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

Videos

Question
Book Icon
Chapter 6, Problem 6.102P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the combustion of kerosene is to be written.

Concept introduction:

A combustion reaction is the reaction in which reactant is reacted with molecular oxygen to form the product. Heat is released and the energy is produced in the reaction. Molecular oxygen is employed as an oxidizing agent in these reactions.

(a)

Expert Solution
Check Mark

Answer to Problem 6.102P

The balanced equation for the combustion of kerosene is,

2C12H26(l)+37O2(g)24CO2(g)+26H2O(g)

Explanation of Solution

Kerosene (C12H26) reacts with molecular oxygen to form carbon dioxide and water molecule. The molecular equation foe the combustion of Kerosene (C12H26) is,

2C12H26(l)+37O2(g)24CO2(g)+26H2O(g)

Conclusion

A combustion reaction is the reaction in which reactant is reacted with molecular oxygen to form carbon dioxide and water molecule.

(b)

Interpretation Introduction

Interpretation:

ΔHf° of kerosene when ΔHrxn° is 1.5×104kJ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=mΔHf (products)°mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(b)

Expert Solution
Check Mark

Answer to Problem 6.102P

ΔHf° of kerosene when ΔHrxn° is 1.5×104kJ is 3.66×102kJ/mol.

Explanation of Solution

The balanced chemical equation for the combustion of kerosene (C12H26) is as follows:

2C12H26(l)+37O2(g)24CO2(g)+26H2O(g)

The standard state of oxygen is O2(g) so the standard enthalpy of formation for O2 is zero.

The formula to calculate the standard enthalpy of a given reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{24ΔHf°[CO2(g)]+26ΔHf°[H2O(g)]}{2ΔHf°[C12H26(l)]+32ΔHf°[O2(g)]}] (1)

Rearrange equation (1) to calculate 2ΔHf° of C2H2(g) is as follows:

2ΔHf°[C12H26(l)]=[(24ΔHf°[CO2(g)]+26ΔHf°[H2O(g)])(ΔHrxn°+32ΔHf°[O2(g)])] (2)

Substitute 393.5kJ/mol for ΔHf°[CO2(g)], 241.826kJ/mol for ΔHf°[H2O(g)] and 1.5×104kJ for ΔHrxn° and 0 for ΔHf°[O2(g)] in the equation (2).

(2mol)ΔHf°[C12H26(l)]=[(24 mol)(393.5kJ/mol)+24(241.826kJ/mol)((1.5×104kJ)+37(0))]ΔHf°[C12H26(l)]=731.476kJ2mol=365.738kJ/mol3.66×102kJ/mol

Conclusion

ΔHf° of kerosene when ΔHrxn° is 1.5×104kJ is 3.66×102kJ/mol.

(c)

Interpretation Introduction

Interpretation:

The heat released by the combustion of 0.50 gal of kerosene (C12H26) is to be determined.

Concept introduction:

Heat (q) is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert gal into qt is:

1gal=4qt

The conversion factor to convert L into qt is:

1L=1.057qt

(c)

Expert Solution
Check Mark

Answer to Problem 6.102P

The heat released by the combustion of 0.50 gal of kerosene (C12H26) is 6.2×104kJ.

Explanation of Solution

The formula to calculate moles of kerosene (C12H26) is as follows:

Moles of C12H26=(volume of C12H26molar mass of C12H26)(density of C12H26) (3)

Substitute 0.50 gal for volume of C12H26, 0.749g/mL for density of C12H26 and 170.33g/mol for molar mass of C12H26 in the equation (1).

Moles of C8H18=(0.50 gal170.33g/mol)(4qt1gal)(1L1.057qt)(1000mL1L)(0.749g/mL)=8.3204282mol

The heat released by 2mol of C12H26 is 1.5×104kJ.

Therefore, the heat released by 8.3204282mol of C12H26 is as follows:

Heatreleased=(8.3204282mol)(1.5×104kJ2mol)=6.2403×104kJ6.2×104kJ

Conclusion

The heat released by the combustion of 0.50 gal of kerosene (C12H26) is 6.2×104kJ.

