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Chapter 6: Problem 129

Which of the following substances have an enthalpy of formation equal to zero? a. \(C l_{2}(g)\) b. \(\mathrm{H}_{2}(g)\) c. \(\mathrm{N}_{2}(l)\) d. \(\mathrm{Cl}(g)\)

Short Answer

Expert verified
The substances with an enthalpy of formation equal to zero are a. \(Cl_{2}(g)\) and b. \(H_{2}(g)\), as they are both in their standard states.

Step by step solution

01

Identify the standard states of the given substances

The standard state of an element is defined as its most stable form at 1 atm of pressure and 298 K (25掳C). For each substance, we will determine if it is in its standard state: a. \(Cl_{2}(g)\): Chlorine is a diatomic molecule and exists as a gas at standard conditions. Therefore, this substance is in its standard state. b. \(H_{2}(g)\): Hydrogen is also a diatomic molecule and exists as a gas at standard conditions. Therefore, this substance is in its standard state. c. \(N_{2}(l)\): Nitrogen is a diatomic molecule, but it is not in liquid form at standard conditions. It exists as a gas. Therefore, this substance is not in its standard state. d. \(Cl(g)\): Chlorine is a diatomic molecule, and it does not exist as a single atom in the gaseous phase at standard conditions. Therefore, this substance is not in its standard state.
02

Apply the enthalpy of formation definition

Now that we have determined the standard states of each substance, we can apply the definition of enthalpy of formation. The enthalpy of formation of an element in its standard state is equal to zero. Based on this definition, we can conclude the following for each substance: a. \(Cl_{2}(g)\): This substance has an enthalpy of formation equal to zero because it is in its standard state. b. \(H_{2}(g)\): This substance has an enthalpy of formation equal to zero because it is in its standard state. c. \(N_{2}(l)\): This substance does not have an enthalpy of formation equal to zero because it is not in its standard state. d. \(Cl(g)\): This substance does not have an enthalpy of formation equal to zero because it is not in its standard state.
03

Final Answer

From the analysis above, we can conclude that the substances with an enthalpy of formation equal to zero are: a. \(Cl_{2}(g)\) b. \(H_{2}(g)\)

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Key Concepts

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

Standard State of Elements
In chemistry, the concept of the standard state of elements helps us understand how elements behave under specific conditions. The standard state is defined as the element's most stable form at a pressure of 1 atmosphere and a temperature of 298 K (25掳C). This is crucial when calculating various thermodynamic properties, such as enthalpy of formation.

Here are some key points to remember about standard states:
  • For gases, the standard state is usually the most stable gaseous form at 1 atm and 298K.
  • For solids and liquids, it is the form that exists most commonly at 1 atm and 298K.
  • Any deviation from this state is not considered a standard state.
When elements are in their standard state, their enthalpy of formation is considered zero. For example, chlorine (Cl2) is stable as a gas in its standard state, which is why its enthalpy of formation is zero. Understanding this concept is essential for correctly interpreting thermodynamic data.
Diatomic Molecules
Diatomic molecules are molecules composed of only two atoms, and they are a common form in which many gases exist naturally. The term 'diatomic' specifically refers to these bonds only involving two atoms, which can be of the same or different chemical elements.

Some elements frequently found as diatomic molecules at standard conditions include:
  • Hydrogen (H2)
  • Nitrogen (N2) (although in the exercise, it should be noted that N2 exists as a gas, not a liquid, in its standard state)
  • Oxygen (O2)
  • Fluorine (F2)
  • Chlorine (Cl2)
The stable nature of these diatomic molecules under standard conditions means that their enthalpy of formation is zero. It's essential to recognize these natural forms when working with chemical reactions and thermodynamics.
Chemical Thermodynamics
Chemical thermodynamics is a branch of chemistry that deals with the relations between chemical reactions and energy changes involving heat and work. A central concept within this field is understanding how energy, specifically enthalpy, changes as substances react or transform.

Key aspects of chemical thermodynamics include:
  • **Enthalpy (\(H\)):** A measure of the total energy of a thermodynamic system.
  • **Enthalpy of Formation (\(\Delta H_f^\circ\)):** The heat change that occurs when one mole of a compound is formed from its elements in their standard states.
  • **Standard States:** Understanding which elements have an enthalpy of formation of zero in their standard states.
The study of these energy transitions helps in predicting how reactions will proceed, how much energy will be absorbed or released, and the overall feasibility of reactions. Applying these principles allows chemists to design processes such as fuel combustion, synthesis of materials, and more. Understanding these core ideas is essential for navigating chemical reactions and their practical applications.

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Most popular questions from this chapter

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

Consider the following equations: $$ \begin{array}{ll}{3 \mathrm{A}+6 \mathrm{B} \longrightarrow 3 \mathrm{D}} & {\Delta H=-403 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{E}+2 \mathrm{F} \longrightarrow \mathrm{A}} & {\Delta H=-105.2 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{C} \longrightarrow \mathrm{E}+3 \mathrm{D}} & {\Delta H=64.8 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6},\) the second and third equations are divided by \(2,\) and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

A system undergoes a process consisting of the following two steps: Step 1: The system absorbs 72 \(\mathrm{J}\) of heat while 35 \(\mathrm{J}\) of work is done on it. Step \(2 :\) The system absorbs 35 \(\mathrm{J}\) of heat while performing 72 \(\mathrm{J}\) of work. Calculate \(\Delta E\) for the overall process.

The bomb calorimeter in Exercise 112 is filled with 987 \(\mathrm{g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C} .\) A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\mathrm{comb}}=\right.\) \(-26.42 \mathrm{kJ} / \mathrm{g}\) ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Photosynthetic plants use the following reaction to produce glucose, cellulose, and so forth: $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \frac{\text { Sunlight }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ How might extensive destruction of forests exacerbate the greenhouse effect?
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