Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 9, Problem 9.73P

(a)

Interpretation Introduction

Interpretation:

The carbon-carbon triple bond energy is to be calculated. Also, the calculated value and the theoretical value are to be compared.

Concept introduction:

The heat of the reaction (ΔHrxn°) is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. ΔHrxn° is negative for exothermic reaction and ΔHrxn° is positive for an endothermic reaction.

The formula to calculate ΔHrxn° of reaction is as follows:

ΔHrxn°=ΔHreactant bond broken°+ΔHproduct bond formed°

Or,

ΔHrxn°=BEreactant bond brokenBEproduct bond formed

The bond energy of reactants is positive and the bond energy of products is negative.

(a)

Expert Solution
Check Mark

Answer to Problem 9.73P

The carbon-carbon triple bond energy is 800kJ/mol and the value is close to the theoretical value.

Explanation of Solution

The given chemical equation for the combustion of ethyne is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.73P

The formula to the enthalpy of the given reaction is as follows:

ΔHrxn°=(2BECH+5/2BEO=O+BEC-C triple bond )(4BEC=O+2BEOH)        (1)

Rearrange the equation (1) to calculate BEC-C triple bond .

BEC-C triple bond =(ΔHrxn°2BECH5/2BEO=O)+(4BEC=O+2BEOH)        (2)

Substitute 413kJ/mol for BECH, 498kJ/mol for BEO=O, 799kJ/mol for BEC=O, 1259 kJ for ΔHrxn° and 467kJ/mol for BEOH in the equation (2).

BEC-C triple bond =[((1259 kJ)(2 mol)(413kJ/mol)(5/2 mol)(498kJ/mol))(4 mol)(799kJ/mol)+(2 mol)(467kJ/mol)]=800kJ/mol

The value is close to the theoretical value 839kJ/mol.

Conclusion

The carbon-carbon triple bond energy is 800kJ/mol and the value is close to the theoretical value.

(b)

Interpretation Introduction

Interpretation:

The heat released when acetylene burn is to be determined.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

(b)

Expert Solution
Check Mark

Answer to Problem 9.73P

The heat released when acetylene burn is 2.417×104kJ.

Explanation of Solution

The formula to calculate the moles of acetylene is as follows:

Moles of acetylene=[given massof acetylenemolecular mass of acetylene]        (3)

Substitute 500g for the mass of acetylene and 26.09 g/mol for the molar mass of acetylene in the equation (3).

Moles of acetylene=[500g26.09 g/mol]=19.2012mol

The heat released by one mole is 1259 kJ/mol. Therefore the heat released by 19.2012mol is calculated as follows:

Heat released=(1259 kJ/mol)(19.2012mol)=2.417×104kJ

Conclusion

The heat released when acetylene burn is 2.417×104kJ.

(c)

Interpretation Introduction

Interpretation:

The mass of CO2 formed is to be determined.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

(c)

Expert Solution
Check Mark

Answer to Problem 9.73P

The mass of CO2 formed is 1690g.

Explanation of Solution

The formula to calculate the mass of CO2 is as follows:

Mass ofCO2=moles ofacetylene(1molCO21molacetylene)(molar mass of CO2)        (4)

Substitute 19.2012mol for moles of acetylene and 44.01 g/mol for molar mass of CO2 in the equation (4).

Mass ofCO2=19.2012mol(2molCO21molacetylene)(44.01 g/mol)=1690.092g1690g

Conclusion

The mass of CO2 formed is 1690g.

(d)

Interpretation Introduction

Interpretation:

The volume of O2 at 298 K and 18.0 atm are consumed is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT        (5)

Here,

P is the pressure.

V is the volume.

T is the temperature.

n is the mole of the gas.

R is the gas constant.

(d)

Expert Solution
Check Mark

Answer to Problem 9.73P

The volume of O2 at 298 K and 18.0 atm are consumed is 65.2L.

Explanation of Solution

The formula to calculate the moles of O2 is as follows:

Moles ofO2=moles ofacetylene(52molO21molacetylene)        (6)

Substitute 19.2012mol for moles of acetylene in the equation (6).

