducting material (like metal).
3. Isolated System: Exchanges neither energy nor matter with the surroundings. Example: Reactants in a sealed, insulated vessel (like a thermos flask).

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The State Of The System

The state of a thermodynamic system is defined by its macroscopic (measurable) properties. For a gas, its state can be described by specifying its pressure ($p$), volume ($V$), temperature ($T$), and amount (number of moles, $n$).

These properties (p, V, T, n, etc.) are called state variables or state functions. Their value depends only on the current state of the system, regardless of how that state was reached (i.e., they are path-independent). Once a minimum number of state variables are fixed, the others are automatically determined.

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The Internal Energy As A State Function

The total energy of a system, encompassing all forms of energy within it (kinetic energy of particles, potential energy due to intermolecular forces, etc.), is called its Internal Energy ($U$). We cannot measure the absolute value of internal energy, but we can measure the change in internal energy ($\Delta U$) when a system transitions from one state to another.

The internal energy of a system can change through the exchange of work ($w$) or heat ($q$) with the surroundings, or by the transfer of matter (in open systems).

Let's consider an adiabatic process, where no heat exchange occurs between the system and surroundings ($q=0$). This requires an adiabatic wall (a boundary that prevents heat flow).

If we do work on an adiabatic system (e.g., stirring water in a thermos flask), its temperature increases. Experiments by J.P. Joule showed that the same amount of work done on the system, regardless of how it was done, produces the same change in temperature and thus the same change in internal energy. The change in internal energy in an adiabatic process ($\Delta U$) is equal to the adiabatic work ($w_{ad}$):

$\Delta U = w_{ad}$ (for an adiabatic change, $q=0$)

This demonstrates that internal energy ($U$) is a state function, as the change $\Delta U$ depends only on the initial and final states, not the process (path) by which work was done.

Conventionally, work done on the system is positive ($w > 0$), increasing the internal energy ($\Delta U > 0$). Work done by the system is negative ($w < 0$), decreasing internal energy ($\Delta U < 0$).

The internal energy can also change by the transfer of heat ($q$) through thermally conducting walls (instead of adiabatic walls) without work being done ($w=0$).

If heat is absorbed by the system at constant volume ($w=0$), its internal energy increases. The change in internal energy is equal to the heat absorbed ($q_V$):

$\Delta U = q_V$ (for a process at constant volume, $w=0$)

Conventionally, heat transferred to the system is positive ($q > 0$), increasing internal energy. Heat transferred from the system is negative ($q < 0$), decreasing internal energy.

(c) The general case - First Law of Thermodynamics

In the most general case, a change in the system's state can involve both work ($w$) and heat ($q$) exchange with the surroundings. The change in internal energy is the sum of the heat absorbed by the system and the work done on the system:

$\Delta U = q + w$

This is the mathematical statement of the First Law of Thermodynamics. It states that the energy of an isolated system is constant, reflecting the law of conservation of energy: energy cannot be created or destroyed, only transformed from one form to another.

For a given change of state, the individual values of $q$ and $w$ depend on the path taken, but their sum ($q+w$) is always equal to $\Delta U$, which is path-independent because $U$ is a state function.

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Applications

Understanding how energy changes occur as heat or work allows us to quantify these changes in chemical reactions and processes.

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Work

In thermodynamics, we often consider pressure-volume (PV) work, which is work done by or on a system due to a change in its volume against an external pressure. Consider a gas in a cylinder with a frictionless piston.

If the gas expands against a constant external pressure ($p_{ex}$), the system does work on the surroundings. If the gas is compressed by a constant external pressure, the surroundings do work on the system.

Let the initial volume be $V_i$ and the final volume be $V_f$. The change in volume is $\Delta V = V_f - V_i$. If the piston moves a distance $l$ and has area $A$, $\Delta V = A \times l$. The force exerted by the external pressure is $F = p_{ex} \times A$.

Work ($w$) is force times distance. Work done on the system by compression is:

$w = F \times (-l)$ (distance moved is against the direction of force exerted by the system)

$w = (p_{ex} \times A) \times (-l)$

$w = -p_{ex} \times (A \times l) = -p_{ex} \Delta V = -p_{ex}(V_f - V_i)$

The negative sign is included so that $w$ has the correct thermodynamic sign convention: if the gas is compressed, $V_f < V_i$, $\Delta V$ is negative, and $w = -p_{ex}(-\Delta V) = +p_{ex}|\Delta V|$ is positive (work done on the system). If the gas expands, $V_f > V_i$, $\Delta V$ is positive, and $w = -p_{ex}(+\Delta V) = -p_{ex}|\Delta V|$ is negative (work done by the system).

$w = -p_{ex}\Delta V$ (for work done against constant external pressure)

If the external pressure is not constant but changes in finite steps, the total work is the sum of work done in each step: $w = -\sum p_{ex} \Delta V$. Graphically, this is the area under the $p_{ex}$ vs $V$ curve.

