ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 4, Problem 4D.21E

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy of the final stage of the production of nitric acid has to be determined.

Concept Introduction:

The standard enthalpy of the reaction is the difference of the sum of standard enthalpy of formation of all the products and the sum of standard enthalpy of formation of all the reactants.  The mathematical equation for the calculation of standard enthalpy of the reaction is shown below.

ΔH°=nΔHf(products)nΔHf(reactants)

(a)

Expert Solution
Check Mark

Answer to Problem 4D.21E

The standard enthalpy of the final stage of the production of nitric acid is -138.18kJ_.

Explanation of Solution

The balanced chemical equation that represents the final stage of the production of nitric acid is shown below.

    3NO2(g)+H2O(l)2HNO3(aq)+NO(g)

The standard enthalpy of formation of products that are HNO3(aq) and NO(g) is 207.36kJmol1 and +90.25kJmol1 respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  nΔHf(products)=nHNO3ΔHf(HNO3,aq)+nNOΔHf(NO,g)        (1)

Where,

  • nΔHf(products) is total enthalpy of formation of products.
  • nHNO3 is the number of moles of HNO3(aq).
  • ΔHf(HNO3,aq) is the standard enthalpy of formation of HNO3(aq).
  • nNO is the number of moles of NO(g).
  • ΔHf(NO,g) is the standard enthalpy of formation of NO(g).

The value of nHNO3 is 2mol.

The value of ΔHf(HNO3,aq) is 207.36kJmol1.

The value of nNO is 1mol.

The value of ΔHf(NO,g) is +90.25kJmol1.

Substitute the value of nHNO3, ΔHf(HNO3,aq), nNO and ΔHf(NO,g) in equation (1).

  nΔHf(products)=(2mol)×(207.36kJmol1)+(1mol)×(+90.25kJmol1)=414.72kJ+90.25kJ=324.47kJ

The standard enthalpy of formation of reactant that is NO2(g) and H2O(l) is 33.18kJmol1 and 285.83kJmol1 respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  nΔHf(reactants)=nNO2ΔHf(NO2,g)+nH2OΔHf(H2O,l)        (2)

Where,

  • nΔHf(reactants) is total enthalpy of formation of reactants.
  • nNO2 is the number of moles of NO2(g).
  • ΔHf(NO2,g) is the standard enthalpy of formation of NO2(g).
  • nH2O is the number of moles of H2O(l).
  • ΔHf(H2O,l) is the standard enthalpy of formation of H2O(l).

The value of nNO2 is 3mol.

The value of ΔHf(NO2,g) is 33.18kJmol1.

The value of nH2O is 1mol.

The value of ΔHf(H2O,l) is 285.83kJmol1.

Substitute the value of nNO2, ΔHf(NO2,g), nH2O and ΔHf(H2O,l) in equation (2).

  nΔHf(reactants)=(3mol)×(33.18kJmol1)+(1mol)×(285.83kJmol1)=99.54kJ285.83kJ=186.29kJ

The standard enthalpy of formation of ethyne is calculated by the expression shown below.

  ΔH°=nΔHf(products)nΔHf(reactants)        (3)

Where,

  • ΔH° is standard enthalpy of reaction.
  • nΔHf(products) is total enthalpy of formation of products.
  • nΔHf(reactants) is total enthalpy of formation of reactants.

The value of nΔHf(products) is 324.47kJ.

The value of nΔHf(reactants) is 186.29kJ.

Substitute the value of nΔHf(products) and nΔHf(reactants) in equation (3).

  ΔH°=324.47kJ(186.29kJ)=324.47kJ+186.29kJ=138.18kJ

Thus, the standard enthalpy of the final stage of the production of nitric acid is -138.18kJ_.

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy of the industrial synthesis of boron trifluoride has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 4D.21E

The standard enthalpy of the industrial synthesis of boron trifluoride is 1593.91kJ_.

Explanation of Solution

The balanced chemical equation that represents the industrial synthesis of boron trifluoride is shown below.

