2.3 The First Law

The first law states that energy is conserved, in other words, energy can be converted from one form to another and can neither be created nor destroyed. Any change in the system’s heat or work must be accounted for in the internal energy of the system.

(6)\[\Delta_{sys}E^\circ = q + w\]

In Eq (6), \(\Delta_{sys}E^\circ\) is the differential change in the internal energy of the system, \(q\) is the heat added or removed from the system, and \(w\) is the work done on or by the system. By convention, \(q\) is positive for an endothermic process and negative for an exothermic process, and \(w\) is positive for work done on the system by the surroundings and negative for work done by the system on the surroundings.

Fig. 30 In a calorimetric determination, either (a) an exothermic process occurs and heat, \(q<0\), indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, \(q>0\), indicating that thermal energy is transferred from the surroundings to the system. Image source: 9.2 Calorimetry - Chemistry: Atoms First | OpenStax

If a chemical reaction occurred in a sealed container (e.g., a combustion reaction) and the product of the reaction was the creation of new products, this would increase the volume and the pressure. Instead of a sealed container, the pressure would remain constant if this reaction occurred in a container with a movable piston that allowed the volume to adjust (\(\Delta V\)). Therefore, the chemical reaction is causing mechanical work to occur in that system. Specifically, this type of work is called pressure-volume (\(PV\)) work. The amount of work done by such a process is given by

(7)\[w = -P\Delta V\]

In Eq (7) \(P\) is the external, opposing pressure, and \(\Delta V\) is the change in the container volume due to the piston being pushed upward. The negative sign in Eq (7) accounts for an increase in volume, resulting in a negative value for \(w\). At constant \(V\), (\(\Delta V = 0\)), \(q_V = _{sys}{E}\).

This condition is not practical as volume is not constant. Rather, in near-Earth’s surface systems, we can assume \(P\) to be constant, resulting in \(q_P\). We can now combine Eqs (6) and (7) as follows:

(8)\[\Delta_{sys}{E} = q_P- P\Delta V\]

Enthalpy

Enthalpy is a state function derived from rearranging Eq (8) as follows:

(9)\[q_P= \Delta_{sys}{E} + P\Delta V\]

The variable \(q_P\) in Eq (9) is called enthalpy (\(\Delta H^\circ\)) and is equal to the heat flow when processes occur at constant \(P\) and the only work done is pressure-volume work. The assumption of constant \(P\) is acceptable for Earth’s near-surface environments, where most systems can be considered open. \(\Delta H^\circ\) is a state function and accounts for both the internal energy (\(\Delta _{sys}E^\circ\)) and work (\(w\)) done on or by the system. It quantifies the heat flow into or out of the system at constant pressure (\(P\)). Enthalpy has the unit of \(\pu{kJ}\), while molar enthalpy has the units of \(\pu{ kJ mol−1}\). In a chemical reaction, (\(\Delta _{rxn}H^\circ\)) accounts for \(\Delta _{prod}H^\circ\) (products in a reaction) and \(\Delta _{reac}H^\circ\) (reactants in a reaction) as shown below:

(10)\[\Delta_{rxn}H^\circ = \Sigma \Delta_{prod}H^\circ - \Sigma \Delta_{reac}H^\circ\]

\(\Delta _{rxn}H^\circ\) can be used to determine if the system (\(sys\)) gains or loses heat. This value can be positive or negative (units: \(\pu{kJ mol-1}\)) and must be applied to a chemical reaction representing the system.

When \(\Delta_{rxn}H^\circ <0\), (\(\Delta_{prod}H^\circ < \Delta_{reac}H^\circ\)), this implies there is more internal energy bound within the reactants and that portion of this energy is now contained within the products. At the same time, the rest is released into the surroundings. The excess energy is released into the surroundings (\(surr\)). Such systems are called exothermic systems.

Example: Exothermic reactions

Virtually any combustion reaction is exothermic as heat is released into the surroundings. Hydrogen gas explosively reacts with oxygen to form water molecules, as shown below:

\[ \ce{2 H2(g) + O2(g) -> 2 H2O(l) + \text{heat}} \]

The heat released into the surroundings is the “product” of these reactions and can be quantified if we know exactly how much internal energy makes up each chemical species.

In the scenarios where \(\Delta_{rxn}H^\circ >0\), (\(\Delta_{prod}H^\circ > \Delta_{reac}H^\circ\)), there is more internal energy in the products than in the reactants. The system has to adjust for the imbalance in the energy by absorbing energy from the surroundings. Such systems are called endothermic systems.

Example: Endothermic reactions

The heating of liquid water from an external heat source (stove) causes water to evaporate or turn into vapor phase water.