In a reversible process, the change occurs in infinite steps, and the system is always in near-equilibrium with the surroundings. The external pressure ($p_{ex}$) is always infinitesimally close to the internal pressure of the gas ($p_{in}$). For an expansion, $p_{ex} = p_{in} - dp$; for compression, $p_{ex} = p_{in} + dp$. In the limit of infinitesimal change ($dV$), $dp \rightarrow 0$, so $p_{ex} \approx p_{in}$. Work done in a reversible process is calculated by integrating the pressure-volume work:

$w_{rev} = -\int_{V_i}^{V_f} p_{ex} dV$

Under reversible conditions, $p_{ex} = p_{in}$ (let's call it $p$), so:

$w_{rev} = -\int_{V_i}^{V_f} p \, dV$

For an ideal gas undergoing an isothermal (constant temperature, $T$) reversible process, $p = nRT/V$. Substituting this into the integral:

$w_{rev} = -\int_{V_i}^{V_f} \frac{nRT}{V} \, dV$

Since $n$ and $T$ are constant for an isothermal process:

$w_{rev} = -nRT \int_{V_i}^{V_f} \frac{1}{V} \, dV = -nRT [\ln V]_{V_i}^{V_f} = -nRT (\ln V_f - \ln V_i)$

$w_{rev} = -nRT \ln \frac{V_f}{V_i} = -2.303 nRT \log \frac{V_f}{V_i}$

Free expansion: Expansion against zero external pressure ($p_{ex}=0$, expansion into vacuum). In this case, $w = -p_{ex}\Delta V = 0$. No work is done during free expansion, regardless of whether the process is reversible or irreversible (as there is no opposing pressure). For an ideal gas, internal energy depends only on temperature ($\Delta U = 0$ for isothermal process). If an ideal gas undergoes free expansion (adiabatic, $q=0$), $\Delta U = q+w = 0+0=0$. If it undergoes isothermal free expansion, $\Delta U=0$ (because it's isothermal), and $w=0$ (because it's free expansion), so $q=\Delta U - w = 0 - 0 = 0$. For an ideal gas undergoing isothermal expansion into vacuum, $\Delta U=0$, $w=0$, and $q=0$.

Summarising $\Delta U = q+w$ for different processes involving ideal gas:

• Isothermal irreversible: $\Delta U = 0$, so $q = -w = p_{ex}(V_f - V_i)$.
• Isothermal reversible: $\Delta U = 0$, so $q = -w = nRT \ln(V_f/V_i)$.
• Adiabatic change: $q = 0$, so $\Delta U = w_{ad}$.

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Enthalpy, H

While constant volume conditions ($V=constant, \Delta V=0, w=0, \Delta U=q_V$) are used in bomb calorimeters, most chemical reactions occur at constant pressure (like atmospheric pressure). To quantify heat changes at constant pressure, we define a new state function called Enthalpy ($H$).

Enthalpy ($H$) is defined as the sum of the internal energy ($U$) and the pressure-volume product ($pV$).

$H = U + pV$

Since $U$, $p$, and $V$ are state functions, $H$ is also a state function; its value depends only on the state of the system.

Consider a change in state at constant pressure:

$\Delta H = \Delta U + \Delta(pV)$

At constant pressure ($p$ is constant):

$\Delta H = \Delta U + p\Delta V$

From the First Law, $\Delta U = q + w$. At constant pressure, work done is $w = -p_{ex}\Delta V$. If the external pressure is equal to the constant internal pressure ($p_{ex}=p$), then $w = -p\Delta V$.

Substituting $w = -p\Delta V$ into $\Delta U = q+w$ gives $\Delta U = q + (-p\Delta V)$, or $\Delta U = q - p\Delta V$.

Rearranging gives $q = \Delta U + p\Delta V$.

Comparing this with the definition of enthalpy change at constant pressure, $\Delta H = \Delta U + p\Delta V$, we see that:

$\Delta H = q_p$

The change in enthalpy ($\Delta H$) is equal to the heat absorbed by the system at constant pressure ($q_p$). Therefore, $\Delta H$ is also path-independent under constant pressure conditions. For exothermic reactions, heat is released by the system, $q_p$ is negative, so $\Delta H$ is negative. For endothermic reactions, heat is absorbed by the system, $q_p$ is positive, so $\Delta H$ is positive.

For reactions involving gases, the volume change ($\Delta V$) at constant temperature and pressure can be related to the change in the number of moles of gas ($\Delta n_g$).

Let the total volume of gaseous reactants be $V_A$ ($n_A$ moles) and the total volume of gaseous products be $V_B$ ($n_B$ moles). At constant $p$ and $T$, using the ideal gas law:

$pV_A = n_A RT$

$pV_B = n_B RT$}

$p(V_B - V_A) = (n_B - n_A) RT$

$p\Delta V = \Delta n_g RT$

where $\Delta n_g = (n_B - n_A) = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.

Substituting this into the enthalpy change equation $\Delta H = \Delta U + p\Delta V$ gives:

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

This equation is useful for converting between $\Delta H$ and $\Delta U$ for reactions involving gases.

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Heat Capacity

When heat is transferred to a system, its temperature usually rises. The amount of heat ($q$) required to change the temperature of a substance by a certain amount ($\Delta T$) is related by the heat capacity ($C$).

$q = C \Delta T$

Heat capacity depends on the amount of substance, its composition, and the process conditions (constant volume or constant pressure).

