Question

For each of the following compounds, write a balanced thermochemical equation depicting the formation of one mole of the compound from its elements in their standard states and then look up \(\Delta H^{\circ} f\) for each substance in Appendix C. (a) \(\mathrm{NO}_{2}(g),(\mathbf{b}) \mathrm{SO}_{3}(g),(\mathbf{c}) \mathrm{NaBr}(s),(\mathbf{d}) \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) .\)

Short answer

The balanced thermochemical equations are: a) \( \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow NO_{2(g)} \) with \(\Delta H^{\circ} f(NO_{2(g)}) = 33.18 \, kJ/mol\) b) \( S_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow SO_{3(g)} \) with \(\Delta H^{\circ} f(SO_{3(g)}) = -395.72 \, kJ/mol\) c) \( Na_{(s)} + \frac{1}{2}Br_{2(g)} \rightarrow NaBr_{(s)} \) with \(\Delta H^{\circ} f(NaBr_{(s)}) = -362.79 \, kJ/mol\) d) \( Pb_{(s)} + 2N_{2(g)} + 6O_{2(g)} \rightarrow Pb(NO_{3})_{2(s)} \) with \(\Delta H^{\circ} f(Pb(NO_{3})_{2(s)}) = -750.9 \, kJ/mol\)

Step-by-step solution

a) Formation of one mole of NO2(g)

The formation of one mole of NO2(g) consists of nitrogen and oxygen in their standard states. Nitrogen's standard state is N2(g), and oxygen's standard state is O2(g). The balanced thermochemical equation is: \( N_{2(g)} + O_{2(g)} \rightarrow 2NO_{2(g)} \) However, we have to depict the formation of one mole of the compound. Therefore, we have to divide the whole equation by 2: \( \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow NO_{2(g)} \) Next, look up the standard enthalpy of formation, \(\Delta H^{\circ} f\), for NO2(g). According to Appendix C, we have: \(\Delta H^{\circ} f(NO_{2(g)}) = 33.18 \, kJ/mol\)

b) Formation of one mole of SO3(g)

The formation of one mole of SO3(g) involves sulfur in its standard state, which is S(s), and oxygen gas. The balanced thermochemical equation is: \( S_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow SO_{3(g)} \) Next, look up the standard enthalpy of formation, for SO3(g). According to Appendix C, we have: \(\Delta H^{\circ} f(SO_{3(g)}) = -395.72 \, kJ/mol\)

c) Formation of one mole of NaBr(s)

The formation of one mole of NaBr(s) involves sodium metal in its standard state, which is Na(s), and Bromine gas in its standard state, Br2(g). The balanced thermochemical equation is: \( Na_{(s)} + \frac{1}{2}Br_{2(g)} \rightarrow NaBr_{(s)} \) Next, look up the standard enthalpy of formation for NaBr(s). According to Appendix C, we have: \(\Delta H^{\circ} f(NaBr_{(s)}) = -362.79 \, kJ/mol\)

d) Formation of one mole of Pb(NO3)2(s)

The formation of one mole of Pb(NO3)2(s) involves lead metal in its standard state, which is Pb(s), nitrogen gas, and oxygen gas. The balanced thermochemical equation is: \( Pb_{(s)} + 2N_{2(g)} + 6O_{2(g)} \rightarrow Pb(NO_{3})_{2(s)} \) However, we have to depict the formation of one mole of the compound. Therefore, we have to divide the whole equation by 2: \( Pb_{(s)} + 2N_{2(g)} + 6O_{2(g)} \rightarrow Pb(NO_{3})_{2(s)} \) Next, look up the standard enthalpy of formation for Pb(NO3)2(s). According to Appendix C, we have: \(\Delta H^{\circ} f(Pb(NO_{3})_{2(s)}) = -750.9 \, kJ/mol\)

Key concepts

Standard State

In chemistry, the standard state of an element or compound is a reference point used to define its properties under specific conditions. The standard state refers to the most stable physical form of the element or compound at one atmosphere of pressure (101.325 kPa) and a specified temperature, usually 25°C (298.15 K). For example, the standard state of oxygen is O₂(g) gas, while for carbon it's graphite, not diamond.

To make thermodynamic calculations simple and consistent, standard states are crucial. When we describe reactions or formations, we're often referring to the process occurring while all reactants and products are in their standard states. This is important for calculating standard enthalpies of formation, as seen in the exercise.

Understanding the standard states is necessary to correctly write thermochemical equations, as it reflects the baseline from which enthalpy changes are measured, ensuring accurate comparison of data across different sources and experiments.

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \(\Delta H^{\circ} f\), is a fundamental quantity in chemical thermodynamics. It is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. For elements in their standard states, the standard enthalpy of formation is zero by definition. This provides a reference point to gauge the energy required to form different substances.

For instance, the positive \(\Delta H^{\circ} f\) value for \( NO_{2(g)} \) suggests that energy is absorbed when NO₂ is formed from nitrogen and oxygen. Conversely, a negative \(\Delta H^{\circ} f\) value, as in the case of \( SO_{3(g)} \) and \( NaBr_{(s)} \), indicates that forming these compounds releases energy, showing an exothermic reaction.

To accurately convey the enthalpy of formation in thermochemical equations, it's essential to consider the stoichiometry. In our exercise, we adjust the coefficients to represent the formation of exactly one mole of product, which allows us to directly apply the \(\Delta H^{\circ} f\) values from the reference tables.

Chemical Thermodynamics

Chemical thermodynamics is the subsection of thermodynamics that deals with the energy and work aspects of chemical reactions. It involves studying the enthalpy changes accompanying chemical processes, which can tell us if a reaction is favorable or spontaneous. This field of study provides insights on reaction conditions, efficiency, and the direction in which a reaction will proceed.

In the context of the exercise, we use chemical thermodynamics to understand the energy changes during the formation of compounds from their elements. When we assess the enthalpy changes, we can predict whether the compounds will form easily or not. This is pivotal for industries like pharmaceuticals and material science, where understanding energy changes leads to optimizing product synthesis and improving efficiency.

Furthermore, applying chemical thermodynamics principles can help us make environmental and cost-effective decisions by choosing reactions that release minimal waste heat and require less energy input.

Enthalpy Change

Enthalpy change, represented as \(\Delta H\), measures the heat energy released or absorbed in a chemical reaction at constant pressure. It is an essential concept in understanding how energy transformations are entwined with chemical reactions. An endothermic process, indicated by a positive enthalpy change, absorbs heat, while an exothermic process, with a negative enthalpy change, releases heat.

The enthalpy change of a reaction can be calculated using the sum of the standard enthalpies of formation of the products minus the reactants. In the exercise provided, \(\Delta H^{\circ} f\) values are given for each compound, which serves as a foundational data for understanding the thermochemical equations of their formation.

Recognizing how to manipulate and interpret enthalpy changes is a key to mastering energy aspects in chemical reactions. This knowledge not only explains reactions on a molecular level but also aids in making predictions about reaction conditions and the amounts of energy that might be harnessed or required in industrial processes.