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Chapter 9: Problem 80

Why are the standard heats of formation of elements in their standard states assigned a value of zero?

Short Answer

Expert verified
Answer: The standard heats of formation of elements in their standard states are assigned a value of zero as a reference point because there is no change in the element's composition when it is in its standard state. In other words, the element is not being converted from one form to another, so no energy is absorbed or released. This simplifies the calculation of the standard heat of formation for compounds and is crucial for thermodynamics calculations.

Step by step solution

01

Definition of Standard Heats of Formation

The standard heat of formation, also known as enthalpy of formation, is the amount of energy (in the form of heat) absorbed or released when a compound is formed from its constituent elements.
02

Standard States of Elements

The standard state of an element refers to the reference state of the element at its established reference pressures (usually 1 atm) and a specific reference temperature (usually 298.15 K or 25 掳C).
03

Heats of Formation for Elements in Standard States: Zero

The standard heat of formation of an element in its standard state is assigned a value of zero as a reference point. This is because there is no change in the element's composition when it is in its standard state. In other words, the element is not being converted from one form to another; thus, no energy is absorbed or released.
04

Heats of Formation for Real Compounds

When we talk about the heat of formation for a real compound, it is the difference in enthalpy between the compound and its constituent elements in their standard states. Assigning a value of zero for the standard state of elements simplifies the calculation of the standard heat of formation for any compound.
05

Importance of Standard Heats of Formation

Standard heats of formation values are essential in thermodynamics calculations, such as determining the overall heat changes during a reaction. By having a baseline (value of zero) for elements in their standard states, we can easily calculate the heat changes caused by reactions involving complex compounds.

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Key Concepts

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

Standard States of Elements
In thermodynamics, the "standard state" of an element is a crucial concept. It defines the physical state of an element under specific conditions that serve as a reference point. Generally, the standard state is determined at a pressure of 1 atm and a temperature of 298.15 K (or 25 掳C).
This concept is vital because it allows chemists to use a common baseline for comparisons.
  • For example, oxygen is typically considered a diatomic gas (O鈧�) in its standard state, while carbon is usually found as graphite.
  • The standard state ensures consistency across scientific measurements and calculations, providing a clear framework for research and analysis.
Essentially, the standard state refers to the most stable form of the element under these conditions, which is why analyzing it is important for understanding material properties and behaviors.
Thermodynamics Calculations
Thermodynamics calculations involve determining various energy changes that occur in chemical reactions. Understanding these calculations is fundamental for predicting how a reaction will proceed and the energy required or released in the process.
One of the main uses of standard heats of formation in these calculations is to determine reaction enthalpies. By using the standard heat of formation values for reactants and products, one can calculate the total heat absorbed or released during a reaction.
  • Standard heats of formation act as a baseline, simplifying the complex calculations needed for various chemical reactions.
  • These calculations provide insights into the feasibility and spontaneity of reactions.
Essentially, thermodynamics calculations with standard heats of formation help chemists and engineers devise practical methods to control and optimize chemical processes.
Enthalpy of Formation
The enthalpy of formation is a key concept that measures the amount of heat energy absorbed or released during the formation of a compound from its elements in their standard states. This is often represented by \ \(螖H_f^鈼媆).
The enthalpy of formation of a compound provides valuable information about the compound's stability. A negative enthalpy of formation indicates that the formation is exothermic (release of heat), and the compound is typically stable.
  • For instance, the formation of water from hydrogen and oxygen is highly exothermic.
  • Enthalpy values are often used to predict the energy requirements or releases during industrial chemical processes.
Understanding the enthalpy of formation allows scientists and engineers to develop methods to synthesize compounds efficiently, manage thermal energy in reactions, and design safer chemical systems.

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Most popular questions from this chapter

Explosives called amatols are mixtures of ammonium nitrate and trinitrotoluene (TNT) introduced during World War I when TNT was in short supply. The mixtures can provide \(30 \%\) more explosive power than TNT alone. Above \(300^{\circ} \mathrm{C},\) ammonium nitrate decomposes to \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O} .\) Write a balanced chemical reaction describing the decomposition of ammonium nitrate, and calculate its \(\Delta H_{\mathrm{rxn}}^{\circ}\) using the appropriate \(\Delta H_{\mathrm{f}}^{\circ}\) values from Appendix 4.

The heavier (more dense) hydrocarbons in camp stove fuel are hexanes \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\). a. Calculate the fuel value of \(\mathrm{C}_{6} \mathrm{H}_{14},\) given that \(\Delta H_{\text {comb }}^{\circ}=\) \(-4163 \mathrm{kJ} / \mathrm{mol}\). b. How much heat is released during the combustion of \(1.00 \mathrm{kg}\) of \(\mathrm{C}_{6} \mathrm{H}_{14} ?\) c. How many grams of \(\mathrm{C}_{6} \mathrm{H}_{14}\) are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\) Assume that all of the heat released during combustion is used to heat the water. d. Assume white gas is \(25 \% \mathrm{C}_{5}\) hydrocarbons (see Problem 9.109) and \(75 \%\) C \(_{6}\) hydrocarbons; how many grams of white gas are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

Use average bond energies to estimate the difference in \(\Delta H_{\mathrm{rxn}}^{\circ}\) values between the incomplete combustion of one mole of ethane to carbon monoxide and water vapor and the complete combustion of ethane to carbon dioxide and water vapor.

If we replace the water in a bomb calorimeter with another liquid, do we need to determine a new heat capacity of the calorimeter?

When \(4.00 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathscr{M}=80.04 \mathrm{g} /\)mol) - the active ingredient in some chemical cold packs is dissolved in \(96.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O},\) the temperature of the resulting solution is \(3.07^{\circ} \mathrm{C}\) colder than the water and ammonium nitrate were before they were mixed together. What is the value of \(\Delta H\) for the following dissolution process? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \quad \Delta H=?$$
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