• Molar heat capacity ($C_m$ or $C$): Heat capacity for one mole of the substance ($C_m = C/n$). It is the heat required to raise the temperature of one mole by one degree Celsius (or Kelvin). Units: J mol$^{-1}$ K$^{-1}$.
• Specific heat capacity ($c$ or $s$): Heat capacity for one unit mass of the substance. It is the heat required to raise the temperature of one unit mass by one degree Celsius (or Kelvin). Units: J g$^{-1}$ K$^{-1}$.

Total heat ($q$) required for a sample of mass ($m$) with specific heat ($c$) is $q = m \times c \times \Delta T$. This is also equal to $C \Delta T$ if $C$ is the heat capacity for the entire sample.

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The Relationship Between Cp And Cv For An Ideal Gas

Heat capacity measured at constant volume is denoted as $C_V$. Heat capacity measured at constant pressure is denoted as $C_p$.

At constant volume: $\Delta U = q_V = C_V \Delta T$

At constant pressure: $\Delta H = q_p = C_p \Delta T$

For an ideal gas, the relationship between $\Delta H$ and $\Delta U$ is $\Delta H = \Delta U + \Delta n_g RT$. For a change involving one mole of ideal gas ($\Delta n_g = 1$ for processes like $A(s) \rightarrow B(g)$ where B is 1 mole, or considering $C_p$ and $C_V$ per mole), $\Delta H = \Delta U + R\Delta T$.

Substituting the expressions for $\Delta H$ and $\Delta U$ in terms of heat capacities:

$C_p \Delta T = C_V \Delta T + R \Delta T$

Dividing by $\Delta T$ (for a non-zero temperature change):

$C_p = C_V + R$

or

$C_p - C_V = R$

This relationship holds for one mole of an ideal gas, where $R$ is the universal gas constant. It shows that $C_p$ is greater than $C_V$ because at constant pressure, some energy is used to do expansion work against the surroundings as the temperature rises, whereas at constant volume, all the energy goes into increasing the internal energy (and thus temperature).

Measurement Of Du And Dh: Calorimetry

Calorimetry is the experimental technique used to measure the heat changes associated with chemical or physical processes. It involves carrying out the process in a device called a calorimeter, which is typically immersed in a liquid (like water) with a known heat capacity. By measuring the temperature change of the liquid and calorimeter, the heat evolved or absorbed can be determined.

Energy changes are usually measured under two specific conditions: constant volume or constant pressure.

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Du Measurements

Heat change at constant volume ($q_V$) is measured using a bomb calorimeter. This apparatus consists of a sealed, rigid steel vessel (the "bomb") where the reaction takes place, immersed in a water bath within an insulated container. The constant volume of the bomb ensures that no PV work is done ($\Delta V=0$, $w=0$).

For combustion reactions, a known amount of the substance is placed in the bomb with excess oxygen and ignited. The heat released by the reaction ($q_V$) is transferred to the water bath and the calorimeter itself, causing a temperature rise ($\Delta T$). The heat absorbed by the calorimeter ($q_{calorimeter}$) is calculated using the heat capacity of the calorimeter system ($C_{calorimeter}$) and the observed temperature change:

$q_{calorimeter} = C_{calorimeter} \Delta T$

Since the system (reaction) and the surroundings (calorimeter + water) are isolated, the heat released by the reaction is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter: $q_{reaction} = -q_{calorimeter}$.

As the reaction is carried out at constant volume, the heat measured ($q_V$) is equal to the change in internal energy of the reaction:

$\Delta_r U = q_V = -q_{calorimeter} = -C_{calorimeter} \Delta T$

The heat capacity of the calorimeter is usually determined by a separate calibration experiment (e.g., burning a substance with known heat of combustion or using electrical heating).

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Dh Measurements

Heat change at constant pressure ($q_p$) is typically measured using a calorimeter open to the atmosphere or maintained at a constant external pressure. This is simpler than a bomb calorimeter and is often done in a coffee-cup calorimeter or similar apparatus.

In this case, the reaction occurs at constant pressure (usually atmospheric pressure). The heat absorbed or evolved during the reaction ($q_p$) causes a temperature change ($\Delta T$) in the solution or system. Assuming the heat capacity of the system ($C_{system}$) is known, the heat absorbed or evolved is calculated:

$q_p = C_{system} \Delta T$

At constant pressure, the heat measured ($q_p$) is equal to the change in enthalpy of the reaction:

$\Delta_r H = q_p = C_{system} \Delta T$

For an exothermic reaction, heat is evolved ($q_p < 0$), so $\Delta_r H < 0$. For an endothermic reaction, heat is absorbed ($q_p > 0$), so $\Delta_r H > 0$.

The heat capacity of the system (e.g., solution) can be estimated from the specific heat capacity and mass of the components.

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Enthalpy Change, Drh Of A Reaction – Reaction Enthalpy

The enthalpy change accompanying a chemical reaction, where reactants are converted into products, is called the reaction enthalpy ($\Delta_r H$). It is defined as the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants, taking into account their stoichiometric coefficients in the balanced chemical equation.

