Question

Using values from Appendix \(C\), calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Br}_{2}(l)\) (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 values of \(\Delta H^{\circ}\) for the four reactions are: a) \(-426.0\) kJ/mol b) \(-782.6\) kJ/mol c) \(-433.828\) kJ/mol d) \(-301.418\) kJ/mol

Step-by-step solution

a) 4 HBr(g) + O2(g) → 2 H2O(l) + 2 Br2(l)

For this reaction, we need the following enthalpy values: - \(\Delta H_{f}^{\circ}\)(HBr(g)) - \(\Delta H_{f}^{\circ}\)(H2O(l)) - \(\Delta H_{f}^{\circ}\)(Br2(l)) From Appendix C, we find the standard enthalpies of formation: - \(\Delta H_{f}^{\circ}(\text{HBr}(g))=-36.4\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{H}_{2}\text{O(l)})=-285.8\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{Br}_{2}\text{(l)})=0\) kJ/mol Applying the formula: \[ \begin{aligned} \Delta H^{\circ} &= \bigl[ 2 \cdot (-285.8) + 2 \cdot 0 \bigr] - \bigl[ 4 \cdot (-36.4) + 1 \cdot 0 \bigr] \\ &= -571.6 + 145.6 \\ &= -426.0 \text{ kJ/mol} \end{aligned} \]

b) 2 NaOH(s) + SO3(g) → Na2SO4(s) + H2O(g)

For this reaction, we need the following enthalpy values: - \(\Delta H_{f}^{\circ}\)(NaOH(s)) - \(\Delta H_{f}^{\circ}\)(SO3(g)) - \(\Delta H_{f}^{\circ}\)(Na2SO4(s)) - \(\Delta H_{f}^{\circ}\)(H2O(g)) From Appendix C, we find the standard enthalpies of formation: - \(\Delta H_{f}^{\circ}(\text{NaOH(s)})=-425.6\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{SO}_3(g))=-395.7\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{Na}_2\text{SO}_4(s))=-1387.1\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{H}_{2}\text{O(g)})=-241.8\) kJ/mol Applying the formula: \[ \begin{aligned} \Delta H^{\circ} &= [(-1387.1) + 1 \cdot (-241.8)] - [2 \cdot (-425.6) + 1 \cdot (-395.7)] \\ &= -1628.9 + 846.3 \\ &= -782.6 \, \text{kJ/mol} \end{aligned} \]

c) CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)

For this reaction, we need the following enthalpy values: - \(\Delta H_{f}^{\circ}\)(CH4(g)) - \(\Delta H_{f}^{\circ}\)(Cl2(g)) - \(\Delta H_{f}^{\circ}\)(CCl4(l)) - \(\Delta H_{f}^{\circ}\)(HCl(g)) From Appendix C, we find the standard enthalpies of formation: - \(\Delta H_{f}^{\circ}(\text{CH}_4(g))=-74.8\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{Cl}_2(g))=0\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{CCl}_4(l))=-139.3\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{HCl(g)})=-92.307\) kJ/mol Applying the formula: \[ \begin{aligned} \Delta H^{\circ} &= [(-139.3) + 4\cdot(-92.307)] - [(-74.8) + 4\cdot 0] \\ &= -508.628 + 74.8 \\ &= -433.828 \, \text{kJ/mol} \end{aligned} \]

d) Fe2O3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2O(g)

For this reaction, we need the following enthalpy values: - \(\Delta H_{f}^{\circ}\)(Fe2O3(s)) - \(\Delta H_{f}^{\circ}\)(HCl(g)) - \(\Delta H_{f}^{\circ}\)(FeCl3(s)) - \(\Delta H_{f}^{\circ}\)(H2O(g)) From Appendix C, we find the standard enthalpies of formation: - \(\Delta H_{f}^{\circ}(\text{Fe}_2\text{O}_3(s))=-824.2\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{HCl(g)})=-92.307\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{FeCl}_3(s))=-400.8\) kJ/mol - \(\Delta H_{f}^{\circ}(\text{H}_{2}\text{O(g)})=-241.8\) kJ/mol Applying the formula: \[ \begin{aligned} \Delta H^{\circ} &= [2\cdot(-400.8) + 3\cdot(-241.8)] - [(-824.2) + 6\cdot(-92.307)] \\ &= -1685.4 + 1383.982 \\ &= -301.418 \, \text{kJ/mol} \end{aligned} \] Thus, the values of \(\Delta H^{\circ}\) for the four reactions are: a) -426.0 kJ/mol b) -782.6 kJ/mol c) -433.828 kJ/mol d) -301.418 kJ/mol

