First Law of Thermodynamics: Work, Heat, Internal Energy, and Enthalpy

The study of energy transformations in chemical and physical processes is the domain of thermodynamics. The First Law of Thermodynamics is a fundamental principle that governs these transformations. It is essentially a statement of the law of conservation of energy, applied to thermodynamic systems. This law helps us understand how energy is exchanged between a system and its surroundings in the form of heat and work, and how this exchange affects the internal energy of the system.

Internal Energy (U)

Internal energy is the total energy contained within a thermodynamic system. It is the sum of all the kinetic and potential energies of the molecules within the system. This includes the translational, rotational, and vibrational kinetic energies of the molecules, as well as the potential energy associated with intermolecular forces and chemical bonds.

For a given amount of substance under specified conditions, the internal energy is a state function. This means that the change in internal energy ($\Delta U$) between two states depends only on the initial and final states, not on the path taken to get from one state to another. Mathematically, if the initial internal energy is $U_1$ and the final internal energy is $U_2$, then the change in internal energy is given by:

$\Delta U = U_2 - U_1$

It is practically impossible to determine the absolute value of internal energy for a system. However, changes in internal energy are measurable and are of great importance in understanding chemical reactions and physical processes.

Heat (q) and Work (w)

Energy can be transferred into or out of a system in two primary ways: heat and work.

Heat (q)

Heat is the transfer of thermal energy between a system and its surroundings due to a temperature difference. Heat flows spontaneously from a region of higher temperature to a region of lower temperature.

• If heat is absorbed by the system from the surroundings, it is considered positive ($q > 0$).
• If heat is released by the system to the surroundings, it is considered negative ($q < 0$).

The unit of heat is typically Joules (J) or calories (cal).

Work (w)

Work is the transfer of energy that occurs when a force acts over a distance. In thermodynamics, work is often associated with volume changes. The most common type of work encountered is pressure-volume (PV) work, which occurs when the volume of a system changes against an external pressure.

For a process where the volume changes from an initial volume ($V_1$) to a final volume ($V_2$) against a constant external pressure ($P_{ext}$), the work done is given by:

$w = -P_{ext} \Delta V$

where $\Delta V = V_2 - V_1$.

• If the system does work on the surroundings (e.g., gas expands), the volume increases ($\Delta V > 0$), and the work done by the system is negative ($w < 0$). This convention is widely used in chemistry.
• If the surroundings do work on the system (e.g., gas is compressed), the volume decreases ($\Delta V < 0$), and the work done on the system is positive ($w > 0$).

The unit of work is also typically Joules (J).

The First Law of Thermodynamics

The First Law of Thermodynamics states that energy cannot be created or destroyed, only converted from one form to another. For a thermodynamic system, this means that the change in its internal energy ($\Delta U$) is equal to the heat added to the system ($q$) plus the work done on the system ($w$).

The mathematical expression for the First Law of Thermodynamics is:

$\Delta U = q + w$

This equation is a cornerstone of thermodynamics. It tells us that the total energy of an isolated system remains constant. If a system is not isolated, its internal energy can change if energy is transferred across its boundary as heat or work.

Applying the First Law: Different Processes

The application of the First Law of Thermodynamics depends on the specific conditions under which the process occurs.

1. Isothermal Process ($\Delta T = 0$)

An isothermal process occurs at constant temperature. In an ideal gas, internal energy is a function of temperature only. Therefore, for an isothermal process involving an ideal gas, $\Delta U = 0$.

From the First Law: $0 = q + w$, which implies $q = -w$. This means that any heat added to the system is entirely used to do work by the system, or any work done on the system is entirely released as heat.

2. Adiabatic Process ($q = 0$)

An adiabatic process occurs when there is no heat exchange between the system and its surroundings. This means $q = 0$.

Applying the First Law: $\Delta U = 0 + w$, so $\Delta U = w$. In an adiabatic process, any change in internal energy is solely due to the work done. If work is done by the system, its internal energy decreases ($\Delta U < 0$). If work is done on the system, its internal energy increases ($\Delta U > 0$).

