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 10, Problem 10.75P

(a)

Interpretation Introduction

Interpretation:

ΔHrxn° for the combustion of gaseous ethanol is to be calculated.

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 10.75P

1267kJ is the heat of combustion of ethanol.

Explanation of Solution

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

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.75P , additional homework tip  1

The number of broken bonds is 5 CH bond, 1 CC bond, 1 CO bond, 1 OH bond, and 3 O=O bonds.

The number of bonds formed is 4 C=O bonds and 6 OH bonds.

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

  ΔHrxn°=[(5BECH+1BECC+1BECO+1BEOH+3BEO=O)(4BEC=O+6BEOH)]        (1)

Substitute 413kJ/mol for BECH, 347kJ/mol for BECC, 358kJ/mol for BECO, 498kJ/mol for BEO=O, 799kJ/mol for BEC=O and 467kJ/mol for BEOH in the equation (1).

  ΔHrxn°=[((5 mol)(413kJ/mol)+(1 mol)(347kJ/mol)+(1 mol)(358kJ/mol)+(1 mol)(467kJ/mol)+(3 mol)(498kJ/mol))(4 mol)(799kJ/mol)+(6 mol)(467kJ/mol)]=1267kJ.

Conclusion

The number of broken bonds is 5 CH bond, 1 CC bond, 1 CO bond, 1 OH bond, and 3 O=O bond while the number of bonds formed is 4 C=O bonds and 6 OH bonds.

(b)

Interpretation Introduction

Interpretation:

The enthalpy of reaction for the combustion of liquid ethanol is to be calculated.

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.

A combustion reaction is that 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.

(b)

Expert Solution
Check Mark

Answer to Problem 10.75P

1226kJ is the enthalpy of reaction for the combustion of liquid ethanol.

Explanation of Solution

The heat of vaporization of ethanol (ΔHvap°) is 40.5kJ/mol. The formula to calculate the enthalpy combustion of liquid ethanol is as follows:

  ΔHcom(liq ethanol)°=ΔHvap°+ΔHrxn°        (2)

Substitute 40.5kJ/mol for ΔHvap° and 1267kJ for ΔHrxn° in the equation (2).

  ΔHcom(liq ethanol)°=(1mol)(40.5kJ/mol)+(1267kJ)=1226.7kJ1226kJ.

Conclusion

The vaporization of a liquid occurs when the intermolecular forces between the molecules break and the molecules are free to vaporize.

(c)

Interpretation Introduction

Interpretation:

The enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies and standard enthalpies of formation is to be compared.

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 reactant at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

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

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

(c)

Expert Solution
Check Mark

Answer to Problem 10.75P

The enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies is close to the enthalpy of reaction for the combustion of liquid ethanol calculated from standard enthalpies of formation.

Explanation of Solution

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

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.75P , additional homework tip  2

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 the reaction (ΔHrxn°) is as follows:

  ΔHrxn°=[{2ΔHf°[CO2(g)]+3ΔHf°[H2O(g)]}{1ΔHf°[C2H5OH(l)]+3ΔHf°[O2(g)]}]        (3)

Substitute 393.5kJ/mol for ΔHf°[CO2(g)], 241.826kJ/mol for ΔHf°[H2O(g)], 277.3kJ/mol for ΔHf°[C2H5OH(l)] and 0 for ΔHf°[O2(g)] in the equation (3).

  ΔHrxn°=[{(2mol)(393.5kJ/mol)+(3mol)(241.826kJ/mol)}{(1mol)(277.3kJ/mol)+(3mol)(0)}]=1234.848kJ1234.8kJ.

The calculated value is close to the value calculated in part (b). Therefore the enthalpy of reaction for the combustion of liquid ethanol calculated from the bond energies and standard enthalpies of formation are in good agreement.

Conclusion

The enthalpy of reaction for the combustion of liquid ethanol calculated from standard enthalpies of formation is 1234.8kJ.