This process can be represented as follows:

\[ \ce{H2O(l) + \text{heat} -> H2O(g)} \]

Heat energy is a “reactant” required to make this reaction possible.

Absolute enthalpies of chemical entities are difficult to measure. These values depend on the chemical composition of the chemical entity, the types of bonds involved, and system conditions (functions such as \(V\), \(P\), \(T\), etc.). Hence, the \(\Delta\) term is used before \(H\) is used to denote the enthalpies of a chemical entity as measured relative to a standard state. The reference point for all enthalpy expressions is called the standard enthalpy of formation (\(\Delta_f H^\circ\)), which is defined as the heat change that results when \(\pu{1 mol}\) of a compound is formed from its constituent elements in their standard states. Standard states refer to the most stable form of an element under standard conditions, meaning at ordinary atmospheric pressure. For example, oxygen can be the most stable form at ordinary atmospheric pressure. Although the standard state does not specify a temperature, we will always use (\(\Delta_f H^\circ\)) values measured at \(\pu{25 ^\circ C}\).

The CRC Handbook of Physics and Chemistry is a good reference for obtaining \(\Delta_f H^\circ\) for most chemicals. The online version is available at Standard Thermodynamic Properties of Chemical Substances (chemnetbase.com)

Example: Determining enthalpy of a reaction

Let’s determine if the following reactions gain heat from or loses heat to the surroundings.

\[\begin{split} \begin{align*} \ce{ 2 H2(g) + O2(g) &-> 2 H2O(g)\\ H2O(l) &-> H2O(g) } \end{align*} \end{split}\]

First, we need to look up \(\Delta_f H^\circ\) for all chemical species in thermodynamic data tables

Chemical	\(\Delta_f H^\circ\),  \(\pu{kJ mol-1}\)	Chemical	\(\Delta_f H^\circ\),  \(\pu{kJ mol-1}\)
\(\ce{O2 (g)}\)	\(0\)	\(\ce{H2O (g)}\)	\(-241.8\)
\(\ce{H2 (g)}\)	\(0\)	\(\ce{H2O (l)}\)	\(-285.8\)
		\(\ce{H2O (s)}\)	\(-291.8\)

Next, we must calculate \(\Delta_{rxn} H^\circ\) for each reaction.

$$

\[\begin{align*} \Delta_{rxn} H^\circ &= [2(-241.8)] - [(2(0) + (0)] = \pu{-483.6 kJ mol-1}\\ \Delta_{rxn} H^\circ &= [(-241.8)] - [(-285.8)] = \pu{+44.0 kJ mol-1} \end{align*}\]

$$

Since \(\Delta_{rxn}H^\circ <0\), the first reaction is exothermic and \(\Delta_{rxn}H^\circ >0\), the second reaction is endothermic.

Since we are working with thermochemical systems where heat is released or captured during a reaction, the following guidelines will help interpret, write, and manipulate thermochemical reactions.

Rules for working with thermochemical data

1. All calculations have to be performed on balanced chemical reactions.

2. Always specify all reactants and products’ physical states of matter.

3. \(\Delta_{rxn}H^\circ\) means that this is the enthalpy change per mole of the reaction (or process) as it is written and does not refer to individual chemical species or the number of moles involved in each reaction.

4. If both sides of a reaction are multiplied by \(n\), then \(\Delta_{rxn}H^\circ\) must be multiplied by \(n\).

5. If reaction is reversed, \(\Delta_{rxn}H^\circ\) magnitude remains same, but sign changes.

Example: Application of thermochemical rules

Let’s calculate the \(\Delta_{rxn}H^\circ\) for the following chemical reactions:

\[\begin{split} \begin{align} \ce{ 4 H2(g) + 2 O2(g) &-> 4 H2O(g) \\ H2O(g) &-> H2O(l) } \end{align} \end{split}\]

If we used \(\Delta_{f}H^\circ\) from the table in the previous example, we could calculate \(\Delta_{rxn}H^\circ\) as \(\pu{−967.2 kJ mol-1}\) and  \(\pu{−44.0 kJ mol-1}\), respectively in these reactions.

Example: Application of thermochemistry and mass balance

Let’s calculate the heat evolved when \(\pu{75 g}\) of \(\ce{SO2(g)}\) is oxidized to \(\ce{SO3 (g)}\) as shown in the reaction below:

\[ \ce{SO2(g) + 1/2 O2(g) -> SO3(g)} \]

\(\Delta_{rxn}H^\circ = \pu{-99.1 kJ mol-1}\)

According to this reaction, \(\pu{1 mole}\) of \(\ce{SO2(g)}\) is oxidized to \(\pu{1 mole}\) of \(\ce{SO3(g)}\) resulting in \(\Delta_{rxn}H^\circ = \pu{-99.1 kJ mol-1}\) (release of energy into surroundings)

The molar mass of \(\ce{SO2(g)}\) is \(\pu{64 g}\) (\(\pu{1 mol} = \pu{64 g}\)) or, in other words, \(\pu{64 g}\) of \(\ce{SO2(g)}\) results in \(\Delta_{rxn}H^\circ = \pu{-99.1 kJ mol-1}\).