    B2O3(s)+3CaF2(s)2BF3(g)+3CaO(s)

The standard enthalpy of formation of products that are BF3(g) and CaO(s) is 1136.8kJmol1 and 635.09kJmol1 respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  nΔHf(products)=nBF3ΔHf(BF3,g)+nCaOΔHf(CaO,s)        (4)

Where,

  • nΔHf(products) is total enthalpy of formation of products.
  • nBF3 is the number of moles of BF3(g).
  • ΔHf(BF3,g) is the standard enthalpy of formation of BF3(g).
  • nCaO is the number of moles of CaO(s).
  • ΔHf(CaO,s) is the standard enthalpy of formation of CaO(s).

The value of nBF3 is 2mol.

The value of ΔHf(BF3,g) is 1136.8kJmol1.

The value of nCaO is 1mol.

The value of ΔHf(CaO,g) is 635.09kJmol1.

Substitute the value of nBF3, ΔHf(BF3,g), nCaO and ΔHf(CaO,s) in equation (4).

  nΔHf(products)=(2mol)×(1136.8kJmol1)+(1mol)×(635.09kJmol1)=2273.6kJ635.09kJmol1=2908.69kJ

The standard enthalpy of formation of reactant that is B2O3(s) and CaF2(s) is 843.8kJmol1 and 1219.6kJmol1 respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  nΔHf(reactants)=nB2O3ΔHf(B2O3,s)+nCaF2ΔHf(CaF2,s)        (5)

Where,

  • nΔHf(reactants) is total enthalpy of formation of reactants.
  • nB2O3 is the number of moles of B2O3(s).
  • ΔHf(B2O3,s) is the standard enthalpy of formation of B2O3(s).
  • nCaF2 is the number of moles of CaF2(s).
  • ΔHf(CaF2,s) is the standard enthalpy of formation of CaF2(s).

The value of nB2O3 is 1mol.

The value of ΔHf(B2O3,s) is 843.8kJmol1.

The value of nCaF2 is 3mol.

The value of ΔHf(CaF2,s) is 1219.6kJmol1.

Substitute the value of nB2O3, ΔHf(B2O3,s), nCaF2 and ΔHf(CaF2,s) in equation (5).

  nΔHf(reactants)=(1mol)×(843.8kJmol1)+(3mol)×(1219.6kJmol1)=843.8kJ3658.8kJ=4502.6kJ

The value of nΔHf(products) is 2908.69kJ.

The value of nΔHf(reactants) is 4502.6kJ.

Substitute the value of nΔHf(products) and nΔHf(reactants) in equation (3).

  ΔH°=2908.69kJ(4502.6kJ)=2908.69kJ+4502.6kJ=1593.91kJ

Thus, the standard enthalpy of the industrial synthesis of boron trifluoride is 1593.91kJ_.

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base has to be determined.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 4D.21E

The standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base is 15.28kJ_.

Explanation of Solution

The balanced chemical equation that represents the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a baseis shown below.

    H2S(aq)+2KOH(aq)K2S(aq)+2H2O(l)

The standard enthalpy of formation of products that are K2S(aq) and H2O(l) is 417.5kJmol1 and 285.83kJmol1 respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  nΔHf(products)=nK2SΔHf(K2S,aq)+nH2OΔHf(H2O,l)        (6)

Where,

  • nΔHf(products) is total enthalpy of formation of products.
  • nK2S is the number of moles of K2S(aq).
  • ΔHf(K2S,aq) is the standard enthalpy of formation of K2S(aq).
  • nH2O is the number of moles of H2O(l).
  • ΔHf(H2O,l) is the standard enthalpy of formation of H2O(l).

The value of nK2S is 1mol.

The value of ΔHf(K2S,aq) is 417.5kJmol1.

The value of nH2O is 2mol.

The value of ΔHf(H2O,l) is 285.83kJmol1.

