Question

Which of the following reaction defines \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\) ? (a) \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HF}(\mathrm{g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

Short answer

Option (b): \( \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g}) \) defines \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}} \).

Step-by-step solution

Understanding Enthalpy of Formation

The standard enthalpy of formation, \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}} \), of a compound is defined as the change in enthalpy when 1 mole of the substance is formed from its constituent elements in their standard states. It is important to note that the formation reaction should produce exactly 1 mole of product.

Analyze Option (a)

The reaction \( \mathrm{C} (\text{diamond}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \) forms carbon dioxide from carbon in the diamond form and oxygen gas. However, \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}} \) requires the reactants to be in their most stable form, which for carbon is graphite, not diamond. Hence, this is not the correct reaction.

Analyze Option (b)

The reaction \( \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g}) \) forms hydrofluoric acid gas from hydrogen gas and fluorine gas, which are both in their standard states. This reaction produces exactly 1 mole of HF, making it the correct choice as it matches the definition.

Analyze Option (c)

The reaction \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \) forms ammonia from nitrogen gas and hydrogen gas, which are in their standard states. However, it produces 2 moles of ammonia, not 1 mole, so it does not fit the definition of \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}} \).

Analyze Option (d)

In the reaction \( \mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \), carbon monoxide reacts with oxygen gas to form carbon dioxide. Although the product is 1 mole of \( \mathrm{CO}_{2}(\mathrm{~g}) \), the reactant \( \mathrm{CO}(\mathrm{g}) \) is not an element in its standard form, thus not fitting the criteria for \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}} \).

Key concepts

Standard State

In chemistry, understanding the concept of the standard state is crucial when discussing enthalpy changes in reactions. The standard state of a substance is its most stable form at a specified temperature, usually 25°C (298 K), at 1 atmosphere of pressure. This standard condition allows for uniformity in measuring and comparing thermodynamic data.

Elements often have a specific form that is considered their standard state.

• For instance, oxygen's standard state is oxygen gas ( O _{ 2 } ).
• For carbon, the most stable form is graphite, not diamond.

When the enthalpy of formation is recorded, it is always relative to these states. For example, the formation of water from hydrogen gas and oxygen gas involves both reactants being in their standard states.

Remember, when evaluating chemical reactions in this context, ensuring that all substances involved are in standard states is vital for determining the correct enthalpy of formation ( ∆ H _f ^ ο ). Unstable or non-standard forms (such as diamond for carbon) would complicate these calculations as they require additional thermodynamic adjustments.

Chemical Reactions

Chemical reactions involve breaking and forming bonds, leading to the transformation of substances. In the context of enthalpy of formation ( ∆ H _f ^ ο ), understanding reaction types is essential because it guides the evaluation of enthalpy changes.

A vital component of these reactions is their balanced representation, indicating that the law of conservation of mass is upheld. For enthalpy of formation specifically, the reaction must:

• Produce exactly 1 mole of the target compound.
• Involve reactants in their standard states.

The selection rule for enthalpy of formation is stringent but straightforward. For instance, forming hydrofluoric acid ( HF gas) from hydrogen and fluorine gases accurately depicts this principle through the equation: ( 1 / 2 H _2 ( g ) + 1 / 2 F _2 ( g ) → HF ( g ) ). This equation satisfies both the production of one mole of the product and the usage of standard state reactants.

Being familiar with these characteristics ensures a better grip on thermochemistry and paves the way for understanding more complex reactions.

Mole Concept

The mole concept is an essential pillar in chemistry that quantifies substances at the molecular level. One mole physically represents 6.022 x 10 ^23 entities, known as Avogadro's number, allowing chemists to "count" atoms, molecules, or ions via measurable mass.

This concept is particularly important when discussing the standard enthalpy of formation, as it depends on the formation of exactly one mole of a compound. By acknowledging this, we ensure consistent and accurate calculations of thermodynamic data.

• For instance, see the enthalpy of formation for water. The formation equation utilizes precisely one mole of water being formed from its elements: H _2 ( g ) + 1 / 2 O _2 ( g ) → H _2 O ( l ).

Grasping how the mole concept relates to reaction equations elevates a student's understanding of chemical compounds and their properties, crucial in both theoretical chemistry and practical laboratory settings.

Ensuring comprehension of this topic is invaluable for anyone pursuing chemistry, as it links the microscopic world of atoms to the macroscopic world we observe, acting as a fundamental bridge for more advanced topics.