Introduction to Enthalpy

Enthalpy, $H$, represents the total heat content of a system and is equivalent to the sum of the internal energy $U$ and the product of pressure and volume $pV$. Enthalpy can be understood as the amount of energy required to create a given system and to make space for it by displacing the environment. It is particularly useful when considering processes occurring at constant pressure.

The First Law of Thermodynamics states that the internal energy, $U$, of a system change due to the heat added to the system, $dQ$, and the work done by or on the system, $dW$.

This relationship shows that any change in a system’s internal energy depends on the heat exchanged and the work done. For processes at constant volume, work done, $dW = 0$, so all heat added to the system changes its internal energy directly:

However, many natural processes occur under constant pressure. In these cases, the work done by the system is expressed as $p \, dV$, where $p$ is the constant pressure and $dV$ is the change in volume. Substituting into the First Law, we get:

To simplify calculations under constant pressure, we define a new thermodynamic quantity called enthalpy, $H$, defined as:

Taking the differential of this equation, we get:

Using the product rule, we expand $d(pV)$ as:

Under constant pressure ($dp = 0$), this simplifies to:

Therefore, at constant pressure, the change in enthalpy ($dH$) represents the heat absorbed by the system. This provides a straightforward way to calculate heat changes in reactions or processes where pressure is held constant.

Exothermic and Endothermic Reactions

Reactions can be classified based on whether they absorb or release heat:

• exothermic reaction: heat is released to the surroundings, resulting in $dH < 0$.
• endothermic reaction, heat is absorbed from the surroundings, giving $dH > 0$.

For example, combustion reactions are typically exothermic, releasing significant amounts of heat, while melting or evaporation are endothermic as they require heat input.

Heat Capacities at Constant Pressure and Volume

Heat capacities measure the amount of heat needed to change a system's temperature. These capacities differ depending on whether the process occurs at constant volume or constant pressure:

Constant Volume Heat Capacity ($C_v$)

At constant volume, there is no expansion work ($dW = 0$), so any heat added directly changes the internal energy. The heat capacity at constant volume, $C_v$, is defined as the rate of change of heat with respect to temperature at constant volume:

Constant Pressure Heat Capacity ($C_p$)

At constant pressure, heat added changes both the internal energy and the enthalpy of the system. The heat capacity at constant pressure, $C_p$, is given by the rate of change of heat with respect to temperature at constant pressure:

Deriving the Relationship between $C_p$ and $C_v$ for Ideal Gases

For an ideal gas, we know:

Using the ideal gas law $pV = nRT$, we rewrite enthalpy as:

Differentiating both sides with respect to temperature at constant pressure, we get:

Since $C_p = \frac{dH}{dT}$ and $C_v = \frac{dU}{dT}$, we find:

This shows that $C_p$ is always greater than $C_v$ for an ideal gas due to the extra work needed to expand at constant pressure.

Thermochemistry

Standard Reaction Enthalpy, $ \Delta_{rxn}H^\circ_{298K} $

The standard reaction enthalpy ($ \Delta_{rxn} H^\circ $) is the enthalpy change that occurs when a chemical reaction is carried out with all reactants and products in their standard states. Standard states are typically at a temperature of 298.15 K and a pressure of 1 bar.

The general formula to calculate the reaction enthalpy is:

Here, $m$ and $n$ represent the stoichiometric coefficients of the reactants and the products, correspondingly. $ \Delta_f H^\circ $ represents the standard enthalpy of formation for each substance.

For a general reaction:

The reaction enthalpy is calculated as:

This formula involves summing the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of the reactants. Remember that $m$ and $n$ are the stoichiometric coefficient of each species in the balanced chemical equation.

Example: Let's consider the combustion of glucose ($C_6H_{12}O_6$) in oxygen:

The standard reaction enthalpy can be calculated using the enthalpies of formation:

Substituting the known values for standard enthalpies of formation:

• $ \Delta_f H^\circ (CO_2) = -393.5 \, \text{kJ/mol} $
• $ \Delta_f H^\circ (H_2O) = -285.8 \, \text{kJ/mol} $
• $ \Delta_f H^\circ (C_6H_{12}O_6) = -1273.3 \, \text{kJ/mol} $
• $ \Delta_f H^\circ (O_2) = 0 \, \text{kJ/mol} $ (since $O_2$ is in its standard state)

Calculating the sums for the products and reactants:

Finally, adding and subtracting these values gives:

This value represents the standard enthalpy change for the combustion of one mole of glucose under standard conditions.

Note: The enthalpy of formation for pure substances in their natural states (e.g. $ O_2 (g) $, $ Al (s) $, $ I_2 (s)$, $ C $ as graphite, $ N_2 (g) $, etc.) is zero:

This means that no enthalpy change is required to form these substances in their most stable, natural states from the elements, as they are already in that form.

Hess’s Law and Reaction Enthalpy Calculation

Hess’s Law states that the enthalpy change for a reaction is independent of the reaction pathway, depending only on the initial and final states. This allows us to calculate reaction enthalpy by summing enthalpy changes for a sequence of intermediate steps that lead from reactants to products.

Where $v_j$ represents the stoichiometric coefficients of the reactants and products.