$\Delta_r H = \sum a_i H_{products} - \sum b_i H_{reactants}$

where $a_i$ and $b_i$ are the stoichiometric coefficients of the products and reactants, respectively, and $H$ represents the molar enthalpy of each substance.

For an exothermic reaction, heat is released, and the products have lower enthalpy than the reactants, so $\Delta_r H$ is negative. For an endothermic reaction, heat is absorbed, and the products have higher enthalpy than the reactants, so $\Delta_r H$ is positive.

Reaction enthalpy is a crucial quantity for understanding the energy balance of chemical processes.

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Standard Enthalpy Of Reactions

To compare reaction enthalpies consistently, standard conditions are defined. The standard enthalpy of reaction ($\Delta_r H^0$) is the enthalpy change for a reaction when all reactants and products are in their standard states at a specified temperature (usually 298 K). The standard state of a substance is its most stable form at 1 bar pressure and the specified temperature. For example, the standard state of oxygen at 298 K is O$_2$(g) at 1 bar; for carbon, it's graphite C(graphite) at 1 bar.

Standard conditions are indicated by the superscript $^0$ (e.g., $\Delta H^0$).

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Enthalpy Changes During Phase Transformations

Changes in the physical state of a substance (phase transformations) also involve enthalpy changes, typically occurring at constant temperature and pressure.

• Standard Enthalpy of Fusion ($\Delta_{fus} H^0$): The enthalpy change when one mole of a solid melts into its liquid form at its melting point and at standard pressure (1 bar). Melting is endothermic ($\Delta_{fus} H^0 > 0$).
• Standard Enthalpy of Vaporization ($\Delta_{vap} H^0$): The enthalpy change when one mole of a liquid vaporizes into its gaseous form at its boiling point and at standard pressure (1 bar). Vaporization is endothermic ($\Delta_{vap} H^0 > 0$).
• Standard Enthalpy of Sublimation ($\Delta_{sub} H^0$): The enthalpy change when one mole of a solid sublimes directly into its gaseous form at a constant temperature and at standard pressure (1 bar). Sublimation is endothermic ($\Delta_{sub} H^0 > 0$).

For a substance, the enthalpy of sublimation is approximately the sum of the enthalpy of fusion and the enthalpy of vaporization at the same temperature: $\Delta_{sub} H^0 \approx \Delta_{fus} H^0 + \Delta_{vap} H^0$.

The magnitude of these phase transition enthalpies reflects the strength of intermolecular forces. Substances with stronger intermolecular forces (like water with hydrogen bonding) have higher $\Delta_{fus} H^0$ and $\Delta_{vap} H^0$ values compared to substances with weaker forces (like acetone with weaker dipole-dipole interactions).

Substance	$\Delta_{fus}H^0$ (kJ mol$^{-1}$)	$T_f$ (K)	$\Delta_{vap}H^0$ (kJ mol$^{-1}$)	$T_b$ (K)
H$_2$O	6.00	273.15	40.79	373.15
NaCl	28.8	1081	170.0	1665
C$_2$H$_5$OH	4.93	159.0	38.56	351.5
CCl$_4$	2.5	250.0	30.4	350.0

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Standard Enthalpy Of Formation

The standard molar enthalpy of formation ($\Delta_f H^0$) of a compound is the standard enthalpy change for the reaction in which one mole of the compound is formed from its constituent elements in their most stable states of aggregation (reference states) at a specified temperature (usually 298 K) and 1 bar pressure.

The reference state of an element is its pure form in its most stable physical state at 1 bar and the specified temperature (e.g., O$_2$(g) for oxygen, C(graphite, s) for carbon, Br$_2$(l) for bromine, H$_2$(g) for hydrogen).

By convention, the standard enthalpy of formation of an element in its reference state is zero ($\Delta_f H^0 = 0$).

Example formation reactions and their standard enthalpies of formation:

• Formation of liquid water: $\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(l)}$; $\Delta_f H^0 = -285.8$ kJ mol$^{-1}$
• Formation of methane gas: $\text{C(graphite, s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$; $\Delta_f H^0 = -74.81$ kJ mol$^{-1}$

Note that $\Delta_f H^0$ refers to the formation of exactly one mole of the product compound from elements in their standard states.

Standard enthalpies of formation are useful for calculating the standard enthalpy of any reaction ($\Delta_r H^0$). The general equation for this calculation is based on Hess's Law:

$\Delta_r H^0 = \sum n_p \Delta_f H^0 (\text{products}) - \sum n_r \Delta_f H^0 (\text{reactants})$

where $n_p$ and $n_r$ are the stoichiometric coefficients of the products and reactants in the balanced equation.

