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 \(\mathrm{C}\). (a) \(\mathrm{NO}_{2}(g),\) (b) \(\mathrm{SO}_{3}(g),\) (c) \(\mathrm{NaBr}(s),\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s).\)

Short answer

(a) \(\frac{1}{2}N_{2}(g) + O_{2}(g) \rightarrow NO_{2}(g)\), \(\Delta H^{\circ}{ }_{f} = +33.2\,\mathrm{kJ\,mol^{-1}}\) (b) \(S(s) + \frac{3}{2}O_{2}(g) \rightarrow SO_{3}(g)\), \(\Delta H^{\circ}{ }_{f} = -395.7\,\mathrm{kJ\,mol^{-1}}\) (c) \(Na(s) + \frac{1}{2}Br_{2}(l) \rightarrow NaBr(s)\), \(\Delta H^{\circ}{ }_{f} = -362.7\,\mathrm{kJ\,mol^{-1}}\) (d) \(Pb(s) + N_{2}(g) + 3\,O_{2}(g)\rightarrow Pb(NO_{3})_{2}(s)\), \(\Delta H^{\circ}{ }_{f} = -765.6\,\mathrm{kJ\,mol^{-1}}\)

Step-by-step solution

(a) Formation of NO2(g)

: From its elements, Nitrogen and Oxygen, we can write the balanced thermochemical equation for the formation of one mole of \(\mathrm{NO}_{2}(g)\) as: \[ \frac{1}{2}N_{2}(g) + O_{2}(g) \rightarrow NO_{2}(g) \] To find the \(\Delta H^{\circ}{ }_{f}\) for one mole of \(\mathrm{NO}_{2}(g)\), refer to Appendix \(\mathrm{C}\) and look for the value. You should find that the \(\Delta H^{\circ}{ }_{f}\) for \(\mathrm{NO}_{2}(g)\) is \( +33.2\,\mathrm{kJ\,mol^{-1}} \).

(b) Formation of SO3(g)

: From its elements, Sulfur and Oxygen, we can write the balanced thermochemical equation for the formation of one mole of \(\mathrm{SO}_{3}(g)\) as: \[ S(s) + \frac{3}{2}O_{2}(g) \rightarrow SO_{3}(g) \] To find the \(\Delta H^{\circ}{ }_{f}\) for one mole of \(\mathrm{SO}_{3}(g)\), refer to Appendix \(\mathrm{C}\) and look for the value. You should find that the \(\Delta H^{\circ}{ }_{f}\) for \(\mathrm{SO}_{3}(g)\) is \( -395.7\,\mathrm{kJ\,mol^{-1}} \).

(c) Formation of NaBr(s)

: From its elements, Sodium and Bromine, we can write the balanced thermochemical equation for the formation of one mole of \(\mathrm{NaBr}(s)\) as: \[ Na(s) + \frac{1}{2}Br_{2}(l) \rightarrow NaBr(s) \] To find the \(\Delta H^{\circ}{ }_{f}\) for one mole of \(\mathrm{NaBr}(s)\), refer to Appendix \(\mathrm{C}\) and look for the value. You should find that the \(\Delta H^{\circ}{ }_{f}\) for \(\mathrm{NaBr}(s)\) is \( -362.7\,\mathrm{kJ\,mol^{-1}} \).

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

: From its elements, Lead, Nitrogen, and Oxygen, we can write the balanced thermochemical equation for the formation of one mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\) as: \[ Pb(s) + 2\,NO_{3}^{-}(aq) \rightarrow Pb(NO_{3})_{2}(s) \] Please note that nitrate ions \(\mathrm{NO}_{3}^{-}\) are formed from Nitrogen and Oxygen: \[ 2\,N(g) + 6\,O_{2}(g) \rightarrow 4\,NO_{3}^{-}(aq) \] Thus, for the balanced thermochemical equation for \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\), we have: \[ Pb(s) + N_{2}(g) + 3\,O_{2}(g)\rightarrow Pb(NO_{3})_{2}(s) \] To find the \(\Delta H^{\circ}{ }_{f}\) for one mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\), refer to Appendix \(\mathrm{C}\) and look for the value. You should find that the \(\Delta H^{\circ}{ }_{f}\) for \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\) is \( -765.6\,\mathrm{kJ\,mol^{-1}} \).

Key concepts

Enthalpy of Formation

The enthalpy of formation, often symbolized as \(\Delta H^{\circ}_{f}\), is an important concept in thermochemistry. It refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This is crucial when understanding how heat energy is absorbed or released during chemical reactions.

In practical terms, the enthalpy of formation helps predict whether a chemical reaction will be exothermic or endothermic. An exothermic reaction releases heat, indicated by a negative \(\Delta H^{\circ}_{f}\), while an endothermic reaction absorbs heat, shown by a positive \(\Delta H^{\circ}_{f}\). Knowing \(\Delta H^{\circ}_{f}\) values allows chemists to calculate the total energy absorbed or released in a reaction.

For example, the formation of \(\text{NO}_2(g)\) from its elements has an enthalpy of formation of \(+33.2\,\text{kJ}\,\text{mol}^{-1}\), indicating that this process absorbs energy from the surroundings.

Chemical Reactions

Chemical reactions are processes where substances transform into new substances by breaking and forming chemical bonds. This transformation involves reactants converting into products.

In thermochemistry, understanding chemical reactions requires writing balanced equations. A balanced thermochemical equation not only shows the reactants and products but also indicates the states of matter and energy changes. For instance, creating \(\text{SO}_3(g)\) from sulfur and oxygen involves a balanced equation: \[ S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \] This equation represents both the process and the energy change, where the heat given off (in this case \(-395.7\,\text{kJ}\,\text{mol}^{-1}\)) is part of the reaction’s thermochemical equation.

Understanding the energetics of chemical reactions allows for applications like designing energy-efficient industrial processes or predicting the stability of compounds during storage.

Standard States

Standard states refer to the physical and chemical properties of elements and compounds under a set of specific conditions, usually at 1 atm pressure and 25°C. These conditions provide a benchmark for measuring properties like enthalpy.

Every element has a specific standard state, often determined by its most stable form under these conditions. For example, for nitrogen, it's \(N_2(g)\), and for oxygen, it's \(O_2(g)\). Standard states are critical when formulating thermochemical equations because they ensure consistency across calculations.

When forming compounds like \(\text{NaBr}(s)\) from sodium and bromine, the elements are in their standard states: \( Na(s) \) and \( \frac{1}{2}Br_2(l) \). The standardization allows chemists to reliably compare enthalpy values across different substances and reactions.

Chemical Compounds

Chemical compounds are substances made up of two or more elements chemically bonded together. They exhibit properties different from the elements they are composed of.

• Compounds like \(\text{NaBr}(s)\) are created through the interaction of their constituent elements, sodium, and bromine.
• The properties of a compound are determined by its chemical structure and the nature of the bonds within it.
• For example, \(\text{Pb(NO}_3)_2(s)\) involves lead, nitrogen, and oxygen, forming a stable compound used in various industrial applications.

Understanding the formation of chemical compounds involves grasping concepts like bond energy, molecular structure, and stoichiometry. These concepts are critical in predicting how compounds will interact in different environments and in creating compounds with specific desired properties.

Overall, studying chemical compounds allows for advancements in fields from materials science to pharmaceuticals.