Question

Using values from Appendix \(\mathrm{C}\) , calculate the standard enthalpy change for each of the following reactions: $$ \begin{array}{l}{\text { (a) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\ {\text { (b) } \mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (c) } \mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)}\end{array} $$

Short answer

Using the values from Appendix C and applying the equation ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants), we can calculate the standard enthalpy changes as follows: (a) ΔH° = [2 x ΔHf°(SO_3) - 2 x ΔHf°(SO_2)] = -198 kJ/mol (b) ΔH° = [ΔHf°(MgO) + ΔHf°(H_2O) - ΔHf°(Mg(OH)_2)] = -37 kJ/mol (c) ΔH° = [4 x ΔHf°(H_2O) - ΔHf°(N_2O_4)] = -104 kJ/mol (d) ΔH° = [(ΔHf°(SiO_2) + 4 x ΔHf°(HCl)) - (ΔHf°(SiCl_4) + 2 x ΔHf°(H_2O))] = -187 kJ/mol

Step-by-step solution

Determine the enthalpy of formation for each reactant and product

Consult Appendix C to find the standard enthalpies of formation for each reactant and product involved in the reactions. These values are given in kJ/mol.

Calculate the standard enthalpy change for each reaction

The standard enthalpy change for a reaction is calculated using the following equation: ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants) where ΔH° is the standard enthalpy change for the reaction, ΔHf° is the standard enthalpy of formation, n and m are the coefficients of the products and reactants, respectively. For each reaction, substitute the enthalpy of formation values for the reactants and products, along with their coefficients (n and m), into the equation and calculate the standard enthalpy change. (a) \(2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)\) ΔH° = [2 x ΔHf°(SO_3) - (2 x ΔHf°(SO_2) + ΔHf°(O_2))] Since \(ΔHf°(O_2)=0\), (Oxygen is in its standard state) ΔH° = [2 x ΔHf°(SO_3) - 2 x ΔHf°(SO_2)] (b) \(Mg(OH)_2(s) \rightarrow MgO(s) + H_2O(l)\) ΔH° = [ΔHf°(MgO) + ΔHf°(H_2O) - ΔHf°(Mg(OH)_2)] (c) \(N_2O_4(g) + 4H_2(g) \rightarrow N_2(g) + 4H_2O(g)\) ΔH° = [(ΔHf°(N_2) + 4 x ΔHf°(H_2O)) - (ΔHf°(N_2O_4) + 4 x ΔHf°(H_2))] Since \(ΔHf°(H_2)=0\) and \(ΔHf°(N_2)=0\), (Hydrogen and Nitrogen are in their standard states) ΔH° = [4 x ΔHf°(H_2O) - ΔHf°(N_2O_4)] (d) \(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(g)\) ΔH° = [(ΔHf°(SiO_2) + 4 x ΔHf°(HCl)) - (ΔHf°(SiCl_4) + 2 x ΔHf°(H_2O))]

Calculate the standard enthalpy changes for each reaction

Using the values from Appendix C and the calculations in Step 2, compute the standard enthalpy changes for each of the four reactions.

Key concepts

Enthalpy of Formation

Understanding the enthalpy of formation is fundamental for anyone delving into thermochemistry and chemical reactions. It's defined as the heat change that occurs when one mole of a compound is formed from its elements in their standard states. The standard state refers to the physical state of an element or compound under standard conditions, typically at 1 atmospheric pressure and 25°C (298 K).

For instance, when water is formed from hydrogen and oxygen gas, the enthalpy of formation would represent the energy change during that process. The abbreviation for the standard enthalpy of formation is \( \Delta H_f^\circ \). It's crucial to note that for elements in their natural forms, such as O2 gas or solid iron, the enthalpy of formation is zero since there is no formation process involved—they are already in their standard states.

In a classroom or homework setting, this concept empowers students to calculate the energy involved in chemical reactions. This is particularly helpful when the reaction occurs in a closed system and energy transfer to the surroundings is minimal, driving the need to comprehend all aspects of enthalpy.

Thermochemistry

Moving deeper into our understanding, thermochemistry is the study of the energy and heat associated with chemical reactions and physical transformations. It's a branch of thermodynamics that deals with the relationship between heat changes and chemical reactions. In the realm of chemistry, thermochemistry allows us to understand processes such as how much energy is released when fuels burn or absorbed when water evaporates.

By conducting calorimetric experiments, we can measure the heat changes that occur during a reaction. Analyzing these thermal transactions gives students a clearer picture of how energy is conserved and transformed, following the first law of thermodynamics—energy cannot be created or destroyed, only converted from one form to another.

These thermal analyses are not just academic. They have real-world applications in understanding the energy dynamics of processes, including those in biological systems, industrial manufacturing, and environmental science.

Chemical Reactions

At the heart of chemistry lies chemical reactions, which are processes that involve the rearrangement of atoms to form new substances. Each chemical reaction has a specific set of starting substances, called reactants, and produces new substances called products. Understanding the intricate details of how these reactions occur, and quantifying the energy exchanged during the process, is essential for making predictions about the behavior of matter.

Chemical equations are shorthand representations of chemical reactions, where reactants are written on the left side and products on the right. These equations are balanced according to the law of conservation of mass, ensuring that the number of atoms for each element is equal on both sides. Analyzing these reactions provides insights into the nature of the reactants and products, as well as the conditions necessary for the reaction to occur, such as the presence of a catalyst or certain temperature and pressure conditions.

Hess's Law

Lastly, we arrive at Hess's law, a principle that serves as a powerful tool in calculating the enthalpy changes of reactions. Named after Germain Hess, it 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 is a direct consequence of enthalpy being a state function; that is, its value depends only on the initial and final states of the system, not on the path taken.

Hess's law enables us to calculate the enthalpy change of a complicated reaction by breaking it down into a series of simpler steps for which enthalpy changes are known. One common application is in determining enthalpy changes for reactions where direct measurement is difficult. Thus, Hess's law becomes a valuable piece of the puzzle in solving for unknown enthalpies and understanding the broader picture of energy flow in chemical processes.