Problem 67 - A Complete Resource Book in Chemistry for JEE Main

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

Answer

Reaction (b) defines \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \).

Step 1: Understanding ΔHf°

The term \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) refers to the standard enthalpy of formation, which is the heat change that occurs when one mole of a compound is formed from its elements in their standard states under standard conditions (1 atm, 25°C, 1 M concentration).

Step 2: Analyzing Option (a)

The reaction \( \mathrm{C} \text{ (diamond) }+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) \) represents the formation of \( \mathrm{CO}_2 \) from carbon in the form of diamond and \( \mathrm{O}_2 \). However, carbon's most stable form under standard conditions is graphite, not diamond.

Step 3: Analyzing Option (b)

The reaction \( \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HF}(\mathrm{g}) \) shows the formation of \( \mathrm{HF} \) from \( \mathrm{H}_2 \) and \( \mathrm{F}_2 \) in their gaseous states, starting from their standard states, producing exactly one mole of product. This fits the definition of \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \).

Step 4: Analyzing Option (c)

The reaction \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \) involves forming 2 moles of \( \mathrm{NH}_3 \). For \( \Delta \mathrm{H}_{f}^{\circ} \), only one mole of a compound should be formed.

Step 5: Analyzing Option (d)

The reaction \( \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) \) involves \( \mathrm{CO} \) and \( \mathrm{O}_2 \), but \( \mathrm{CO} \) is not an element in its standard state, thus this is not a standard formation reaction.

Step 6: Selecting the Correct Option

From our analysis, option (b) is the only reaction that meets the criteria for defining \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) because it forms one mole of \( \mathrm{HF} \) directly from its elements in their standard states.

Key Concepts

Standard Enthalpy Change

The standard enthalpy change, denoted as \( \Delta H^{\circ} \), is a key concept in chemical thermodynamics. It represents the heat exchange involved in a chemical reaction occurring under standard conditions—pressure at 1 atm and temperature at 25°C (298 K). These standardized conditions ensure that enthalpy data is consistent and comparable.

The measurement of \( \Delta H^{\circ} \) provides insight into the energy profile of a reaction. A negative \( \Delta H^{\circ} \) signifies that the reaction is exothermic, releasing heat to the surroundings. Conversely, a positive \( \Delta H^{\circ} \) indicates an endothermic reaction, which absorbs heat.

For students studying chemistry, understanding standard enthalpy change is crucial because it helps predict reaction feasibility and energy efficiency which are vital for both laboratory experiments and industrial applications.

Formation Reactions

Formation reactions are specific chemical reactions where one mole of a compound is created from its elemental components, each in their standard states. The standard state of an element is its most stable form at 1 atm and 25°C. For instance, carbon's most stable form is graphite, not diamond.

In the exercise provided, option (b) \( \frac{1}{2} \text{H}_{2}(\text{g}) + \frac{1}{2} \text{F}_{2}(\text{g}) \longrightarrow \text{HF}(\text{g}) \) is a perfect example of a formation reaction. It forms exactly one mole of the compound (\( \text{HF} \)) directly from hydrogen and fluorine gases, which are in their most stable forms under standard conditions.

Understanding formation reactions is fundamental for calculating the standard enthalpy of formation, \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \), which provides a basis for estimating the heat changes during larger or more complex reactions using Hess’s Law.

Chemical Thermodynamics

Chemical thermodynamics is an extensive field of chemistry that investigates the relationships and conversions between different forms of energy in chemical processes. It provides the tools necessary to understand not only how much energy is exchanged but also in what direction energy flows, as well as determining spontaneity of reactions.

At the core of chemical thermodynamics is the concept of enthalpy, including the standard enthalpy of formation \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \), which helps simplify the calculation of entire reaction enthalpies by using tabulated values for individual substances.

• Thermodynamic principles guide the prediction of reaction spontaneity using the Gibbs Free Energy, which itself relies on enthalpy changes and entropy considerations.
• The concepts of chemical equilibrium and reaction rates are also explored within this discipline, providing a full picture on how reactions proceed and settle.

Overall, grasping chemical thermodynamics principles empowers students to tackle more advanced studies in chemistry and realize their essential applications in real-world problems and industries.