Endothermic and Exothermic Processes

Define Endothermic and Exothermic Processes

Let's start of by defining some key background terms:

There are three main types of Thermodynamic processes:

• System change - the passage of a system from some initial state to a final state of thermodynamic equilibrium.

• Cycles within a system - a process that begins with an initial state then moves the system through cyclic stages and ends with a final state.

• Flow processes - for a given system, flow processes account for the flow of matter into and out of the system and account for the transfer of heat, work, kinetic and potential energies across the walls of the container.

Distinguish between Exothermic and Endothermic Processes

What is an endothermic and an exothermic process? How do endothermic and exothermic processes differ?

Both endothermic and exothermic processes involve the transfer of energy in the form of heat.

• Endothermic processes are those which absorb energy as heat.
• Exothermic processes are those which release energy as heat.

Examples of Endothermic and Exothermic Processes

Let's classify the additional examples of processes as exothermic and endothermic.

1. Almost all of the machines currently in use, and the engines that drive them, are a result of the application of thermodynamics and involve endothermic and exothermic processes.

Figure 1: Thermodynamic model of an engine.

2. Chemistry labs are actually factories that manipulate heat to create substances with new properties and possibilities. Heat is a form of energy that is absorbed or released in chemical reactions. It is this energy, in the form of heat, that transforms one substance into another.

Figure 2: Modern chemistry laboratory

Energy Changes in Chemical Reactions: Exothermic and Endothermic Processes

Consider the following enthalpy, or potential energy, diagrams for a set of hypothetical chemical reactions:

1. An endothermic chemical reaction:

Figure 3: Enthalpy diagram, endothermic reaction.

2. An exothermic chemical reaction:

Figure 4: Enthalpy diagram, exothermic reaction.

The above energy diagrams graph the enthalpy, or potential energies, associated with products and reactants. Reactants change to products through a reaction pathway that involves the addition of kinetic energy, in the form of heat, to the system. The enthalpy of the formation of a chemical substance can be viewed as being equivalent to the potential energy that is stored as heat within the chemical bonds of a compound. We note that:

• In an exothermic chemical reaction, reactants have higher potential energy than products, and the enthalpy of the reaction is negative, -ΔH. (Energy is released to the surroundings)

• In an endothermic chemical reaction, products have more potential energy than reactants, and the enthalpy of the reactions is positive, +ΔH. (Energy is absorbed from the surroundings)

Why are some solvation processes endothermic and others exothermic?

When we look at the process of solvation, the process can be endothermic or exothermic.

Figure 5: Model of how a solvent dissolves a solute to form a solution.

In the above figure, we note that the solute is often a salt which is held together by ionic bonds in a regularly ordered array (crystal lattice). On the other hand, within the solvent there are comparatively weaker interactions, such as hydrogen bonds, dipole-dipole interactions and London forces. In addition, the solvent is less ordered in some sense when compared to the solute.

The first step in the process of dissolving the solute into the solvent would involve the breaking of an ionic bond between a cation (positively charged) and an anion (negatively charged). This is accomplished by replacing each ionic bond with multiple solvent-solute interactions. This step in the solvation process involves the breaking of a strong bond by multiple weak interactions with the solvent and as a result is heat-releasing, or an exothermic reaction.

1. For example, consider the dissolution of sodium hydroxide, NaOH, in water.

$$NaOH_{(s)} + H_2O_{(l)} \rightarrow Na^+_{(aq)} + OH^-_{(aq)} + H_2O_{(l)}$$

ii. This is a strongly exothermic process and results from the breaking of the sodium hydroxide ionic bond by multiple interactions with water:

Figure 6: Model of dissolution of sodium hydroxide in water.

iii. The overall process of the dissolution of sodium hydroxide in water can also be depicted in an enthalpy diagram:

Figure 7: Enthalpy diagram for the dissolution of sodium hydroxide.

Now, let's consider the thermochemistry for the balanced chemical equation for the dissolution of sodium hydroxide in water:

$$NaOH_{(s)} + H_2O_{(l)} \rightarrow Na^+_{(aq)} + OH^-_{(aq)} + H_2O_{(l)}$$

Or just,

$$NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^-_{(aq)}$$

Let's calculate the final temperature of the water for this reaction.