Key concepts

Standard Enthalpy of Formation

The concept of standard enthalpy of formation, denoted by \text{\(\Delta H_f^\circ\)}\text{, is a fundamental principle in thermochemistry. It refers to the change in enthalpy when one mole of a compound is formed from its elements in their most stable states under standard conditions (1 atm and 298.15 K). For an element in its standard state, like \text{\(O_2(g)\)} or \text{\(Br_2(l)\)}, the standard enthalpy of formation is zero, since the element is already in its most stable form.

To determine the enthalpy change for a reaction using standard enthalpies of formation, we apply the following equation:}\text{\[\Delta H^\circ = \sum (\Delta H_f^\circ \text{ of products}) \times \text{ their coefficients} - \sum (\Delta H_f^\circ \text{ of reactants}) \times \text{ their coefficients}\]}\text{In this equation, we sum the products' standard enthalpies of formation (with their stoichiometric coefficients in the reaction) and subtract the sum of the reactants' standard enthalpies, also considering their stoichiometric coefficients. This method is directly applied in solving the exercise provided, leading to an easy-to-understand calculation of the overall enthalpy change for the reaction.

One piece of advice to improve understanding is to always double-check the balance of the reaction before applying the values of \text{\(\Delta H_f^\circ\)}. A balanced equation ensures that the stoichiometry is accurate, which is essential for correct calculations.}

Thermochemical Equations

Thermochemical equations are representations of chemical reactions that include the change in enthalpy (\text{\(\Delta H\)}) as part of the equation. They provide crucial information on heat exchange during the reaction, allowing scientists and engineers to understand the energy implications of reactions.

Each thermochemical equation is associated with a specific quantity of reactants and products, indicating that the enthalpy change is dependent on the amounts in which substances react and form. It's important to note that if the coefficients in a thermochemical equation are multiplied by a factor, the value of \text{\(\Delta H\)} must be multiplied by the same factor. Similarly, when a reaction is reversed, the sign of \text{\(\Delta H\)} is also reversed. This characteristic is captured in exercises like the one mentioned, where the \text{\(\Delta H^\circ\)} of a reaction is deduced from known enthalpies of formation.

To boost comprehension, it's helpful for students to practice writing and balancing thermochemical equations themselves, taking care to correctly signify whether heat is released (exothermic, \text{\(\Delta H < 0\)}) or absorbed (endothermic, \text{\(\Delta H > 0\)}) in the course of the reaction.}

Hess's Law

A pinnacle concept in understanding enthalpy changes is Hess's Law, which asserts that the total enthalpy change for a chemical reaction is the same, no matter how the reaction occurs, as long as the initial and final conditions are identical. In other words, enthalpy is a state function—it only depends on the initial and final states, not on the path taken to get there.

Hess's Law is particularly useful when the enthalpy change of a reaction is difficult to measure directly. By combining a series of intermediate steps for which \text{\(\Delta H\)} is known, the overall enthalpy change for the reaction can be determined. In the context of the exercise, this law underpins the logic for calculating \text{\(\Delta H^\circ\)} from standard enthalpies of formation, as it involves a hypothetical series of steps leading to the formation of products from the most stable forms of elements.

For a better grasp of Hess's Law, students can benefit from practicing with imaginary intermediate steps or reactions, seeing how they add up to the same total enthalpy change for the direct reaction. This will reinforce the concept that the route taken in a chemical transformation does not alter the quantity of energy involved, only the distribution of steps.