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Which of the following standard enthalpy of formation values is not zero at \(25^{\circ} \mathrm{C}: \mathrm{Na}(\) monoclinic \(), \mathrm{Ne}(g)\) \(\mathrm{CH}_{4}(g), \mathrm{S}_{8}(\) monoclinic \(), \mathrm{Hg}(l), \mathrm{H}(g) ?\)

Short Answer

Expert verified
\( \mathrm{CH}_{4}(g) \) and \( \mathrm{S}_{8} \) monoclinic both have non-zero standard enthalpy of formation.

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01

Understanding Enthalpy of Formation

The standard enthalpy of formation for an element in its most stable form is zero at 25°C and 1 atm pressure. This is a key principle of thermodynamics.
02

Identifying Elements

Determine which substances are elements and which are compounds. In the exercise, \( \mathrm{Na} \), \( \mathrm{Ne} \), \( \mathrm{Hg} \), and \( \mathrm{H} \) are elements, whereas \( \mathrm{CH}_{4} \) is a compound.
03

Comparing Elemental States

For elements, determine if the given state is the most stable at 25°C. \( \mathrm{Na} \) is normally solid, \( \mathrm{Ne} \) is gas, \( \mathrm{Hg} \) is liquid, and \( \mathrm{H} \) exists as \( \mathrm{H}_2 \) gas.
04

Recognizing Compounds

Compounds, like \( \mathrm{CH}_{4} \), do not have a standard enthalpy of formation of zero, as they are not in their elemental form. Each compound’s formation involves creating it from its constituent elements.
05

Identifying Most Common State

Verify the most common and stable form of elements at 25°C. \( \mathrm{S}_{8} \) normally exists as rhombic, not monoclinic. Therefore, \( \mathrm{S}_{8} \) monoclinic has a non-zero standard enthalpy of formation.

Key Concepts

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

Understanding Thermodynamics
Thermodynamics is a foundational concept in chemistry and physics that helps us understand how energy is transferred and transformed in a system. It deals with concepts such as energy, work, heat, and how they interact in various processes. The standard enthalpy of formation is an important thermodynamic quantity. It is defined as the heat absorbed or released when one mole of a compound is formed from its elements in their most stable forms at a pressure of 1 atm and a temperature of 25°C (standard conditions).

One key aspect of thermodynamics is the conservation of energy principle, which states that energy cannot be created or destroyed but can be transferred from one form to another. This principle is used to calculate enthalpy changes in reactions and understand how systems reach stability.
Identifying the Most Stable Form of Elements
The term 'most stable form' refers to the physical state and molecular configuration of an element that is most energetically favorable under standard conditions (25°C and 1 atm). The standard enthalpy of formation for an element in its most stable form is defined as zero because no energy is required to form the element from itself.

For example:
  • Sodium (\(\mathrm{Na}\)) is most stable as a solid metal.
  • Helium (\(\mathrm{Ne}\)) is most stable in its gaseous form.
  • Mercury (\(\mathrm{Hg}\)) is an exception among metals, being most stable as a liquid.
  • Hydrogen naturally forms \(\mathrm{H}_2\) molecules and is a gas at room temperature.
Understanding the stable form is crucial for determining the standard enthalpy of formation for both elements and compounds.
Distinguishing Compounds and Elements
A major part of thermodynamics is understanding the distinction between compounds and elements. Elements are pure substances consisting of only one type of atom, and they serve as the building blocks for compounds. Compounds are formed when two or more elements chemically bond together.

Since elements in their most stable forms have a standard enthalpy of formation of zero, it is important to recognize these states to correctly assess the energy changes in compound formation. Chemical compounds, however, always have a non-zero standard enthalpy of formation because they are created from their elements.

For instance, methane (\(\mathrm{CH}_4\)) is a compound, not an element, so it does not have a zero enthalpy of formation. This highlights the process of elements reaching a more stable state through the formation of compounds, often releasing or absorbing energy in the process.
Exploring Phase Stability
Phase stability refers to the ability of a substance to remain in a particular physical state (solid, liquid, or gas) under specific conditions of temperature and pressure. It relates closely to the concept of the most stable form.

For most elements, the phase stability is quite intuitive:
  • Solids like sodium (\(\mathrm{Na}\)) tend to retain their structure until a high temperature is reached.
  • Gases such as neon (\(\mathrm{Ne}\)) are stable at room temperature under standard pressure conditions.
  • Mercury (\(\mathrm{Hg}\)) is unique due to its liquid phase stability at standard conditions.
In the case of sulfur (\(\mathrm{S}_8\)), the rhombic form is more stable than the monoclinic form at 25°C, hence the monoclinic form has a non-zero enthalpy of formation.

Recognizing these stability characteristics is essential for correctly identifying the energy changes involved when substances transition between phases or form new compounds.

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Most popular questions from this chapter

Explain what is meant by a state function. Give two examples of quantities that are state functions and two that are not state functions.

Consider the following data:$$\begin{array}{lcc}\text { Metal } & \text { Al } & \text { Cu } \\\\\hline \text { Mass }(\mathrm{g}) & 10 & 30 \\\\\text { Specific heat }\left(\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\\\\text { Temnerature }{ }^{\circ}{ }^{\circ} \mathrm{C} \text { ) } & 40 & 60\end{array}$$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from \(\mathrm{Al}\) to Cu because \(\mathrm{Al}\) has a larger specific heat. (b) Heat will flow from \(\mathrm{Cu}\) to \(\mathrm{Al}\) because \(\mathrm{Cu}\) has a larger mass. (c) Heat will flow from \(\mathrm{Cu}\) to \(\mathrm{Al}\) because \(\mathrm{Cu}\) has a larger heat capacity (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\Delta H_{\mathrm{f}}^{\circ}\) ? (a) \(\mathrm{H}_{2}(g)+\mathrm{S}(\) rhombic \() \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)

Consider this reaction: $$\begin{array}{l}2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\ \Delta H=-1452.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ What is the value of \(\Delta H\) if (a) the equation is multiplied throughout by \(2 ;(b)\) the direction of the reaction is reversed so that the products become the reactants, and vice versa; (c) water vapor instead of liquid water is formed as the product?

The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting; that is, the conversion of \(\mathrm{ZnS}\) to \(\mathrm{ZnO}\) by heating:$$\begin{aligned}2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g) & \Delta H=-879 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$$ Calculate the heat evolved (in kJ) per gram of \(\mathrm{ZnS}\) roasted.

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