Therefore, \(\pu{75 g}\) of \(\ce{SO2(g)}\) results in:

\[ (\pu{75 g}\, \ce{SO2}) \cdot \left(\dfrac{\pu{1 mol}\, \ce{SO2}}{\pu{64 g}\, \ce{SO2}}\right) = \pu{1.2 mol}\, \ce{SO2} \]

\(\therefore \Delta_\ce{SO2} H^\circ = (\pu{1.2 mol}) \cdot (\pu{-99.1 kJ mol-1}) = \pu{116 kJ}\).

Hess’s Law

Because \(\Delta_{rxn}H^\circ\) is a state function, the change in enthalpy that occurs when reactants are converted to products in a reaction is the same whether the reaction occurs in one step or a series of steps. This observation is called Hess’s law. This is an application of the concept that in thermodynamics, the system does not care how many steps (individual reactions) are required to reach equilibrium – just the final reaction.

Another practical concern is that many compounds cannot directly synthesize from their elements. Sometimes, the reaction proceeds too slowly, or side reactions produce substances other than the desired compound. In these cases, \(\Delta_f H^\circ\) can be determined indirectly using Hess’s law.

Example: Hess’s law

Example 1

Combustion of graphite (\(\ce{C}\)) to can occur as follows:

\[ \ce{ C(s) + 1/2 O2(g) -> CO(g)} \]

This reaction can be created by combining two separate reactions shown below:

Reaction	\(\Delta_{rxn}H^\circ\), \(\pu{kJ mol-1}\)
\(\ce{CO2(g) -> CO(g) + 1/2 O2(g)}\)	\(-283.0\)
\(\ce{C(s) + O2(g) -> CO2(g)}\)	\(-393.5\)

We can add balanced chemical equations similar to adding algebraic equalities and canceling identical items on opposite sides of the equation arrow. To combine the above two reactions, we must reverse the first reaction to get common terms on opposite sides. Also, remember to change the sign of \(\Delta_{rxn}H^\circ\) when reversing reactions.

Reaction	\(\Delta_{rxn}H^\circ\), \(\pu{kJ mol-1}\)
\(\ce{CO2(g) -> CO(g) + 1/2 O2(g) }\)	\(-283.0\)
\(\ce{C(s) + O2(g) -> CO2(g)}\)	\(-393.5\)
—	—
\(\ce{C(s) + 1/2 O2(g) -> CO(g)}\)	\(-110.5\)

Therefore, \(\Delta_{rxn}H^\circ = \pu{-110.5 kJ mol-1}\) for the final reaction.

Example 2

Let’s determine \(\Delta_{rxn}H^\circ\) for the reaction below using \(\Delta_{rxn}H^\circ\) data for the following reactions:

\[ \ce{ C(s) + 2 H2(g) + 1/2 O2(g) -> CH3OH(l) }\]

Reaction	\(\Delta_{rxn}H^\circ\), \(\pu{kJ mol-1}\)
\(\ce{CH3OH(l) + 3/2 O2(g) -> CO2(g) + 2 H2O(l)}\)	\(-726.4\)
\(\ce{C(s) + O2(g) -> CO2(g)}\)	\(-393.5\)
\(\ce{H2(g) + 1/2 O2(g) -> H2O(l)}\)	\(-285.8\)

The following changes were made to each of the above reactions to get common reactants and products to match our required reaction:

1. Rxn 1: Reversed, the sign of \(\Delta_{rxn}H^\circ\) changed

2. Rxn 2: No change

3. Rxn 3: Multiplied by 2, including \(\Delta_{rxn}H^\circ\) value

Reaction	\(\Delta_{rxn}H^\circ\), \(\pu{kJ mol-1}\)
\(\ce{CO2(g) + 2 H2O(l) -> CH3OH(l) + 3/2 O2(g) }\)	\(726.4\)
\(\ce{C(s) + O2(g) -> CO2(g)}\)	\(-393.5\)
\(\ce{2 H2(g) + O2(g) -> 2 H2O(l)}\)	\(-571.6\)
—	—
\(\ce{C(s) + 2 H2 (g) + 1/2 O2(g) -> CH3OH(l)}\)	\(-238.7\)

Therefore, \(\Delta_{rxn}H^\circ = \pu{-238.7 kJ mol-1}\) for the final reaction.