Moles ofO2=19.2012mol(52molO21molacetylene)=48.0030722 mol 

Rearrange the equation (5) to calculate the volume of O2.

VO2=nRTP        (7)

Substitute the value 18atm for P, 298K for T, 48.0030722 mol  for n and 0.0821 Latm/Kmol for R in the equation (7).

VO2=(48.0030722 mol )(0.0821 Latm/Kmol)(298K)18atm=65.2145 L65.2L

Conclusion

The volume of O2 at 298 K and 18.0 atm are consumed is 65.2L.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 9 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 9.5 - Prob. 9.5AFPCh. 9.5 - Prob. 9.5BFPCh. 9 - Prob. 9.1PCh. 9 - Prob. 9.2PCh. 9 - What is the relationship between the tendency of a...Ch. 9 - Prob. 9.4PCh. 9 - Prob. 9.5PCh. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - Prob. 9.9PCh. 9 - Prob. 9.10PCh. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - Give the group number and condensed electron...Ch. 9 - Give the group number and condensed electron...Ch. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - For each pair, choose the compound with the larger...Ch. 9 - Prob. 9.27PCh. 9 - For each pair, choose the compound with the...Ch. 9 - Prob. 9.29PCh. 9 - Use the following to calculate of NaCl: Compared...Ch. 9 - Use the following to calculate of MgF2: Compared...Ch. 9 - Prob. 9.32PCh. 9 - Born-Haber cycles were used to obtain the first...Ch. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - How does the energy of the bond between a given...Ch. 9 - When liquid benzene (C6H6) boils, does the gas...Ch. 9 - Prob. 9.39PCh. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - The text points out that, for similar types of...Ch. 9 - Why is there a discrepancy between an enthalpy of...Ch. 9 - Which of the following gases would you expect to...Ch. 9 - Which of the following gases would you expect to...Ch. 9 - Use bond energies to calculate the enthalpy of...Ch. 9 - Prob. 9.48PCh. 9 - Prob. 9.49PCh. 9 - Prob. 9.50PCh. 9 - Prob. 9.51PCh. 9 - What is the general relationship between IE1 and...Ch. 9 - Is the H—O bond in water nonpolar covalent, polar...Ch. 9 - Prob. 9.54PCh. 9 - How is the partial ionic character of a bond in a...Ch. 9 - Using the periodic table only, arrange the...Ch. 9 - Prob. 9.57PCh. 9 - Prob. 9.58PCh. 9 - Prob. 9.59PCh. 9 - Prob. 9.60PCh. 9 - Use Figure 9.21 to indicate the polarity of each...Ch. 9 - Prob. 9.62PCh. 9 - Prob. 9.63PCh. 9 - Prob. 9.64PCh. 9 - Prob. 9.65PCh. 9 - Prob. 9.66PCh. 9 - Prob. 9.67PCh. 9 - Prob. 9.68PCh. 9 - Prob. 9.69PCh. 9 - Prob. 9.70PCh. 9 - Prob. 9.71PCh. 9 - Geologists have a rule of thumb: when molten rock...Ch. 9 - Prob. 9.73PCh. 9 - Use Lewis electron-dot symbols to represent the...Ch. 9 - Prob. 9.75PCh. 9 - Prob. 9.76PCh. 9 - By using photons of specific wavelengths, chemists...Ch. 9 - Prob. 9.78PCh. 9 - Prob. 9.79PCh. 9 - Prob. 9.80PCh. 9 - Prob. 9.81PCh. 9 - Prob. 9.82PCh. 9 - Prob. 9.83PCh. 9 - Find the longest wavelengths of light that can...Ch. 9 - The work function (ϕ) of a metal is the minimum...Ch. 9 - Prob. 9.86PCh. 9 - Prob. 9.87PCh. 9 - Prob. 9.88PCh. 9 - In a future hydrogen-fuel economy, the cheapest...Ch. 9 - Prob. 9.90PCh. 9 - Prob. 9.91P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Calorimetry Concept, Examples and Thermochemistry | How to Pass Chemistry; Author: Melissa Maribel;https://www.youtube.com/watch?v=nSh29lUGj00;License: Standard YouTube License, CC-BY