Substance	$\Delta_fH^0$ (kJ mol$^{-1}$)	Substance	$\Delta_fH^0$ (kJ mol$^{-1}$)
H$_2$O(l)	–285.83	CO(g)	–110.53
H$_2$O(g)	–241.82	CO$_2$(g)	–393.51
HF(g)	–273.3	CH$_4$(g)	–74.81
HCl(g)	–92.32	C$_2$H$_6$(g)	–84.68
HBr(g)	–36.40	C$_3$H$_8$(g)	–103.85
HI(g)	26.48	CH$_3$OH(l)	–238.86
NaCl(s)	–384.1	C$_2$H$_5$OH(l)	–277.69
KCl(s)	–436.7	C$_6$H$_6$(l)	49.04
KBr(s)	–393.8	Glucose(s)	–1260
KI(s)	–327.0	Al$_2$O$_3$(s)	–1675.7
AgCl(s)	–127.07	Fe$_2$O$_3$(s)	–824.2
AgBr(s)	–100.84	CaCO$_3$(s)	–1206.9
AgI(s)	–55.92	CaO(s)	–635.1
		SO$_2$(g)	–296.8
		SO$_3$(g)	–395.7

Example: Decomposition of CaCO$_3$(s) at standard conditions:

CaCO$_3$(s) $\rightarrow$ CaO(s) + CO$_2$(g); $\Delta_r H^0 = ?$

Using the values from Table 6.2:

$\Delta_r H^0 = [\Delta_f H^0(\text{CaO, s}) + \Delta_f H^0(\text{CO}_2\text{, g})] - [\Delta_f H^0(\text{CaCO}_3\text{, s})]$

$\Delta_r H^0 = [(-635.1 \text{ kJ mol}^{-1}) + (-393.5 \text{ kJ mol}^{-1})] - [-1206.9 \text{ kJ mol}^{-1}]$

$\Delta_r H^0 = (-1028.6 + 1206.9) \text{ kJ mol}^{-1} = 178.3 \text{ kJ mol}^{-1}$.

This positive value confirms that the decomposition of CaCO$_3$(s) is an endothermic reaction.

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Thermochemical Equations

A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products (including allotropic forms for elements) and the standard enthalpy change ($\Delta_r H^0$) for the reaction as written.

Example: $\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}$; $\Delta_r H^0 = -1367$ kJ mol$^{-1}$.

Conventions for thermochemical equations:

1. Coefficients refer to the number of moles of reactants and products.
2. The $\Delta_r H^0$ value corresponds to the reaction as written (per mole of reaction). The unit kJ mol$^{-1}$ refers to 'per mole of reaction', where one mole of reaction corresponds to the stoichiometric amounts shown. Doubling the coefficients doubles the $\Delta_r H^0$ value.
3. Reversing the equation reverses the sign of $\Delta_r H^0$.

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Hess’s Law Of Constant Heat Summation

Based on enthalpy being a state function, Hess's Law states that the total enthalpy change for a reaction is the same, whether the reaction is carried out in one step or in a series of steps. If a reaction can be expressed as the sum of several steps, the standard enthalpy change for the overall reaction is the sum of the standard enthalpy changes for the individual steps at the same temperature.

If Reaction (Overall): A $\rightarrow$ B with $\Delta_r H$

Can occur via steps: A $\rightarrow$ C ($\Delta_r H_1$), C $\rightarrow$ D ($\Delta_r H_2$), D $\rightarrow$ B ($\Delta_r H_3$)

Then, $\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3$.

Hess's law is extremely useful because it allows us to calculate enthalpy changes for reactions that are difficult or impossible to measure directly by manipulating thermochemical equations (reversing, multiplying by coefficients, adding them). For example, to find $\Delta_r H^0$ for C(graphite, s) + $\frac{1}{2}$O$_2$(g) $\rightarrow$ CO(g), which is hard to measure directly, we can use the known enthalpies of combustion of C and CO to CO$_2$ and combine them using Hess's Law.

(i) $\text{C(graphite, s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$; $\Delta_r H^0_1 = -393.5$ kJ mol$^{-1}$

(ii) $\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$; $\Delta_r H^0_2 = -283.0$ kJ mol$^{-1}$

Reverse equation (ii) to get CO as a product: $\text{CO}_2\text{(g)} \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)}$; $\Delta_r H^0_3 = +283.0$ kJ mol$^{-1}$

Add equation (i) and (iii):

$\text{C(graphite, s)} + \text{O}_2\text{(g)} + \text{CO}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)}$

Cancel CO$_2$(g) and $\frac{1}{2}$O$_2$(g) from both sides:

$\text{C(graphite, s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$

The enthalpy change for this reaction is $\Delta_r H^0 = \Delta_r H^0_1 + \Delta_r H^0_3 = (-393.5 \text{ kJ mol}^{-1}) + (283.0 \text{ kJ mol}^{-1}) = -110.5$ kJ mol$^{-1}$.

Enthalpies For Different Types Of Reactions

Specific names are given to enthalpy changes associated with particular types of chemical reactions or processes.

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Standard Enthalpy Of Combustion

The standard enthalpy of combustion ($\Delta_c H^0$) is the standard enthalpy change when one mole of a substance undergoes complete combustion (reacts with excess oxygen) at a specified temperature (usually 298 K) and 1 bar pressure. Combustion reactions are generally exothermic ($\Delta_c H^0 < 0$).

Example: Complete combustion of one mole of methane:

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$; $\Delta_c H^0 = -890.3$ kJ mol$^{-1}$.

Standard enthalpies of combustion are useful for calculating enthalpies of formation using Hess's Law.