Substitute the value of nK2S, ΔHf(K2S,aq), nH2O and ΔHf(H2O,l) in equation (6).

  nΔHf(products)=(1mol)×(417.5kJmol1)+(2mol)×(285.83kJmol1)=417.5kJ571.66kJ=989.16kJ

The standard enthalpy of formation of reactant that is H2S(aq) and KOH(aq) is 39.7kJmol1 and 482.37kJmol1 respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  nΔHf(reactants)=nH2SΔHf(H2S,aq)+nKOHΔHf(KOH,aq)        (7)

Where,

  • nΔHf(reactants) is total enthalpy of formation of reactants.
  • nH2S is the number of moles of H2S(aq).
  • ΔHf(H2S,aq) is the standard enthalpy of formation of H2S(aq).
  • nKOH is the number of moles of KOH(aq).
  • ΔHf(KOH,aq) is the standard enthalpy of formation of KOH(aq).

The value of nH2S is 1mol.

The value of ΔHf(H2S,aq) is 39.7kJmol1.

The value of nKOH is 2mol.

The value of ΔHf(KOH,aq) is 482.37kJmol1.

Substitute the value of nH2S, ΔHf(H2S,aq), nKOH and ΔHf(KOH,aq) in equation (7).

  nΔHf(reactants)=(1mol)×(39.7kJmol1)+(2mol)×(482.37kJmol1)=39.7kJ964.74kJ=1004.44kJ

The value of nΔHf(products) is 989.16kJ.

The value of nΔHf(reactants) is 1004.44kJ.

Substitute the value of nΔHf(products) and nΔHf(reactants) in equation (3).

  ΔH°=989.16kJ(1004.44kJ)=989.16kJ+1004.44kJ=15.28kJ

Thus, the standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base is 15.28kJ_.

