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Chapter 34: Problem 8

Why is \(\Delta H_{\mathrm{ac}}^{\circ}\) of \(\mathrm{C}(\mathrm{g})=0 ?\) Why is \(\Delta H_{\mathrm{ac}}^{\circ}\) of \(\mathrm{H}(\mathrm{g})=0 ?\)

Short Answer

Expert verified
The standard enthalpy of atomisation (\(\Delta H_{\mathrm{ac}}^{\circ}\)) of both C(g) and H(g) is zero because they are already in their atomic states and require no energy to be atomized.

Step by step solution

01

Explaining \(\Delta H_{\mathrm{ac}}^{\circ}\) of Carbon

Carbon in its standard state (graphite) is a solid, not a gas. However, when the term 'C(g)' is used, it refers to Carbon in its atomic gaseous state - essentially one mole of Carbon atoms in gas form. Because C(g) is already in its atomic state (ie. it is a single atom and doesn't need to be broken down further), no energy is required to atomize it. Therefore, the enthalpy of atomisation (\(\Delta H_{\mathrm{ac}}^{\circ}\)) of C(g) is zero.
02

Explaining \(\Delta H_{\mathrm{ac}}^{\circ}\) of Hydrogen

Similar to Carbon, Hydrogen in its atomic gaseous state H(g) doesn't require any energy to be atomized since it is already in its atomic state. Thus, the \(\Delta H_{\mathrm{ac}}^{\circ}\) of H(g) is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Enthalpy Change
The concept of the standard enthalpy change is foundational in understanding chemical thermodynamics. It is defined as the enthalpy change that occurs when a reaction takes place under standard conditions (298 K, 1 atm pressure), and the substances involved are in their standard states.

To put it simply, it's the heat energy change accompanying a chemical process, like combustion or evaporation, when everything is at 'standard' lab conditions. This concept is essential because it allows chemists to compare the intrinsic energy changes of reactions without the variables of temperature or pressure confounding the comparison.
Atomic Gaseous State
An atom in the atomic gaseous state is a single atom that exists as a gas, rather than bonded to other atoms as in a molecule or solid structure. This state is important because substances in their gaseous state have different properties compared to when they’re solid or liquid.

Understanding the energy required to convert a substance into its atomic gaseous state is key in areas such as spectroscopy and predicting reaction behavior. The enthalpy change to achieve this state from the standard state is known as the enthalpy of atomization.
Chemical Thermodynamics
The field of chemical thermodynamics deals with the relationship between chemical reactions and the laws of heat and temperature. It provides a framework for predicting the direction of chemical processes and the energy exchanged during these reactions.

Concepts like enthalpy, entropy, and free energy are cornerstones of this discipline, and they help scientists and engineers to understand how energy transformations within a chemical reaction influence the feasibility and efficiency of industrial and biochemical processes.
Enthalpy Change
The term enthalpy change represents the heat exchange in a chemical reaction at constant pressure. An exothermic reaction, for instance, releases heat, resulting in a negative enthalpy change. In contrast, an endothermic reaction, which requires heat from its surroundings, has a positive enthalpy change.

This principle is crucial in understanding why some reactions are spontaneous—like the burning of natural gas on a stove—while others require continual heat—like the baking of bread. It's also the reason why the enthalpy change for atomization is zero when a substance is already in its atomic gaseous state; no additional energy is required or released.

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Most popular questions from this chapter

The net attractive potential energy in Figure 2 is \(-6.93 \times 10^{-18} \mathrm{~J}\). Will energy be required to separate the nuclei or will energy be released upon separation? Explain.

For molecules, why are all of the values for enthalpies of atom combination negative?

a) Write the Lewis structures for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). b) Explain how these Lewis structures are consistent with the relative enthalpies of atom combination for \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\).

What is the value of \(\Delta H^{\circ}\) for the overall process of separating one mole of \(\mathrm{CH}_{4}\) into its constituent atoms, and then reforming one mole of \(\mathrm{CH}_{4}\) ?
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