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Enthalpy Of Atomization

The standard enthalpy of atomization ($\Delta_a H^0$) is the standard enthalpy change when one mole of a substance is completely broken down into individual gaseous atoms. This process involves breaking all the bonds in the substance and converting the elements into their atomic gaseous state.

Example: Atomization of methane gas:

$\text{CH}_4\text{(g)} \rightarrow \text{C(g)} + 4\text{H(g)}$; $\Delta_a H^0 = 1665$ kJ mol$^{-1}$.

Example: Atomization of sodium metal (solid to gaseous atoms):

$\text{Na(s)} \rightarrow \text{Na(g)}$; $\Delta_a H^0 = 108.4$ kJ mol$^{-1}$. For metals, this is also the enthalpy of sublimation.

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Bond Enthalpy

Chemical reactions involve breaking existing bonds in reactants and forming new bonds in products. Energy is required to break bonds (endothermic), and energy is released when new bonds form (exothermic).

For diatomic molecules, the bond dissociation enthalpy is the enthalpy change when one mole of a specific type of bond in a gaseous diatomic molecule is broken to form gaseous atoms. This is the same as the enthalpy of atomization for a diatomic molecule.

Example: $\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$; $\Delta_{H-H} H^0 = 435.0$ kJ mol$^{-1}$.

For polyatomic molecules, the energy required to break a specific type of bond can vary slightly depending on the molecule and the location of the bond within the molecule. For example, the four C-H bonds in methane (CH$_4$) require different amounts of energy to break successively.

In such cases, the concept of mean bond enthalpy is used. It is the average value of bond dissociation enthalpies for a given type of bond across different molecules or in successive bond breaking steps within the same molecule.

For CH$_4$, the mean C-H bond enthalpy is calculated by dividing the total atomization enthalpy by the number of C-H bonds (4):

Mean $\Delta_{C-H} H^0 = \frac{\Delta_a H^0(\text{CH}_4)}{4} = \frac{1665 \text{ kJ mol}^{-1}}{4} \approx 416$ kJ mol$^{-1}$.

Mean bond enthalpy values can be used to estimate the enthalpy change for a gas phase reaction using the relationship:

$\Delta_r H^0 \approx \sum (\text{bond enthalpies of reactants}) - \sum (\text{bond enthalpies of products})$

This equation represents the energy required to break all bonds in the reactants minus the energy released when all bonds in the products are formed. Note this is an approximation, especially if the reaction is not entirely in the gas phase or if mean bond enthalpies are used.

	H	C	N	O	F	Si	P	S	Cl	Br	I
H	435.8	414	389	464	569	293	318	339	431	368	297
C		347	293	351	439	289	264	259	330	276	238
N			159	201	272	-	209	-	201	243	-
O				138	184	368	351	-	205	-	201
F					155	540	490	327	255	197	-
Si						176	213	226	360	289	213
P							213	230	331	272	213
S								213	251	213	-
Cl									243	218	209
Br										192	180
I											151

Bond	Enthalpy (kJ mol$^{-1}$)
N = N	418
C = C	611
O = O	498
N $\equiv$ N	946
C $\equiv$ C	837
C = N	615
C = O	741
C $\equiv$ N	891
C $\equiv$ O	1070

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Lattice Enthalpy

The Lattice Enthalpy ($\Delta_{lattice}H^0$) of an ionic compound is the standard enthalpy change when one mole of the solid ionic compound completely dissociates into its constituent gaseous ions.

Example: $\text{Na}^+\text{Cl}^-\text{(s)} \rightarrow \text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)}$; $\Delta_{lattice}H^0 = +788$ kJ mol$^{-1}$.

Lattice enthalpy is a measure of the strength of the electrostatic forces holding the ions together in the crystal lattice. It is difficult to measure directly. It is usually determined indirectly using a Born-Haber cycle, which is an enthalpy diagram based on Hess's Law.

A Born-Haber cycle links the enthalpy of formation of an ionic compound to various steps, including atomization of the elements, ionization enthalpy of the metal, electron gain enthalpy of the non-metal, and lattice enthalpy.

Summing the enthalpy changes around the cycle allows calculation of the unknown lattice enthalpy (e.g., $\Delta_f H^0(\text{NaCl, s}) = \Delta_{sub} H^0(\text{Na, s}) + \Delta_i H^0(\text{Na, g}) + \frac{1}{2}\Delta_{bond} H^0(\text{Cl}_2\text{, g}) + \Delta_{eg} H^0(\text{Cl, g}) + \Delta_{lattice} H^0(\text{NaCl, s})$). By convention, the cycle usually adds steps from the elements in their standard states to the gaseous ions, and then from the gaseous ions to the solid ionic compound (lattice enthalpy, which is exothermic from ions to solid, or endothermic from solid to ions).

$\Delta_{lattice}H^0$ from solid to gaseous ions is typically a large positive value.

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Enthalpy Of Solution

The enthalpy of solution ($\Delta_{sol}H^0$) is the standard enthalpy change when one mole of a solute dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution refers to dissolving in so much solvent that interactions between solute particles are negligible.