Example Calculation of Reaction Enthalpy Using Hess’s Law

Problem: Calculate the enthalpy for acetylene formation using the following reactions and enthalpy values:

Solution: The balanced equation for acetylene is:

To find the overall reaction enthalpy change ($\Delta H_{comb}$), we use Hess’s Law by first balancing the chemical equation and adjusting stoichiometric coefficients where necessary.:

1. Multiply the first reaction by 2 to match the stoichiometric needs of the overall balanced equation:
   \[ 2C(s) + 2O_2(g) \rightarrow 2CO_2(g) \quad \Delta H^\circ = 2 \times (-395.5) = -787.0 \, \text{kJ/mol} \]
2. The second reaction remains unchanged:
   \[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H^\circ = -285.8 \, \text{kJ/mol} \]
3. The third reaction is reversed:
   \[ 2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g) \quad \Delta H^\circ = +1299.6 \, \text{kJ/mol} \]

Now, we sum the enthalpy changes for all reactions:

This results in:

The total enthalpy change for the overall reaction is $ +226.8 \, \text{kJ/mol} $, and we conclude that the enthalpy of formation for acetylene ($C_2H_2$) is $ +226.8 \, \text{kJ/mol} $ under these conditions.

Kirchhoff's Law - Temperature Dependence of Enthalpy

Kirchhoff's Law describes how the enthalpy of a chemical reaction changes with temperature. This is especially useful in thermodynamics, where the reaction enthalpy is often measured at one temperature (usually 298.15 K), but the reaction might occur at a different temperature. Kirchhoff’s Law helps adjust the enthalpy to this new temperature by taking into account the heat capacities of the reactants and products.

Case 1: Constant Heat Capacity

If the heat capacities $C_p$ of the reactants and products are assumed to be constant over the temperature range of interest, then the relationship between the enthalpy changes at two different temperatures ($T_1$ and $T_2$) is given by:

Since $C_p$ is constant, this simplifies to:

This version of Kirchhoff’s Law is straightforward when the heat capacities of the reactants and products are independent of temperature, which is often a reasonable approximation over small temperature ranges. It tells us that the enthalpy at the new temperature, $T_2$, is the enthalpy at the reference temperature, $T_1$, adjusted by the product of the heat capacity and the temperature difference.

Example for Constant $C_p$: Suppose we want to calculate the enthalpy change for a reaction at $T_2 = 400 \, K$ given that the enthalpy change at $T_1 = 298.15 \, K$ is $ \Delta H(T_1) = -100 \, \text{kJ/mol} $, and the average heat capacity of the system is $C_p = 75 \, \text{J/mol·K}$. Applying Kirchhoff’s Law:

Convert the units of $C_p$ to kJ/mol·K for consistency:

Now plug in the values:

Simplifying the equation:

Thus, the enthalpy change at 400 K is $ \Delta H(T_2) = -92.36 \, \text{kJ/mol} $.

Case 2: Temperature-Dependent Heat Capacity

In many cases, the heat capacity $C_p$ is not constant and varies with temperature. When this happens, Kirchhoff's Law is extended to include a more complex relationship where $C_p$ is expressed as a function of temperature:

In this case, the enthalpy change is calculated by integrating over the temperature range:

Performing the integration gives:

This equation allows us to calculate the enthalpy change more accurately when heat capacity varies with temperature, especially for reactions that occur over large temperature ranges.

Example for Temperature-Dependent $C_p$: Suppose the heat capacity of a substance is given by:

We want to calculate the enthalpy change for a reaction between $T_1 = 298.15 \, K$ and $T_2 = 600 \, K$. The enthalpy change at $T_1$ is $ \Delta H(T_1) = -200 \, \text{kJ/mol} $.

Using Kirchhoff’s Law:

After calculating the terms, we get:

Simplifying the terms gives the final value for the enthalpy change at $T_2 = 600 \, K$.

Kirchhoff’s Law allows us to adjust the enthalpy of a reaction from one temperature to another, accounting for changes in heat capacity. When $C_p$ is constant, the adjustment is simple, but when $C_p$ varies with temperature, the relationship becomes more complex, requiring integration over the temperature range.

Phase Transitions – Enthalpy of Transformation (Latent Energy of Transformation)

Phase transitions occur at constant temperature, meaning that the temperature does not change during the transformation from one phase to another. The energy required to effect these phase changes is known as the enthalpy of transformation or latent heat. Each phase transition (e.g., melting, vaporization, sublimation) has a characteristic enthalpy change associated with it.

Enthalpy of Condensation and Vaporization

The enthalpy of vaporization ($ \Delta H_{vap}^\circ $) is the energy required to convert one mole of a substance from liquid to gas at constant temperature, while the enthalpy of condensation ($ \Delta H_{cond}^\circ $) is the energy released when a gas condenses into a liquid at the same temperature. These phase changes occur at equilibrium at the boiling point.

For water at 373 K (100°C):

Enthalpy of Melting

The enthalpy of melting (also called the enthalpy of fusion), $ \Delta H_m^\circ $, is the energy required to convert one mole of a substance from solid to liquid at constant temperature. For water, this transition occurs at 273 K (0°C):

Enthalpy of Sublimation

Sublimation is the process by which a solid converts directly into a gas without passing through the liquid phase. The enthalpy of sublimation ($ \Delta H_{sub}^\circ $) is the sum of the enthalpy of fusion and the enthalpy of vaporization:

For water, the enthalpy of sublimation can be calculated by summing the enthalpies of melting and vaporization.