(d)

Interpretation Introduction

Interpretation:

ΔHrxn° for the formation of gaseous ethanol from ethylene gas with water vapor is to be calculated.

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.

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

(d)

Expert Solution
Check Mark

Answer to Problem 10.75P

37kJ is the heat of formation of gaseous ethanol from ethylene gas.

Explanation of Solution

The Lewis structures for the reaction of the formation of gaseous ethanol are as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.75P , additional homework tip  3

The number of broken bonds is 4 CH bonds, 1 C=C bond, and 2 OH bond.

The number of bonds formed is 1 CC bond, 5 CH bonds, 1 CO bond, and 1 OH bond.

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

  ΔHrxn°=[(4BECH+1BEC=C+2BEOH)(5BECH+1BECC+1BECO+1BEOH)]        (4)

Substitute 413kJ/mol for BECH, 347kJ/mol for BECC, 358kJ/mol for BECO, 467kJ/mol for BEOH and 614kJ/mol for BEC=C in the equation (4).

  ΔHrxn°=[((4 mol)(413kJ/mol)+(1 mol)(614kJ/mol)+(2 mol)(467kJ/mol))(5 mol)(413kJ/mol)+(1 mol)(347kJ/mol)+(1 mol)(358kJ/mol)+(1 mol)(467kJ/mol)]=37kJ.

Conclusion

The number of broken bonds is 4 CH bonds, 1 C=C bond, and 2 OH bond while the number of bonds formed is 1 CC bond, 5 CH bonds, 1 CO bond and 1 OH bond.

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Chapter 10 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 10.2 - Prob. 10.6AFPCh. 10.2 - Prob. 10.6BFPCh. 10.2 - Prob. 10.7AFPCh. 10.2 - Prob. 10.7BFPCh. 10.2 - Prob. 10.8AFPCh. 10.2 - Prob. 10.8BFPCh. 10.3 - Prob. 10.9AFPCh. 10.3 - Prob. 10.9BFPCh. 10.3 - Prob. B10.1PCh. 10.3 - Prob. B10.2PCh. 10 - Prob. 10.1PCh. 10 - When is a resonance hybrid needed to adequately...Ch. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Draw a Lewis structure for (a) SiF4; (b) SeCl2;...Ch. 10 - Draw a Lewis structure for (a) ; (b) C2F4; (c)...Ch. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.11PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - Molten beryllium chloride reacts with chloride ion...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Phosgene is a colorless, highly toxic gas that was...Ch. 10 - If you know the formula of a molecule or ion, what...Ch. 10 - In what situation is the name of the molecular...Ch. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Consider the following molecular shapes. (a) Which...Ch. 10 - Use wedge-bond perspective drawings (if necessary)...Ch. 10 - Prob. 10.33PCh. 10 - Determine the electron-group arrangement,...Ch. 10 - Determine the electron-group arrangement,...Ch. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Determine the shape, ideal bond angle(s), and the...Ch. 10 - Prob. 10.41PCh. 10 - Determine the shape around each central atom in...Ch. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Arrange the following ACln species in order of...Ch. 10 - State an ideal value for each of the bond angles...Ch. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - How can a molecule with polar covalent bonds not...Ch. 10 - Prob. 10.54PCh. 10 - Consider the molecules SCl2, F2, CS2, CF4, and...Ch. 10 - Consider the molecules BF3, PF3, BrF3, SF4, and...Ch. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - When SO3 gains two electrons, forms. (a) Which...Ch. 10 - The actual bond angle in NO2 is 134.3°, and in it...Ch. 10 - Prob. 10.69PCh. 10 - Propylene oxide is used to make many products,...Ch. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - A gaseous compound has a composition by mass of...Ch. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Pure HN3 (atom sequence HNNN) is explosive. In...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Oxalic acid (H2C2O4) is found in toxic...Ch. 10 - Prob. 10.87PCh. 10 - Hydrazine (N2H4) is used as a rocket fuel because...Ch. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Consider the following molecular shapes: Match...Ch. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Phosphorus pentachloride, a key industrial...
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