When an ionic compound dissolves in water, it involves two main steps:

1. Breaking the crystal lattice: This requires energy, equal to the lattice enthalpy ($\Delta_{lattice}H^0$, positive).
2. Hydrating the ions: Gaseous ions are surrounded by water molecules, releasing energy (enthalpy of hydration, $\Delta_{hyd}H^0$, negative).

The enthalpy of solution is the sum of these enthalpy changes:

$\Delta_{sol}H^0 = \Delta_{lattice}H^0 + \Delta_{hyd}H^0$ (for ionic compounds dissolving in water)

If $\Delta_{hyd}H^0$ is more negative than $\Delta_{lattice}H^0$ is positive, the dissolution is exothermic ($\Delta_{sol}H^0 < 0$). If $\Delta_{lattice}H^0$ is larger, the dissolution is endothermic ($\Delta_{sol}H^0 > 0$). For many salts, $\Delta_{sol}H^0$ is positive, and solubility increases with temperature.

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Enthalpy Of Dilution

The enthalpy of dilution is the enthalpy change that occurs when a solution of a particular concentration is diluted by adding more solvent. This process involves separating solute particles and solvent molecules and forming new interactions upon dilution. It depends on the initial concentration and the amount of solvent added.

Example: Dissolving HCl gas in water.

$\text{HCl(g)} + 10 \text{ aq.} \rightarrow \text{HCl.10 aq.}$ $\Delta H = -69.01$ kJ/mol

$\text{HCl(g)} + 40 \text{ aq.} \rightarrow \text{HCl.40 aq.}$ $\Delta H = -72.79$ kJ/mol

The enthalpy change for diluting HCl solution from containing 10 mol water to 40 mol water is the difference between these two steps, effectively representing $\text{HCl.10 aq.} + 30 \text{ aq.} \rightarrow \text{HCl.40 aq.}$. Using Hess's Law, we reverse the first reaction and add the second:

$\text{HCl.10 aq.} \rightarrow \text{HCl(g)} + 10 \text{ aq.}$ $\Delta H = +69.01$ kJ/mol

$\text{HCl(g)} + 40 \text{ aq.} \rightarrow \text{HCl.40 aq.}$ $\Delta H = -72.79$ kJ/mol

Adding them: $\text{HCl.10 aq.} + 30 \text{ aq.} \rightarrow \text{HCl.40 aq.}$ $\Delta H = 69.01 - 72.79 = -3.78$ kJ/mol.

The enthalpy of dilution from 10 aq to 40 aq is -3.78 kJ/mol.

Spontaneity

The First Law of Thermodynamics deals with energy conservation but does not predict the direction of a process. Many naturally occurring processes (like heat flow from hot to cold, gases expanding) proceed spontaneously in one direction and do not reverse on their own. Spontaneity in thermodynamics refers to the potential of a process to occur without external assistance. It doesn't imply the speed of the process. A spontaneous process is generally irreversible; reversing it requires external intervention.

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Is Decrease In Enthalpy A Criterion For Spontaneity ?

Many spontaneous processes, like a ball rolling downhill or the combustion of fuel, are exothermic (release energy, $\Delta H < 0$). This might suggest that a decrease in enthalpy is the driving force for spontaneity.

Indeed, most exothermic reactions are spontaneous (e.g., formation of NH$_3$, HCl, H$_2$O from elements). However, there are spontaneous processes that are endothermic ($\Delta H > 0$), such as the dissolution of many salts (e.g., KNO$_3$ in water makes the water colder) or the reaction:

$\frac{1}{2}\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{NO}_2\text{(g)}$; $\Delta_r H^0 = +33.2$ kJ mol$^{-1}$ (spontaneous endothermic)

Since spontaneous endothermic reactions exist, a decrease in enthalpy ($\Delta H < 0$) is a driving factor for spontaneity but not the sole criterion.

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Entropy And Spontaneity

What drives spontaneous processes that don't involve a decrease in enthalpy, or even involve an increase? Consider the spontaneous diffusion of two gases when a partition is removed.

Initially, the gases are ordered (A on one side, B on the other). After diffusion, they are mixed; the system is less ordered, more random, or more chaotic. This increase in disorder is a characteristic of spontaneous processes in isolated systems.

This leads to the introduction of another thermodynamic state function: Entropy ($S$). Entropy is a measure of the degree of randomness or disorder in a system. Higher disorder corresponds to higher entropy. For a given substance, $S_{solid} < S_{liquid} < S_{gas}$.

Entropy ($S$), like $U$ and $H$, is a state function ($\Delta S$ is path-independent). The change in entropy for a reversible process ($\Delta S_{sys}$) is defined as the heat transferred reversibly ($q_{rev}$) divided by the absolute temperature ($T$):

$\Delta S_{sys} = \frac{q_{sys, rev}}{T}$

The Second Law of Thermodynamics provides a criterion for spontaneity based on entropy: For any spontaneous process, the total entropy of the universe (system + surroundings) increases.

$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$ (for a spontaneous process)

At equilibrium, $\Delta S_{total} = 0$. A process is non-spontaneous if $\Delta S_{total} < 0$.

Thus, an increase in the total entropy of the universe is the driving force for spontaneous processes.