(d)

Interpretation Introduction

Interpretation:

The volume of kerosene in gal that must be burned for a kerosene furnace to produce 1250Btu is to be calculated.

Concept introduction:

Heat (q) is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert Btu into kJqt is:

1Btu=1.055kJ

(d)

Expert Solution
Check Mark

Answer to Problem 6.102P

The volume of kerosene in gal is 1.1×102gal.

Explanation of Solution

The heat released by the combustion of 0.50 gal of kerosene (C12H26) is 6.2×104kJ.

The formula to calculate the volume is as follows:

Volume=(energy of kerotene)(0.50 gal6.2×104kJ) (4)

Substitute 1250Btu for energy of kerotene in the equation (4).

Volume=(1250Btu)(1.055kJ1Btu)(0.50 gal6.2×104kJ)1.1×102gal

Conclusion

The volume of kerosene in gal is 1.1×102gal.

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 6 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 6.3 - When 25.0 mL of 2.00 M HNO3 and 50.0 mL of 1.00 M...Ch. 6.3 - Prob. 6.6BFPCh. 6.3 - Prob. 6.7AFPCh. 6.3 - Prob. 6.7BFPCh. 6.4 - Prob. 6.8AFPCh. 6.4 - Prob. 6.8BFPCh. 6.5 - Prob. 6.9AFPCh. 6.5 - Prob. 6.9BFPCh. 6.6 - Prob. 6.10AFPCh. 6.6 - Prob. 6.10BFPCh. 6.6 - Prob. 6.11AFPCh. 6.6 - Prob. 6.11BFPCh. 6.6 - Prob. B6.1PCh. 6.6 - Prob. B6.2PCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - A system releases 255 cal of heat to the...Ch. 6 - What is the change in internal energy (in J) of a...Ch. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Thermal decomposition of 5.0 metric tons of...Ch. 6 - Prob. 6.15PCh. 6 - The external pressure on a gas sample is 2660...Ch. 6 - The nutritional calorie (Calorie) is equivalent to...Ch. 6 - If an athlete expends 1950 kJ/h, how long does it...Ch. 6 - Prob. 6.19PCh. 6 - Hot packs used by skiers produce heat via the...Ch. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - For each process, state whether ΔH is less than...Ch. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - What data do you need to determine the specific...Ch. 6 - Is the specific heat capacity of a substance an...Ch. 6 - Prob. 6.35PCh. 6 - Both a coffee-cup calorimeter and a bomb...Ch. 6 - Find q when 22.0 g of water is heated from 25.0°C...Ch. 6 - Calculate q when 0.10 g of ice is cooled from...Ch. 6 - A 295-g aluminum engine part at an initial...Ch. 6 - Prob. 6.40PCh. 6 - Two iron bolts of equal mass—one at 100.°C, the...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - A 30.5-g sample of an alloy at 93.0°C is placed...Ch. 6 - When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL...Ch. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - A chemist places 1.750 g of ethanol, C2H6O, in a...Ch. 6 - High-purity benzoic acid (C6H5COOH; ΔH for...Ch. 6 - Two aircraft rivets, one iron and the other...Ch. 6 - A chemical engineer burned 1.520 g of a...Ch. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Consider the following balanced thermochemical...Ch. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - When 1 mol of KBr(s) decomposes to its elements,...Ch. 6 - Prob. 6.61PCh. 6 - Compounds of boron and hydrogen are remarkable for...Ch. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Write the balanced overall equation (equation 3)...Ch. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Calculatefor each of the following: SiO2(s) +...Ch. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - The common lead-acid car battery produces a large...Ch. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - The following scenes represent a gaseous reaction...Ch. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - Prob. 6.102PCh. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Liquid methanol (CH3OH) canbe used as an...Ch. 6 - Prob. 6.108P
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
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY