Problem 67
Question
Which of the following reaction defines \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) ? (a) \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HF}(\mathrm{g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\)
Step-by-Step Solution
Verified Answer
Reaction (b) defines \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \).
1Step 1: Understanding the Concept of Standard Enthalpy of Formation
The standard enthalpy of formation, \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \), is defined as the heat change that results when one mole of a compound is formed from its elements in their standard states under standard conditions \((1 \text{ atm}, 25^\circ\text{C})\). The elements are in their most stable form.
2Step 2: Analyzing Each Reaction
Let's consider each reaction:**(a)** \( \mathrm{C} \text{(diamond)} + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \): This reaction is not a formation reaction since diamond is not the standard form of carbon.**(b)** \( \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g}) \): This reaction shows formation of 1 mole of \( ext{HF} \) from its elements in their standard states.**(c)** \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \): This forms 2 moles of \( \text{NH}_3 \), but \( \text{NH}_3 \) is not formed from elements in 1 mole.**(d)** \( \mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \): This is not the formation from elemental forms, as CO is not an element.
3Step 3: Identifying the Correct Reaction
The correct reaction must show the formation of exactly one mole of a compound from its elements in their most stable form at standard conditions. From the analysis in Step 2, reaction (b) satisfies this condition.
Key Concepts
ThermochemistryChemical ReactionsHeat Change
Thermochemistry
Thermochemistry is the branch of chemistry that deals with the energy changes, particularly the heat exchange, that occur during chemical reactions. Whether you're burning wood, cooking food, or even metabolizing food energy in your body, thermochemistry helps explain the energetic transformations happening behind the scenes.
- In thermochemistry, the focus is often on understanding how energy in the form of heat is absorbed or released during a reaction.
- It is essential for predicting how reactions proceed and for determining the feasibility of a reaction.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. These reactions are governed by the laws of conservation of mass and energy. When considering enthalpy, it is the heat content of a system at constant pressure that is often crucial.
- The standard enthalpy of formation refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
- For an element, the standard enthalpy of formation is zero because no formation is needed.
Heat Change
Heat change during a chemical reaction is a vital measure in thermochemistry. The concept is intimately tied to the first law of thermodynamics, stating that energy cannot be created or destroyed, only transformed.
- The heat change is denoted as \( \Delta H \) and indicates whether a reaction is endothermic or exothermic.
- Standard enthalpies of formation, \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \), offer a simplified way to calculate the heat change for a reaction.
Other exercises in this chapter
Problem 65
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