Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 21, Problem 45PS

When boron hydrides burn in air, the reactions are very exothermic

  1. (a) Write a balanced equation for the combustion of B5H9(g) in air to give B2O3(s) and H2O(g).
  2. (b) Calculate the enthalpy of combustion for B5H9(g) (ΔfH° = 73.2 kJ/mol), and compare it with the enthalpy of combustion of B2H6 (−2038 kJ/mol). (The enthalpy of formation of B2O3(s) is −1271.9 kJ/mol.)
  3. (c) Compare the enthalpy of combustion of C2H6(g) with that of B2H6(g). Which transfers more energy as heat per gram?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The balanced chemical equation for the combustion of B5H9(g) should be written.

Concept introduction:

  • Balanced chemical equation of a reaction is written according to law of conservation of mass.
  • Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation

Answer to Problem 45PS

The balanced chemical equation for the combustion of B5H9(g) in air is,

  2B5H9(g)+12O2(g)5B2O3(s)+9H2O(g)

Explanation of Solution

Boron hydride of the formula B5H9(g) reacts with the oxygen present in air to give oxide of boron. The equation is said to be balanced if it has an equal number of atoms on both product and reactant side. This is done by multiplying the compounds with the stoichiometric coefficient.

The balanced chemical equation for the combustion of B5H9(g) in air is,

  2B5H9(g)+12O2(g)5B2O3(s)+9H2O(g)

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The enthalpy of combustion of B5H9(g) should be determined.

Concept introduction:

Enthalpy of reaction is the energy required or energy released when a chemical reaction occurs. If the process is endothermic then the energy is absorbed on the formation of the product. In such a case the sign of enthalpy change is positive. If the process is exothermic, then the energy is released on the formation of the product. In such a case the sign of enthalpy change is negative. In general, combustion reactions are exothermic in nature.

The standard enthalpy change is expressed as,

ΔrH°=nΔfH°(products)nΔfH°(reactants)

Answer to Problem 45PS

The ΔrH° for the combustion of B5H9(g) in air is 4341.2 kJ/mol-rxn.

Explanation of Solution

The ΔrH° for the combustion of B5H9(g) in air is calculated below.

Given:

 Refer to Appendix L for the values of standard enthalpies.

The standard enthalpy of B5H9(g) is 73.2 kJ/mol.

The standard enthalpy of O2(g) is 0 kJ/mol.

The standard enthalpy of B2O3(s) is 1271.9 kJ/mol.

The standard enthalpy of H2O(g) is 241.83 kJ/mol.

The balanced reaction is,

  2B5H9(g)+12O2(g)5B2O3(s)+9H2O(g)

The standard enthalpy change is expressed as,

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(5 mol B2O3(s)/mol-rxn)ΔfH°[B2O3(s)]+(9 mol H2O(g)/mol-rxn)ΔfH°[H2O(g)]] [(2 mol B5H9(g)/mol-rxn)ΔfH°[B5H9(g)]+(12 mol O2(g)/mol-rxn)ΔfH°[O2(g)]]] 

Substitute the values,

ΔrH°=[[(5 mol B2O3(s)/mol-rxn)(1271.9 kJ/mol)+(9 mol H2O(g)/mol-rxn)(241.83 kJ/mol)] [(2 mol B5H9(g)/mol-rxn)(73.2 kJ/mol)+(12 mol O2(g)/mol-rxn)(0 kJ/mol)]]  =4341.2 kJ/mol-rxn

The enthalpy of combustion of B2H6(g) is 2038 kJ/mol-rxn.

Thus, the enthalpy of combustion of B5H9(g) is double to that B2H6(g).

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The enthalpy of combustion of C2H6(g) should be given and compare the value with the enthapy of combustion of B5H9(g).

Concept introduction:

Enthalpy of reaction is the energy required or energy released when a chemical reaction occurs. If the process is endothermic then the energy is absorbed on the formation of the product. In such a case the sign of enthalpy change is positive. If the process is exothermic, then the energy is released on the formation of the product. In such a case the sign of enthalpy change is negative. In general, combustion reactions are exothermic in nature.

The standard enthalpy change is expressed as,

ΔrH°=nΔfH°(products)nΔfH°(reactants)

Answer to Problem 45PS

The enthalpy of combustion of C2H6(g) is 1428.7 kJ/mol-rxn.

The enthalpy of combustion for C2H6(g) per gram is 47.6 kJ/g.

The enthalpy of combustion for B2H6(g) per gram is 73.6 kJ/g.

Explanation of Solution

The enthalpy of combustion of C2H6(g) is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpies.

The standard enthalpy of C2H6(g) is 83.85 kJ/mol.

The standard enthalpy of O2(g) is 0 kJ/mol.

The standard enthalpy of CO2(g) is 393.509 kJ/mol.

The standard enthalpy of H2O(g) is 241.83 kJ/mol.

The balanced chemical equation is,

  C2H6(g)+72O2(g)2CO2(g)+3H2O(g)

The standard enthalpy change is expressed as,

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(2 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]+(3 mol H2O(g)/mol-rxn)ΔfH°[H2O(g)]] [(1 mol C2H6(g)/mol-rxn)ΔfH°[C2H6(g)]+(3.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]]] 

Substitute the values,

ΔrH°=[[(2 mol CO2(g)/mol-rxn)(393.509 kJ/mol)+(3 mol H2O(g)/mol-rxn)(241.83 kJ/mol)] [(1 mol C2H6(g)/mol-rxn)(83.85 kJ/mol)+(3.5 mol O2(g)/mol-rxn)(0 kJ/mol)]]=1428.7 kJ/mol-rxn

The molar mass of C2H6(g) is 30 g/mol.

Thus, the enthalpy of combustion for C2H6(g) per gram is 47.6 kJ/g.

The molar mass of B2H6(g) is 27.66 g/mol.

Thus, the enthalpy of combustion for B2H6(g) per gram is 73.6 kJ/g.

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Chapter 21 Solutions

Chemistry & Chemical Reactivity

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