Skip to main content
Ch.8 - Basic Concepts of Chemical Bonding
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBrown 14th EditionCh.8 - Basic Concepts of Chemical BondingProblema 118
Capitolo 8, Problema 118

(a) Use average bond enthalpies (Table 8.3) to estimate H for the atomization of naphthalene, C10H8:

Guida verificata passo dopo passo
1
Identify the bonds present in naphthalene, C_{10}H_{8}. Naphthalene consists of carbon-carbon (C-C) single bonds, carbon-carbon (C=C) double bonds, and carbon-hydrogen (C-H) bonds.
Determine the number of each type of bond in a single molecule of naphthalene. Typically, naphthalene has 5 C-C single bonds, 6 C=C double bonds, and 8 C-H bonds.
Use the average bond enthalpies from Table 8.3 to find the energy required to break each type of bond. For example, use the bond enthalpy for C-C, C=C, and C-H bonds.
Calculate the total energy required to break all the bonds in one molecule of naphthalene by multiplying the number of each type of bond by its respective bond enthalpy and summing these values.
The total energy calculated represents the enthalpy change (\( \Delta H \)) for the atomization of naphthalene, as it is the energy required to break all bonds in the molecule.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
1m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Bond Enthalpy

Bond enthalpy, or bond dissociation energy, is the energy required to break a specific bond in a molecule in the gas phase. It is an average value derived from various compounds and reflects the strength of a bond. Understanding bond enthalpy is crucial for estimating the energy changes during chemical reactions, such as atomization, where bonds are broken to form individual atoms.
Video consigliato:
Percorso guidato
02:34
Enthalpy of Formation

Atomization

Atomization refers to the process of converting a compound into its individual atoms, typically requiring energy input to break all chemical bonds. In the context of naphthalene (C10H8), atomization involves breaking the carbon and hydrogen bonds to yield separate carbon and hydrogen atoms. The enthalpy change associated with this process can be calculated using the bond enthalpies of the bonds broken.
Video consigliato:
Percorso guidato
02:10
Atom Structure

Enthalpy Change (ΔH)

Enthalpy change (ΔH) is a measure of the heat content change in a system during a chemical reaction at constant pressure. It can be calculated using the formula ΔH = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed). For the atomization of naphthalene, the total bond enthalpies of the bonds broken must be summed to estimate the overall energy change associated with the process.
Video consigliato:
Percorso guidato
02:34
Enthalpy of Formation
Pratica correlata
Domanda del libro di testo
Use average bond enthalpies from Table 8.4 to estimate the enthalpies of the following gas-phase reactions: Reaction 1: HF1g2 + H2O1g2 Δ F-1g2 + H3O+1g2 Reaction 2: HCl1g2 + H2O1g2 Δ Cl-1g2 + H3O+1g2 Are both reactions exothermic? How do these values relate to the different strengths of hydrofluoric and hydrochloric acid?
1372
views
1
rank
Domanda del libro di testo

Ammonium chloride, NH4Cl, is a very soluble salt in water. (c) If you dissolve 14 g of ammonium chloride in 500.0 mL of water, what is the molar concentration of the solution?

1481
views
Domanda del libro di testo

(a) Compare the bond enthalpies (Table 8.3) of the carbon– carbon single, double, and triple bonds to deduce an average pi bond contribution to the enthalpy. What fraction of a single bond does this quantity represent?

1106
views
Domanda del libro di testo

Ammonium chloride, NH4Cl, is a very soluble salt in water. (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?

1084
views