Given that the molar heat of solution (ΔH_solution ) of sodium hydroxide (NaOH) is ΔH_solution = -44.51 kJ/mol, we want to calculate the final temperature of the water (T_f ) contained in an insulated container, after 45 grams of NaOH is dissolved into it.

i. Let the initial temperature of the water be 20.0°C and a the volume of the water be 1L. The formula for the temperature difference (ΔT ) for the dissolution of NaOH in water is:

$$\Delta T=\frac{\Delta H_{solution\,NaOH}}{C_{p\,(H_2O)}*m_{tot}}$$

Where:

• ΔT is the temperature difference,
• ΔH_{solution, NaOH} is the molar heat of solution for sodium hydroxide
• c_p (H₂O) is the specific heat of water
• m_tot is the mass of water plus the mass of NaOH.

First, we calculate the molar heat of solution for 45 grams of sodium hydroxide:

$$\Delta H_{solution\,NaOH}=(45\,g\,NaOH)*\frac{1\,mol\,NaOH}{40.00\frac{g}{mol}}*\frac{-44.51\,kj}{1\,mol\,NaOH}*\frac{1000\,J}{1\,kJ}=-5.564x10^4\,J$$

Where the molar mass of sodium hydroxide is 40.00 g/mol.

Notice, that the molar heat of solution for this reaction (ΔH_{solution, NaOH} = -55.64 kJ/mol) is negative, which shows that the reaction is exothermic.

ii. Now we insert this value of the molar heat of solution for 45 grams of sodium hydroxide into the temperature difference formula:

$$\Delta T=\frac{5.564x10^4\,J}{(4.18\frac{J}{g\,^\circ C})*(1000\,g\,H_2O+45\,g\,NaOH)}=12.7^\circ C$$

The thermochemistry data² is included in the table below:

$$\Delta H^\circ =[2*\Delta H^\circ_{f\,(g)\,NH_3}+10*\Delta H^\circ_{f\,(l)\,H_2O}+\Delta H^\circ_{f\,(s)\,BaCl_2}]-[\Delta H^\circ_{f\,(s)\,Ba(OH)_2*8H_2O}+2*\Delta H^\circ_{f\,(s)\,NH_4Cl}]$$

Then, inserting the table values:

$$\Delta H^\circ=[2*(-46\frac{kJ}{mol})+10*(-286\frac{kJ}{mol})+(-859\frac{kJ}{mol})]-[(-3345\frac{kJ}{mol})+2*(-314\frac{kJ}{mol}]=162\frac{kJ}{mol}$$

Notice, that the positive value for the enthalpy of this reaction, ΔH° = +162 kJ/mol, shows that the reaction is endothermic.

ii. Calculation of the reaction entropy, ΔS° :

$$\Delta S^\circ = [2*\Delta S^\circ_{(g)\,NH_3}+10*\Delta S^\circ_{(l)\,H_2O}+\Delta S^\circ_{(s)\,BaCl_2}]-[\Delta S^\circ_{(s)\,Ba(OH)_2*8H_2O}+2*\Delta S^\circ_{(s)\,NH_4Cl}]$$

Then, inserting the table values:

$$\Delta S^\circ = [2*(0.192\frac{kJ}{K*mol}+10*(0.070\frac{kJ}{K*mol})+(0.124\frac{kJ}{K*mol})]-[(0.427\frac{kJ}{K*mol})+2*(0.095\frac{kJ}{K*mol})]=0.591\frac{kJ}{K*mol}$$

iii. Lastly, we calculate the Gibbs free energy difference, ΔGº :

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 162\frac{kJ}{mol}-(298\,K*0.591\frac{kJ}{K*mol})=-14.12\frac{kJ}{mol}$$

Notice, that the negative value for the Gibbs free energy difference of this reaction, ΔG° = -14.12 kJ/mol, shows that the reaction is spontaneous, or favorable.

Changes in the Enthalpy of Exothermic and Endothermic Reactions

Can a process be exothermic and endothermic? Answer: Yes, a process can contain a series of chemical reactions that are individually either endothermic or exothermic. For example, the biochemical process of glycolysis includes at least 12 different reactions, a few of which are endothermic with the majority of the other reactions in process being exothermic.