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 use Appendix \(C\) to obtain the value of \(\Delta H_{f}^{\circ}:\) (a) \(\mathrm{NH}_{3}(g)\), (b) \(\mathrm{SO}_{2}(g)\) (c) \(\mathrm{RbClO}_{3}(s)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)\).

Short answer

For the formation of one mole of each compound from its elements in their standard states, we have the following balanced thermochemical equations and values of ∆Hf°: (a) \( 1/2 N_{2}(g) + 3/2 H_{2}(g) \rightarrow NH_{3}(g), \space \Delta H_{f}^{\circ} = -22.95 \, kJ/mol \) (b) \( S(s) + O_{2}(g) \rightarrow SO_{2}(g), \space \Delta H_{f}^{\circ} = -296.8 \, kJ/mol \) (c) \( Rb(s) + \frac{1}{2}Cl_{2}(g) + \frac{3}{2}O_{2}(g) \rightarrow RbClO_{3}(s), \space \Delta H_{f}^{\circ} = -432.0 \, kJ/mol \) (d) \( NH_{4}^{+}(aq) + NO_{3}^{-}(aq) \rightarrow NH_{4} NO_{3}(s), \space \Delta H_{f}^{\circ} = -365.1 \, kJ/mol \)

Step-by-step solution

(a) NH3(g) Formation

1. Write the balanced thermochemical equation. \( N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g) \) Now, you need to find the enthalpy of formation for NH3 from the given data table (Appendix C): ∆Hf° (NH₃(g)) = -45.9 kJ/mol Note: Since we only need one mole of the compound, we need to adjust the equation and the enthalpy value: \( 1/2 N_{2}(g) + 3/2 H_{2}(g) \rightarrow NH_{3}(g), \space \Delta H_{f}^{\circ} = -22.95 \, kJ/mol \)

(b) SO2(g) Formation

1. Write the balanced thermochemical equation. \( S(s) + O_{2}(g) \rightarrow SO_{2}(g) \) From the given data (Appendix C), find the enthalpy of formation for SO2: ∆Hf° (SO₂(g)) = -296.8 kJ/mol

(c) RbClO3(s) Formation

1. Write the balanced thermochemical equation. \( Rb(s) + \frac{1}{2}Cl_{2}(g) + \frac{3}{2}O_{2}(g) \rightarrow RbClO_{3}(s) \) From the given data (Appendix C), find the enthalpy of formation for RbClO3: ∆Hf° (RbClO₃(s)) = -432.0 kJ/mol

(d) NH4NO3(s) Formation

1. Write the balanced thermochemical equation. Due to NH4NO3 being a compound composed of two ions, the equation can be broken down in three steps (a) \( N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g), \space \Delta H_{f}^{\circ} = -45.9 \, kJ/mol \) (b) \( 2NH_{3}(g) \rightarrow 2NH_{4}^{+}(aq), \space \Delta H_{f}^{\circ} = -153.4 \, kJ/mol \) (c) \( N_{2}(g) + 2O_{2}(g) \rightarrow 2NO_{3}^{-}(aq), \space \Delta H_{f}^{\circ} = -207.1 \, kJ/mol \) Now add (b) and (c) to obtain the enthalpy of formation for NH4NO3: \( NH_{4}^{+}(aq) + NO_{3}^{-}(aq) \rightarrow NH_{4} NO_{3}(s), \space \Delta H_{f}^{\circ} = -365.1 \, kJ/mol \)

Key concepts

Thermochemical Equations

A thermochemical equation represents the change in energy, often heat, during a chemical reaction. These equations are very similar to balanced chemical equations but with a significant addition: they include the enthalpy change, commonly denoted as ΔH. For reactions involving the formation of a compound from its elements in their standard states, the enthalpy change is the Enthalpy of Formation (ΔH_f°).

When writing a thermochemical equation for the formation of a compound, such as ammonia (NH₃(g)), it is important to show the reactants (elements in standard states) and the product (the compound formed), as well as the corresponding enthalpy change. An arrow (→) is used to separate reactants and products, and the enthalpy change is typically written to the right of the equation. If the enthalpy change is negative, it indicates that the reaction releases heat, and if it is positive, the reaction absorbs heat.

Standard States

The standard state of a substance refers to its thermodynamically stable state under specific standard conditions of pressure and temperature, which is 1 atmosphere of pressure and 25°C (298 K). For different phases of matter, the standard states are well-defined: for solids and liquids, it is the pure substance in its most stable form, and for gases, it is ideally behaved at 1 atm pressure. Elements in their standard states are considered to have an Enthalpy of Formation of zero because they are in their most stable forms.

In our exercise examples, this is demonstrated for elements like nitrogen, N₂(g), hydrogen, H₂(g), and sulfur, S(s), which are all in their standard states. It’s vital to recognize that the use of standard states allows us to use tabulated values of ΔH_f°, such as those given in Appendix C, to calculate enthalpy changes for chemical reactions.

ΔHf° Values

The ΔH_f° value, also known as the standard enthalpy of formation, is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. ΔH_f° values are essential thermodynamic quantities often tabulated for convenience, as seen in Appendix C. These values are given in units of energy per amount of substance, usually kilojoules per mole (kJ/mol).

For a compound like NH₃(g), the ΔH_f° value is -22.95 kJ/mol, which means it releases 22.95 kJ of energy when one mole of ammonia is formed from nitrogen and hydrogen gas under standard conditions. If the value is negative, the formation is exothermic; if it is positive, the process requires input of energy, making it endothermic. By using these values, students can predict heat flow for reactions and gauge the stability of compounds.

Balanced Equations

Chemical equations must be balanced to obey the Law of Conservation of Mass, meaning the number of atoms of each element should be the same on the reactant and product sides of the equation. A properly balanced thermochemical equation also reflects the stoichiometry needed to calculate the standard enthalpy change for the reaction. For example, when balancing the thermochemical equation for the formation of NH₃(g) from elemental nitrogen and hydrogen, the coefficients need to be adjusted for the reaction to produce exactly one mole of NH₃(g), ensuring the proper ΔH_f° value corresponds to that amount.

The exercise demonstrates how to balance complex formulas and adjust coefficients and enthalpy values accordingly. The equation becomes more meaningful for calculations when balanced correctly, as it shows the exact amounts of starting materials and the associated enthalpy change for forming the desired product.