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
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Chapter 6, Problem 6.91P

(a)

Interpretation Introduction

Interpretation:

ΔHrxn° for the three reactions involved in the combustion of candle (C21H44) is to be determined.

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.

(a)

Expert Solution
Check Mark

Answer to Problem 6.91P

ΔHrxn° for the three reactions are 13108kJ, 7165kJ and 4844kJ respectively.

Explanation of Solution

The balanced chemical equations for the combustion of paraffin (C21H44) are as follows:

C21H44(s)+32O2(g)21CO2(g)+22H2O(g)C21H44(s)+432O2(g)21CO(g)+22H2O(g)C21H44(s)+11O2(g)21C(s)+22H2O(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 for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{21ΔHf°[CO2(g)]+22ΔHf°[H2O(g)]}{1ΔHf°[C21H44(s)]+32ΔHf°[O2(g)]}] (1)

Substitute 393.5kJ/mol for ΔHf°[CO2(g)], 241.826kJ/mol for ΔHf°[H2O(g)], 476kJ/mol for ΔHf°[C21H44(s)] and 0 for ΔHf°[O2(g)] in the equation (1).

ΔHrxn°=[{(21mol)(393.5kJ/mol)+(22mol)(241.826kJ/mol)}{(1mol)(476kJ/mol)+(22mol)(0)}]=13107.672kJ13108kJ

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{21ΔHf°[CO(g)]+22ΔHf°[H2O(g)]}{1ΔHf°[C21H44(s)]+432ΔHf°[O2(g)]}] (2)

Substitute 110.5kJ/mol for ΔHf°[CO(g)], 241.826kJ/mol for ΔHf°[H2O(g)], 476kJ/mol for ΔHf°[C21H44(s)] and 0 for ΔHf°[O2(g)] in the equation (2).

ΔHrxn°=[{(21mol)(110.5kJ/mol)+(22mol)(241.826kJ/mol)}{(1mol)(476kJ/mol)+(432mol)(0)}]=7164.672kJ7165kJ

The formula to calculate the standard enthalpy for third reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{21ΔHf°[C(s)]+22ΔHf°[H2O(g)]}{1ΔHf°[C21H44(s)]+11ΔHf°[O2(g)]}] (3)

Substitute 0 for ΔHf°[C(s)], 241.826kJ/mol for ΔHf°[H2O(g)], 476kJ/mol for ΔHf°[C21H44(s)] and 0 for ΔHf°[O2(g)] in the equation (3).

ΔHrxn°=[{(21mol)(0)+(22mol)(241.826kJ/mol)}{(1mol)(476kJ/mol)+(11mol)(0)}]=4844.172kJ4844kJ

Conclusion

ΔHrxn° for the three reactions are 13108kJ, 7165kJ and 4844kJ respectively.

(b)

Interpretation Introduction

Interpretation:

The heat released when a 254g candle burns 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.

(b)

Expert Solution
Check Mark

Answer to Problem 6.91P

The heat released when a 254g candle burns is 1.12×104kJ.

Explanation of Solution

The formula to calculate heat released when 254g candle burns is as follows:

Heat released=(mass of C21H44molar mass of C21H44)(heat released per mole) (4)

Substitute 254g for mass of C21H44, 296.56g/mol for molar mass of C21H44 and 13107.672kJ/mol for heat released per mole in the equation (4).

Heat released=(254g296.56g/mol)(13107.672kJ/mol)=1.12266×104kJ1.12×104kJ

Conclusion

The heat released when a 254g candle burns is 1.12×104kJ.

(c)

Interpretation Introduction

Interpretation:

The heat released when 8.00% by mass of the candle burns incompletely and 5.00% by mass of it undergoes soot formation 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.

(c)

Expert Solution
Check Mark

Answer to Problem 6.91P

The heat released is 1.05×104kJ.

Explanation of Solution

Consider 87.00% by mass of the candle undergoes complete combustion.

The formula to calculate moles of candle (C21H44) is as follows:

Moles of C21H44=(mass of C21H44molar mass of C21H44) (5)

Substitute 254g for mass of C21H44 and 296.56g/mol for molar mass of C21H44 in the equation (5).

Moles of C21H44=(254g296.56g/mol)=0.856488mol

The formula to calculate heat released when candle burns is as follows:

Heat released=[(mass % of complete combustion100%)(moles of C21H44)(ΔHrxn°of complete combustion)(mass % of incomplete combustion100%)(moles of C21H44)(ΔHrxn°of incomplete combustion)(mass % of soot formation100%)(moles of C21H44)(ΔHrxn°of soot formation)] (6)

Substitute 87.00% for mass % of complete combustion, 0.856488mol for moles of C21H44, 13107.672kJ/mol for ΔHrxn° of complete combustion, 8.00% for mass % of incomplete combustion, 7164.672kJ/mol for ΔHrxn° of incomplete combustion, 5.00% for mass % of soot formation and 4844.172kJ/mol for ΔHrxn° of soot formation in the equation (6).

Heat released=[(87.00%100%)(0.856488mol)(13107.672kJ/mol)(8.00%100%)(0.856488mol)(7164.672kJ/moln)(5.00%100%)(0.856488mol)(4844.172kJ/mol)]=[(0.87)(0.856488mol)(13107.672kJ/mol)(0.080)(0.856488mol)(7164.672kJ/moln)(0.050)(0.856488mol)(4844.172kJ/mol)]=1.04655×104kJ1.05×104kJ

Conclusion

The heat released is 1.05×104kJ.

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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
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