3. Isochoric Process ($\Delta V = 0$ or $w = 0$)

An isochoric process occurs at constant volume. If the volume is constant, no PV work is done, so $w = 0$.

Applying the First Law: $\Delta U = q + 0$, so $\Delta U = q$. In a constant volume process, all the heat added to the system goes into increasing its internal energy.

4. Isobaric Process ($P = \text{constant}$)

An isobaric process occurs at constant pressure. In this case, work done is $w = -P \Delta V$.

The First Law becomes: $\Delta U = q - P \Delta V$.

This is a very common scenario in chemistry, such as reactions carried out in open beakers or flasks.

Example of Applying the First Law

Consider a gas in a cylinder fitted with a piston. Suppose 100 J of heat is added to the gas ($q = +100$ J), and the gas expands, doing 30 J of work on the piston ($w = -30$ J). What is the change in the internal energy of the gas?

Using the First Law:

$\Delta U = q + w$ $\Delta U = (+100 \text{ J}) + (-30 \text{ J})$ $\Delta U = +70 \text{ J}$

The internal energy of the gas increases by 70 J.

Enthalpy (H)

While internal energy ($U$) is a very useful concept, it is often more convenient to work with another thermodynamic property called enthalpy ($H$). Enthalpy is defined as the sum of the internal energy of a system plus the product of its pressure and volume.

The mathematical definition of enthalpy is:

$H = U + PV$

Like internal energy, enthalpy is a state function. Therefore, changes in enthalpy ($\Delta H$) depend only on the initial and final states.

Enthalpy Change ($\Delta H$)

The change in enthalpy for a process is given by:

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

For processes occurring at constant pressure ($P = \text{constant}$), this simplifies significantly. If the pressure is constant, then $\Delta (PV) = P \Delta V$.

So, at constant pressure:

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

Now, let's substitute the First Law ($\Delta U = q + w$) into this equation:

$\Delta H = (q + w) + P \Delta V$

At constant pressure, the work done is $w = -P \Delta V$. Substituting this:

$\Delta H = (q - P \Delta V) + P \Delta V$ $\Delta H = q$

This is a crucial result: For a process occurring at constant pressure, the change in enthalpy ($\Delta H$) is equal to the heat absorbed or released by the system ($q_P$).

This makes enthalpy particularly useful for studying chemical reactions, as many reactions are carried out under constant atmospheric pressure.

Relationship between $\Delta H$ and $\Delta U$

The relationship $\Delta H = \Delta U + P \Delta V$ (at constant pressure) or $\Delta H = \Delta U + \Delta (PV)$ (general case) shows how enthalpy and internal energy are related.

For reactions involving gases, where there can be a significant change in the number of moles of gas, the difference between $\Delta H$ and $\Delta U$ can be substantial.

Consider a reaction where $\Delta n_g$ is the change in the number of moles of gas:

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

Assuming ideal gas behavior, $PV = nRT$. For a change in state, $P \Delta V + V \Delta P = \Delta n_g RT$.

If the process is at constant pressure ($ \Delta P = 0 $), then $P \Delta V = \Delta n_g RT$.

Substituting this into the enthalpy equation:

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

This equation is extremely important for relating enthalpy changes to internal energy changes for reactions involving gases. The gas constant $R$ is $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.

Types of Enthalpy Changes

When we talk about $\Delta H$, it's often associated with specific types of processes:

• Enthalpy of Reaction ($\Delta H_{rxn}$): The heat change associated with a chemical reaction carried out at constant pressure.
• Enthalpy of Combustion ($\Delta H_c$): The heat released when one mole of a substance undergoes complete combustion under standard conditions.
• Enthalpy of Formation ($\Delta H_f$): The heat change when one mole of a compound is formed from its constituent elements in their standard states.
• Enthalpy of Fusion ($\Delta H_{fus}$) and Vaporization ($\Delta H_{vap}$): Heat changes associated with melting and boiling, respectively.

Exothermic and Endothermic Processes

Based on the sign of heat flow ($q$) or enthalpy change ($\Delta H$), processes are classified as:

• Exothermic Process: Releases heat to the surroundings. For these processes, $q < 0$ and $\Delta H < 0$. The system's enthalpy decreases. Example: Combustion of fuel, neutralization reactions.
• Endothermic Process: Absorbs heat from the surroundings. For these processes, $q > 0$ and $\Delta H > 0$. The system's enthalpy increases. Example: Melting ice, photosynthesis, dissolving ammonium nitrate in water.

Example: Combustion of Methane

The combustion of methane is a classic exothermic reaction:

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

The standard enthalpy of combustion for methane is $\Delta H_c^\circ = -890.4 \text{ kJ/mol}$. This means that when one mole of methane burns completely under standard conditions (constant pressure), 890.4 kJ of heat is released to the surroundings.

Let's calculate the change in internal energy ($\Delta U$) for this reaction at 298 K (25 °C) and 1 atm pressure, assuming the water produced is liquid.

First, calculate $\Delta n_g$: Moles of gaseous products = 1 (for $CO_2$) Moles of gaseous reactants = 1 (for $CH_4$) + 2 (for $O_2$) = 3 $\Delta n_g = 1 - 3 = -2$ mol

Using the gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ and $T = 298 \text{ K}$:

$\Delta H = \Delta U + \Delta n_g RT$ $-890.4 \text{ kJ/mol} = \Delta U + (-2 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$

Convert R to kJ: $R = 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}$

$-890.4 \text{ kJ/mol} = \Delta U + (-2) \times (8.314 \times 10^{-3}) \times (298) \text{ kJ/mol}$ $-890.4 \text{ kJ/mol} = \Delta U + (-4.95) \text{ kJ/mol}$

$\Delta U = -890.4 \text{ kJ/mol} + 4.95 \text{ kJ/mol}$ $\Delta U = -885.45 \text{ kJ/mol}$

As expected, $\Delta H$ is slightly more negative than $\Delta U$ because the reaction involves a decrease in the number of gas moles, meaning the system does less work on the surroundings.

Cyclic Processes and State Functions

A cyclic process is one in which a system returns to its original state. For any state function (like internal energy $U$ and enthalpy $H$), the change over a complete cycle is zero.

$\Delta U_{\text{cycle}} = U_{\text{final}} - U_{\text{initial}} = 0$ (since $U_{\text{final}} = U_{\text{initial}}$) $\Delta H_{\text{cycle}} = H_{\text{final}} - H_{\text{initial}} = 0$ (since $H_{\text{final}} = H_{\text{initial}}$)

However, heat ($q$) and work ($w$) are path functions. Their values depend on the specific process. Therefore, for a cyclic process:

$\Delta U = q + w = 0 \implies q = -w$

This means that in a cyclic process, the total heat absorbed by the system must be equal to the total work done by the system. This is the principle behind heat engines.

Joule's Experiment

James Prescott Joule performed a series of experiments to demonstrate the equivalence of heat and work and to show that internal energy is a state function. In one famous experiment, he measured the temperature rise of water in an insulated container when a paddle wheel, driven by falling weights, stirred the water.

The work done by the falling weights ($w$) was converted into heat ($q$) within the water, causing its temperature to rise. Joule found that the amount of heat produced was directly proportional to the amount of work done, establishing the mechanical equivalent of heat. Crucially, he showed that the change in the internal energy of the water ($\Delta U$) depended only on the initial and final temperatures, not on whether the energy was added as heat or work. This reinforced the concept of internal energy as a state function.

Importance in Chemistry

The First Law of Thermodynamics and the concept of enthalpy are indispensable in chemistry.

• Predicting Energy Changes: It allows chemists to calculate the energy released or absorbed in reactions, which is vital for designing chemical processes and understanding reaction feasibility.
• Calorimetry: Experiments measuring heat changes (calorimetry) directly use the First Law. For example, measuring $q$ at constant volume gives $\Delta U$, and measuring $q$ at constant pressure gives $\Delta H$.
• Understanding Reaction Energetics: Enthalpy changes help determine if a reaction is exothermic or endothermic, providing insights into bond energies and molecular stability.
• Industrial Applications: Designing efficient industrial processes, from power generation to chemical synthesis, relies heavily on thermodynamic principles to manage energy inputs and outputs.