/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Define formation reaction and gi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

Define formation reaction and give an example.

Short Answer

Expert verified
A formation reaction is where a compound forms from its elements in their standard states; e.g., forming water from hydrogen and oxygen: \(\text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)\).

Step by step solution

01

Understanding Formation Reactions

A formation reaction is a chemical reaction in which a single compound forms from its constituent elements in their standard states. The standard state of an element is its most stable form at 1 atmosphere and 25掳C (298 K). These reactions are important in calculating changes in enthalpy using standard enthalpies of formation.
02

Example of Formation Reaction

Consider the formation of water from hydrogen and oxygen. In its formation reaction, hydrogen gas (\( ext{H}_2\)) and oxygen gas (\( ext{O}_2\)) react to form water (\( ext{H}_2 ext{O}\)). The balanced chemical equation for this reaction is:\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)\]This equation illustrates that water forms from hydrogen and oxygen in their standard states.
03

Properties of Formation Reactions

Formation reactions are characterized by forming one mole of a compound from its elements in their standard states. They are used extensively in thermochemistry to compute reaction enthalpies using Hess's Law. The enthalpy change of a formation reaction is the standard enthalpy of formation (螖H掳f), typically expressed in kilojoules per mole (kJ/mol).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Standard State
In thermochemistry, the concept of the standard state is essential for understanding how chemical reactions progress and for calculating their associated energy changes. The standard state of a substance is the form it normally takes under one atmosphere of pressure and a temperature of 25掳C (298 K). This provides a reference point to ensure consistency when reporting and calculating thermodynamic properties.

A few key points about standard states include:
  • For gases, the standard state is a pure gas at 1 atm pressure.
  • Liquids and solids are in their pure form under the same conditions.
  • For solutions, the standard state is a concentration of exactly 1 mol/L.
Using standard states allows scientists to predict and compare energy changes in chemical reactions, as all measurements and calculations are made under standardized conditions.
Standard Enthalpy of Formation
The standard enthalpy of formation, denoted as 螖H掳f, describes the change in enthalpy when one mole of a compound forms from its elements, each in their standard state. It is a pivotal concept in thermochemistry because it gives insight into the energy requirement or release during the formation of compounds. Here are some important details:

  • For any element in its standard state, the standard enthalpy of formation is defined as zero. This serves as a baseline for measuring energy changes.
  • The standard enthalpy of formation is often expressed in kilojoules per mole (kJ/mol). A positive value indicates an endothermic reaction, whereas a negative value suggests an exothermic process.
  • These values are essential for calculating the overall enthalpy change in reactions using Hess's Law, which allows us to add enthalpies of formation to find a net change for complex reactions.
In conclusion, understanding 螖H掳f helps predict whether a reaction will be energetically favorable and how much energy it will exchange.
Thermochemistry
Thermochemistry is the study of the energy and heat associated with chemical reactions and changes in state. This field is critical for understanding how energy moves in the form of heat, which is important for both chemical industry and academic research.

Here are some of the key aspects of thermochemistry:
  • It involves the measurement of energy changes that occur during chemical and physical processes.
  • Thermochemistry uses concepts like enthalpy, entropy, and Gibbs free energy to predict the direction and extent of chemical reactions.
  • This branch of chemistry employs Hess's Law, which allows the calculation of enthalpy changes for reactions that aren't easily measured by summing the enthalpies of formation for intermediate steps.
Thermochemistry, by tracking the flow of heat, helps scientists and engineers to harness reactions efficiently, whether in creating new materials, generating energy, or understanding biological processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the work when a gas contracts from \(3.45 \mathrm{~L}\) to \(0.97 \mathrm{~L}\) under an external pressure of \(0.985 \mathrm{~atm} ?\)

If the \(\Delta H\) for \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}\) is \(-65.6 \mathrm{~kJ}\), what is the \(\Delta H\) for this reaction? \(2 \mathrm{C}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{6}\)

What is the enthalpy change for the unknown reaction? $$ \begin{array}{l} \mathrm{Pb}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{~s}) \Delta H=-359 \mathrm{~kJ} \\ \mathrm{PbCl}_{2}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PbCl}_{4}(\ell) \Delta H=? \\ \mathrm{~Pb}(\mathrm{~s})+2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PbCl}_{4}(\ell) \Delta H=-329 \mathrm{~kJ} \end{array} $$

The \(\Delta H\) for \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}\) is \(181 \mathrm{~kJ}\). What is the \(\Delta H\) for this reaction? \(\mathrm{NO} \rightarrow 1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2}\)

The \(\Delta H\) for \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is \(-44 \mathrm{~kJ}\). What is the \(\Delta H\) for this reaction? \(2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.