Problem 69 - Chemistry The Central Science In Si Units

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 \(\Delta H^{\circ}\) for each substance in Appendix \(\mathrm{C}\). (a) \(\mathrm{NO}_{2}(g)\), (b) \(\mathrm{SO}_{3}(g)\) (c) \(\mathrm{NaBr}(s)\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) .\)

Step-by-Step Solution

Answer

Formation equations and standard enthalpies: (a) NO2: +33.18 kJ/mol; (b) SO3: -395.7 kJ/mol; (c) NaBr: -360.1 kJ/mol; (d) Pb(NO3)2: -451.9 kJ/mol.

Step 1: Identify Elements in Standard States for NO2(g)

For the compound \( NO_2(g) \), the elements in their standard states are nitrogen \( N_2(g) \) and oxygen \( O_2(g) \). Both are gases at 1 atm and 25°C.

Step 2: Write Formation Equation for NO2(g)

The balanced thermochemical equation for the formation of one mole of \( NO_2(g) \) from its elements in their standard states is: \[\frac{1}{2} N_2(g) + O_2(g) \rightarrow NO_2(g) \]This reaction represents the formation of \( NO_2(g) \) from nitrogen and oxygen gases.

Step 3: Look up ΔH° for NO2(g)

The standard enthalpy of formation \( \Delta H^{\circ} \) for \( NO_2(g) \) is +33.18 kJ/mol.

Step 4: Identify Elements in Standard States for SO3(g)

For the compound \( SO_3(g) \), the elements in their standard states are sulfur \( S(s) \) and oxygen \( O_2(g) \). Sulfur is usually in a solid form (rhombic sulfur), and oxygen is a gas.

Step 5: Write Formation Equation for SO3(g)

The balanced thermochemical equation for the formation of one mole of \( SO_3(g) \) is: \[S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \]This reflects the composition of \( SO_3 \) from solid sulfur and gaseous oxygen.

Step 6: Look up ΔH° for SO3(g)

The standard enthalpy of formation \( \Delta H^{\circ} \) for \( SO_3(g) \) is -395.7 kJ/mol.

Step 7: Identify Elements in Standard States for NaBr(s)

For \( NaBr(s) \), the standard state elements are sodium \( Na(s) \) and bromine \( Br_2(l) \). Sodium is solid, and bromine is liquid at room temperature.

Step 8: Write Formation Equation for NaBr(s)

The balanced thermochemical equation for the formation of one mole of \( NaBr(s) \) is: \[Na(s) + \frac{1}{2} Br_2(l) \rightarrow NaBr(s) \]This illustrates the formation of sodium bromide from sodium metal and bromine liquid.

Step 9: Look up ΔH° for NaBr(s)

The standard enthalpy of formation \( \Delta H^{\circ} \) for \( NaBr(s) \) is -360.1 kJ/mol.

Step 10: Identify Elements in Standard States for Pb(NO3)2(s)

For \( Pb(NO_3)_2(s) \), the elements are lead \( Pb(s) \), nitrogen (as \( N_2(g) \)), and oxygen \( O_2(g) \). Lead is solid, nitrogen is gas, and oxygen is gas.

Step 11: Write Formation Equation for Pb(NO3)2(s)

The balanced thermochemical equation is laid out as follows:\[Pb(s) + N_2(g) + 3O_2(g) \rightarrow Pb(NO_3)_2(s) \]This demonstrates the compound's creation from elemental lead, nitrogen, and oxygen.

Step 12: Look up ΔH° for Pb(NO3)2(s)

The standard enthalpy of formation \( \Delta H^{\circ} \) for \( Pb(NO_3)_2(s) \) is -451.9 kJ/mol.

Key Concepts

Enthalpy of Formation

The enthalpy of formation, often represented as \( \Delta H^{\circ}_f \), is a crucial figure in chemistry that tells us the heat change when one mole of a compound is formed from its elements in their standard states.

It's essentially a snapshot of the energy landscape for chemical formation. When we talk about the enthalpy of formation, it's important to remember that it has units of kilojoules per mole (kJ/mol).

A positive \( \Delta H^{\circ}_f \) indicates that energy is absorbed during formation (endothermic process), while a negative value signifies energy release (exothermic process).

In thermochemical equations, the values for enthalpy of formation are critical in predicting whether a reaction will require or release energy.

Understanding these values helps chemists to design efficient chemical processes, understanding their environmental impact, and developing sustainable solutions.

Standard States of Elements

When elements are used in reactions, we refer to their standard states, which are their most stable forms at 1 atmosphere of pressure and 25°C (298 K).

This provides a common reference point when comparing energies and reactions. For example, oxygen is a gas \( O_2(g) \), nitrogen is also a gas \( N_2(g) \), and sulfur appears as solid rhombic sulfur \( S(s) \) in its standard state.

Recognizing these states is essential in writing accurate thermochemical equations, as these involve comparing reactions based on their elemental sources.

• Nitrogen: gas \( N_2(g) \)
• Oxygen: gas \( O_2(g) \)
• Sulfur: solid \( S(s) \)
• Sodium: solid \( Na(s) \)
• Bromine: liquid \( Br_2(l) \)

This consistency allows for accurate enthalpy comparisons and calculation, improving the precision of thermodynamic predictions in chemical reactions.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the same number of atoms for each element is present on both sides of the equation. This respects the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system.

When writing a balanced thermochemical equation for the formation of a compound, it's crucial to check:

• The total number of each type of atom is the same on both sides.
• The compounds and elements are correctly represented using their chemical formulas.
• Fractional coefficients can be used to accurately show the stoichiometry, especially when handling substances with diatomic molecules like \( \frac{1}{2} O_2(g) \) for oxygen.

Balancing is the first step in comparing the enthalpies of reactions, allowing for accurate thermodynamic calculations.

Formation Reactions

Formation reactions involve combining elements in their standard states to form a compound. This process is described by a balanced thermochemical equation and is fundamental when calculating enthalpy changes.

For example, to form \( NO_2(g) \), nitrogen and oxygen are used: \[ \frac{1}{2} N_2(g) + O_2(g) \rightarrow NO_2(g) \]

Each reaction carefully reflects the stoichiometry needed to produce exactly one mole of the desired compound, offering insights into the energy dynamics of the reaction.

• Each formation equation must yield exactly one mole of the product.
• Elements must be in their standard states as per IUPAC conventions.
• Accurately balanced equations allow for direct comparison of \( \Delta H^{\circ}_f \) values across different substances.

Formation reactions thus help chemists understand how elements interact to create new substances, using energy considerations to appreciate the efficiency and feasibility of industrial and laboratory processes.