Animated Solution for Chemistry - Chemical Thermodynamics: Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.

This question tests a fundamental concept in chemical thermodynamics: the precise definition of the standard enthalpy of formation (${\Delta }_f{H}^{\circ }$). It's a classic trap for students who rush through the options without checking all the necessary conditions.

The standard enthalpy of formation is defined as the enthalpy change that occurs when exactly one mole of a compound is formed from its constituent elements, with all substances in their standard states at a specified temperature (usually $298\text{ K}$) and $1\text{ bar}$ pressure.

To determine if a given reaction's enthalpy change (${\Delta }_r{H}^{\circ }$) is equal to the enthalpy of formation of the product, we must rigorously check two conditions: 1. The Product Check: Is there exactly $1\text{ mole}$ of the product being formed? 2. The Reactant Check: Are all reactants elements, and are they in their most stable standard reference states?

Let's put each option through this two-step test.

Option (A): $\frac{3}{2}{\text{O}}_2(\text{g})\to {\text{O}}_3(\text{g})$ - Product Check: We are forming $1\text{ mole}$ of ozone (${\text{O}}_3$). This passes. - Reactant Check: The reactant is oxygen gas (${\text{O}}_2$), which is the most stable standard state of the element oxygen. This passes. - Verdict: This reaction perfectly represents the standard enthalpy of formation of ozone.

Option (B): $\frac{1}{8}{\text{S}}_8(\text{s})+{\text{O}}_2(\text{g})\to {\text{SO}}_2(\text{g})$ - Product Check: We are forming $1\text{ mole}$ of sulfur dioxide (${\text{SO}}_2$). This passes. - Reactant Check: The reactants are oxygen gas (${\text{O}}_2$) and solid sulfur (${\text{S}}_8$). The most stable allotrope of sulfur at standard conditions is rhombic sulfur, represented as ${\text{S}}_8(\text{s})$. This passes. - Verdict: This reaction represents the standard enthalpy of formation of sulfur dioxide.

Option (C): $2{\text{H}}_2(\text{g})+{\text{O}}_2(\text{g})\to 2{\text{H}}_2\text{O}(\text{l})$ - Product Check: Look closely! We are forming $2\text{ moles}$ of liquid water. The definition strictly demands $1\text{ mole}$. This fails immediately. - Verdict: The enthalpy of this reaction is actually twice the standard enthalpy of formation of water: ${\Delta }_r{H}^{\circ }=2\times {\Delta }_f{H}^{\circ }({\text{H}}_2\text{O})$.

Option (D): $2\text{C}(\text{g})+3{\text{H}}_2(\text{g})\to {\text{C}}_2{\text{H}}_6(\text{g})$ - Product Check: We are forming $1\text{ mole}$ of ethane (${\text{C}}_2{\text{H}}_6$). This passes. - Reactant Check: The reactants are hydrogen gas (${\text{H}}_2$) and gaseous carbon ($\text{C}(\text{g})$). Wait, gaseous carbon? The most stable standard state of carbon at $298\text{ K}$ and $1\text{ bar}$ is solid graphite, not gas. This fails. - Verdict: Because the carbon is not in its standard state, this is not a standard formation reaction.

By systematically applying the definition, we find that only options (A) and (B) meet all the strict criteria. This highlights the importance of paying attention to stoichiometric coefficients and physical states in thermochemical equations.