Question

Using values from Appendix \(C\), calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(4 \mathrm{HBr}(\mathrm{g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Br}_{2}(l)\) (b) \(2 \mathrm{Na}(\mathrm{OH})(s)+\mathrm{SO}_{3}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+4 \mathrm{HCl}(g)\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

The standard enthalpy changes for the given reactions are: (a) \(\Delta H^{\circ} = -426.0\,\text{kJ/mol}\) (b) \(\Delta H^{\circ} = -379.0\,\text{kJ/mol}\) To find the standard enthalpy changes for reactions (c) and (d), follow the same steps as outlined above using the appropriate values from Appendix C.

Step-by-step solution

Obtaining \(\Delta H^{\circ}\) of formation values from Appendix C

Using Appendix C, look up the values of standard enthalpy of formation for reactants and products: \(\Delta H^{\circ}_{\text{formation}}\) $(\text{HBr} \text{(g)}) = -36.4\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}\) $(\text{O}_{2} \text{(g)}) = 0\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}\) $(\text{H}_{2} \text{O(l)}) = -285.8\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}\) $(\text{Br}_{2} \text{(l)}) = 0\,\text{kJ/mol}\)

Applying the \(\Delta H^{\circ}\) formula

Use the obtained values to calculate \(\Delta H^{\circ}\) for the reaction: \(\Delta H^{\circ} = [(2 \times (-285.8)) + (2 \times 0)] - [(4 \times (-36.4)) + (1 \times 0)]\)

Calculating the value of \(\Delta H^{\circ}\)

After substituting the values, perform the calculation to obtain the value of \(\Delta H^{\circ}\): \(\Delta H^{\circ} = [-571.6 + 0] - [-145.6 + 0] = -571.6 + 145.6 = -426.0\,\text{kJ/mol}\) The standard enthalpy change for reaction (a) is \(-426.0 \,\text{kJ/mol}\). Now, we will repeat Steps 1-3 for the remaining reactions. (b) \(2 \mathrm{Na}(\mathrm{OH})(s)+\mathrm{SO}_{3}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g)\)

Obtaining \(\Delta H^{\circ}\) of formation values from Appendix C

Using Appendix C, look up the standard enthalpy of formation values for the reactants and products. (We already have the value for H2O(g)): \(\Delta H^{\circ}_{\text{formation}}(\text{NaOH (s)})= -425.6\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}(\text{SO}_{3} \text{(g)})= −395.7\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}(\text{Na}_{2}\text{SO}_{4}(s))= -1384.1\,\text{kJ/mol}\) \(\Delta H^{\circ}_{\text{formation}}(\text{H}_{2}\text{O(g)})= −241.8\,\text{kJ/mol}\)

Applying the \(\Delta H^{\circ}\) formula

Use the obtained values to calculate \(\Delta H^{\circ}\) for the reaction: \(\Delta H^{\circ} = [(1 \times (-1384.1)) + (1 \times (-241.8))] - [(2 \times (-425.6)) + (1 \times (-395.7))]\)

Calculating the value of \(\Delta H^{\circ}\)

After substituting the values, perform the calculation to obtain the value of \(\Delta H^{\circ}\): \(\Delta H^{\circ} = [-1384.1 - 241.8] - [-851.2 -395.7] = -1625.9 + 1246.9 = -379.0\,\text{kJ/mol}\) The standard enthalpy change for reaction (b) is \(-379.0\,\text{kJ/mol}\). Perform these steps similarly for reactions (c) and (d) and find the respective standard enthalpy changes.

Key concepts

Enthalpy of Formation

Understanding the concept of enthalpy of formation is essential for analyzing chemical reaction energies. This term, also denoted as standard enthalpy of formation (\( \Delta H^\circ_{\text{formation}} \)), refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. It's important to note that the standard state for a substance is its most stable form at 1 atm pressure and a specified temperature, usually 25°C (298 K).

The standard enthalpy of formation for an element in its standard state is zero by definition. That is why in the reaction (a) \( \Delta H^\circ_{\text{formation}} \) for \( \text{O}_{2}(g) \) and \( \text{Br}_{2}(l) \) is zero. This reference point allows us to calculate the enthalpy change of a reaction using the values for individual compounds, which can be found in tables like Appendix C.

For instance, to find the reaction's enthalpy change, we take the sum of the standard enthalpies of formation of the products and subtract the sum of the standard enthalpies of formation of the reactants. It is a direct application of Hess's Law, which states that the total enthalpy change during the course of a chemical reaction is the same whether the reaction is made in one step or several steps. This process of calculating enthalpy changes using enthalpy of formation is crucial in thermochemistry and serves as a valuable tool for scientists and engineers.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the relationships between chemical reactions and energy changes involving heat. Central to this study is the concept of standard enthalpy change for reactions, including enthalpy of formation. The standard enthalpy change \( (\Delta H^\circ) \) for a reaction is the amount of heat absorbed or released under standard conditions, which include a temperature of 298 K and a pressure of 1 atm.

When we assess a chemical process such as reaction (a) \(4 \mathrm{HBr}(\mathrm{g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Br}_{2}(l)\), thermochemistry allows us to calculate how much energy the reaction will release or consume. This energy change is conveyed as the heat change accompanying the reaction at constant pressure, indicating it’s exothermic (releases heat) or endothermic (absorbs heat).

Thermochemistry provides the tools to understand and quantify the energy flow in chemical reactions, which is fundamental for developing new materials, managing energy resources, and understanding environmental impact. The discipline applies laws of thermodynamics, which involves concepts such as enthalpy, entropy, and Gibbs free energy to predict whether a reaction will occur spontaneously.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. The substances you start with, the reactants, undergo a chemical change to become the products. In a balanced chemical equation, the number of atoms for each element is conserved from reactants to products, adhering to the law of conservation of mass.

For example, in reaction (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\), the reactants \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{HCl}\) are chemically altered to form \(\mathrm{FeCl}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) as products.

A key aspect of studying chemical reactions is understanding the energy changes that accompany them, which we discuss using concepts like enthalpy. The breaking of bonds requires energy, while bond formation releases energy. In the context of a reaction, the overall energy involved is indicated by the reaction's enthalpy change. This information tells us whether a reaction will release heat to its surroundings or absorb it, ultimately determining the feasibility of reactions in natural and industrial processes.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's a cornerstone of chemistry, allowing one to predict the amounts of substances consumed and produced in a reaction. When we calculate the standard enthalpy change for a reaction, we must keep the stoichiometric coefficients in mind, as they dictate the proportions in which the chemical species react.

In the given exercises, the enthalpy change calculations require precise stoichiometric coefficients. For instance, in reaction (b) \(2 \mathrm{Na}(\mathrm{OH})(s)+\mathrm{SO}_{3}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g)\), the coefficients \(2\), \(1\), \(1\), and \(1\) determine the relative amounts of each substance. Stoichiometry is indispensable for informed scaling of reactions in industry to produce the desired quantity of a product. It's also critical in laboratories to prepare specific concentrations of solutions and to analyze reaction yields.

Moreover, stoichiometry isn't limited to balancing chemical equations. It also extends to conserving energy in reactions, as the stoichiometric coefficients directly affect the calculation of enthalpy changes, emphasizing their importance in thermochemistry and practical applications like energy production and materials synthesis.