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Gibbs Energy And Spontaneity

The total entropy change ($\Delta S_{total}$) is the criterion for spontaneity for the universe. However, calculating $\Delta S_{surr}$ (which involves heat exchange) is not always convenient, especially for closed or open systems. A new thermodynamic function, the Gibbs energy ($G$), is introduced to predict spontaneity based on the properties of the system alone (at constant temperature and pressure).

Gibbs energy ($G$) is defined as:

$G = H - TS$

$G$ is an extensive property and a state function. For a process occurring at constant temperature ($T$), the change in Gibbs energy ($\Delta G$) is related to the changes in enthalpy and entropy of the system by the Gibbs equation:

$\Delta G = \Delta H - T\Delta S$ (at constant $T$)

If the process also occurs at constant pressure, then $\Delta G_{sys} = \Delta H_{sys} - T\Delta S_{sys}$. Dropping the 'sys' subscript, $\Delta G = \Delta H - T\Delta S$.

The relationship between $\Delta G$ and spontaneity at constant pressure and temperature is:

• If $\Delta G < 0$: The process is spontaneous.
• If $\Delta G > 0$: The process is non-spontaneous.
• If $\Delta G = 0$: The system is at equilibrium.

$\Delta G$ represents the maximum amount of non-PV work (useful work) that can be obtained from a process at constant temperature and pressure. Hence, it's also known as free energy.

The Gibbs equation $\Delta G = \Delta H - T\Delta S$ shows how enthalpy, entropy, and temperature influence spontaneity:

• If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in disorder): $\Delta G = (-) - T(+)$ = negative at all temperatures. Process is always spontaneous.
• If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in disorder): $\Delta G = (+) - T(-)$ = positive at all temperatures. Process is never spontaneous.
• If $\Delta H < 0$ (exothermic) and $\Delta S < 0$ (decrease in disorder): $\Delta G = (-) - T(-)$. $\Delta G$ is negative if $| \Delta H | > |T\Delta S |$. Spontaneous at low temperatures.
• If $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in disorder): $\Delta G = (+) - T(+)$. $\Delta G$ is negative if $| T\Delta S | > |\Delta H |$. Spontaneous at high temperatures.

This is why many endothermic reactions that increase disorder become spontaneous at high temperatures.

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Entropy And Second Law Of Thermodynamics

As discussed, the Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe increases ($\Delta S_{total} > 0$). This explains why spontaneous exothermic reactions (which increase surroundings' entropy) are common, and why processes naturally move towards greater disorder.

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Absolute Entropy And Third Law Of Thermodynamics

Particles in a substance have various types of motion (translational, rotational, vibrational). Entropy increases with temperature as these motions become more vigorous. The Third Law of Thermodynamics establishes a reference point for entropy:

The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero (0 K).

At absolute zero, a perfect crystal has maximum order, and its entropy is considered zero. This law allows for the calculation of absolute entropy values for pure substances by integrating heat capacity data from 0 K up to a desired temperature. Standard molar entropies ($S^0$) are tabulated, allowing calculation of standard entropy changes for reactions ($\Delta_r S^0 = \sum n_p S^0(\text{products}) - \sum n_r S^0(\text{reactants})$) using a similar approach to Hess's Law for enthalpy.

Gibbs Energy Change And Equilibrium

For a reaction at equilibrium, the forward and reverse processes occur at equal rates, and there is no net change in the system's state. The criterion for equilibrium at constant temperature and pressure using Gibbs energy is:

$\Delta G = 0$ (at equilibrium)

The standard Gibbs energy change for a reaction ($\Delta_r G^0$) is related to the equilibrium constant ($K$) of the reaction. For a reaction $\text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}$, the equilibrium constant $K$ is given by $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ (in terms of concentrations) or $K_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}$ (in terms of partial pressures for gases). In thermodynamics, $K$ refers to $K_p$ for gas-phase reactions or $K_c$ for solution-phase reactions, often using activities.

The relationship between $\Delta_r G^0$ and $K$ is:

$\Delta_r G^0 = -RT \ln K$

or $\Delta_r G^0 = -2.303 RT \log K$

where $R$ is the gas constant and $T$ is the absolute temperature.

This equation is fundamental as it connects thermodynamics (predicting spontaneity, $\Delta G^0$) with chemical kinetics/equilibrium (predicting the extent of reaction, $K$).

• If $\Delta_r G^0 < 0$, $\ln K$ is positive, $K > 1$. The equilibrium lies towards the products, and the reaction proceeds significantly in the forward direction under standard conditions.
• If $\Delta_r G^0 > 0$, $\ln K$ is negative, $K < 1$. The equilibrium lies towards the reactants, and the reaction does not proceed significantly in the forward direction under standard conditions.
• If $\Delta_r G^0 = 0$, $\ln K = 0$, $K = 1$. The reaction is at equilibrium under standard conditions.

The relationship $\Delta_r G^0 = \Delta_r H^0 - T\Delta_r S^0$ allows us to calculate $\Delta_r G^0$ from standard enthalpy and entropy changes. Then, using $\Delta_r G^0 = -RT \ln K$, we can determine the equilibrium constant $K$ and predict the extent of the reaction.

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