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Chapter 4 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 4 - Prob. 4A.3ECh. 4 - Prob. 4A.4ECh. 4 - Prob. 4A.5ECh. 4 - Prob. 4A.6ECh. 4 - Prob. 4A.7ECh. 4 - Prob. 4A.8ECh. 4 - Prob. 4A.9ECh. 4 - Prob. 4A.10ECh. 4 - Prob. 4A.11ECh. 4 - Prob. 4A.12ECh. 4 - Prob. 4A.13ECh. 4 - Prob. 4A.14ECh. 4 - Prob. 4B.1ASTCh. 4 - Prob. 4B.1BSTCh. 4 - Prob. 4B.2ASTCh. 4 - Prob. 4B.2BSTCh. 4 - Prob. 4B.3ASTCh. 4 - Prob. 4B.3BSTCh. 4 - Prob. 4B.1ECh. 4 - Prob. 4B.2ECh. 4 - Prob. 4B.3ECh. 4 - Prob. 4B.4ECh. 4 - Prob. 4B.5ECh. 4 - Prob. 4B.6ECh. 4 - Prob. 4B.7ECh. 4 - Prob. 4B.8ECh. 4 - Prob. 4B.9ECh. 4 - Prob. 4B.10ECh. 4 - Prob. 4B.11ECh. 4 - Prob. 4B.12ECh. 4 - Prob. 4B.13ECh. 4 - Prob. 4B.14ECh. 4 - Prob. 4B.15ECh. 4 - Prob. 4B.16ECh. 4 - Prob. 4C.1ASTCh. 4 - Prob. 4C.1BSTCh. 4 - Prob. 4C.2ASTCh. 4 - Prob. 4C.2BSTCh. 4 - Prob. 4C.3ASTCh. 4 - Prob. 4C.3BSTCh. 4 - Prob. 4C.4ASTCh. 4 - Prob. 4C.4BSTCh. 4 - Prob. 4C.1ECh. 4 - Prob. 4C.2ECh. 4 - Prob. 4C.3ECh. 4 - Prob. 4C.4ECh. 4 - Prob. 4C.5ECh. 4 - Prob. 4C.6ECh. 4 - Prob. 4C.7ECh. 4 - Prob. 4C.8ECh. 4 - Prob. 4C.9ECh. 4 - Prob. 4C.10ECh. 4 - Prob. 4C.11ECh. 4 - Prob. 4C.12ECh. 4 - Prob. 4C.13ECh. 4 - Prob. 4C.14ECh. 4 - Prob. 4C.15ECh. 4 - Prob. 4C.16ECh. 4 - Prob. 4D.1ASTCh. 4 - Prob. 4D.1BSTCh. 4 - Prob. 4D.2ASTCh. 4 - Prob. 4D.2BSTCh. 4 - Prob. 4D.3ASTCh. 4 - Prob. 4D.3BSTCh. 4 - Prob. 4D.4ASTCh. 4 - Prob. 4D.4BSTCh. 4 - Prob. 4D.5ASTCh. 4 - Prob. 4D.5BSTCh. 4 - Prob. 4D.6ASTCh. 4 - Prob. 4D.6BSTCh. 4 - Prob. 4D.7ASTCh. 4 - Prob. 4D.7BSTCh. 4 - Prob. 4D.1ECh. 4 - Prob. 4D.2ECh. 4 - Prob. 4D.3ECh. 4 - Prob. 4D.4ECh. 4 - Prob. 4D.5ECh. 4 - Prob. 4D.6ECh. 4 - Prob. 4D.7ECh. 4 - Prob. 4D.8ECh. 4 - Prob. 4D.10ECh. 4 - Prob. 4D.11ECh. 4 - Prob. 4D.13ECh. 4 - Prob. 4D.14ECh. 4 - Prob. 4D.15ECh. 4 - Prob. 4D.16ECh. 4 - Prob. 4D.17ECh. 4 - Prob. 4D.18ECh. 4 - Prob. 4D.19ECh. 4 - Prob. 4D.20ECh. 4 - Prob. 4D.21ECh. 4 - Prob. 4D.22ECh. 4 - Prob. 4D.23ECh. 4 - Prob. 4D.24ECh. 4 - Prob. 4D.25ECh. 4 - Prob. 4D.26ECh. 4 - Prob. 4D.29ECh. 4 - Prob. 4D.30ECh. 4 - Prob. 4E.1ASTCh. 4 - Prob. 4E.1BSTCh. 4 - Prob. 4E.2ASTCh. 4 - Prob. 4E.2BSTCh. 4 - Prob. 4E.5ECh. 4 - Prob. 4E.6ECh. 4 - Prob. 4E.7ECh. 4 - Prob. 4E.8ECh. 4 - Prob. 4E.9ECh. 4 - Prob. 4E.10ECh. 4 - Prob. 4F.1ASTCh. 4 - Prob. 4F.1BSTCh. 4 - Prob. 4F.2ASTCh. 4 - Prob. 4F.2BSTCh. 4 - Prob. 4F.3ASTCh. 4 - Prob. 4F.3BSTCh. 4 - Prob. 4F.4ASTCh. 4 - Prob. 4F.4BSTCh. 4 - Prob. 4F.5ASTCh. 4 - Prob. 4F.5BSTCh. 4 - Prob. 4F.6ASTCh. 4 - Prob. 4F.6BSTCh. 4 - Prob. 4F.7ASTCh. 4 - Prob. 4F.7BSTCh. 4 - Prob. 4F.8ASTCh. 4 - Prob. 4F.8BSTCh. 4 - Prob. 4F.9ASTCh. 4 - Prob. 4F.9BSTCh. 4 - Prob. 4F.1ECh. 4 - Prob. 4F.2ECh. 4 - Prob. 4F.3ECh. 4 - Prob. 4F.4ECh. 4 - Prob. 4F.5ECh. 4 - Prob. 4F.6ECh. 4 - Prob. 4F.7ECh. 4 - Prob. 4F.9ECh. 4 - Prob. 4F.10ECh. 4 - Prob. 4F.11ECh. 4 - Prob. 4F.12ECh. 4 - Prob. 4F.13ECh. 4 - Prob. 4F.14ECh. 4 - Prob. 4F.15ECh. 4 - Prob. 4F.16ECh. 4 - Prob. 4F.17ECh. 4 - Prob. 4G.1ASTCh. 4 - Prob. 4G.1BSTCh. 4 - Prob. 4G.2ASTCh. 4 - Prob. 4G.2BSTCh. 4 - Prob. 4G.1ECh. 4 - Prob. 4G.2ECh. 4 - Prob. 4G.3ECh. 4 - Prob. 4G.5ECh. 4 - Prob. 4G.7ECh. 4 - Prob. 4G.8ECh. 4 - Prob. 4G.9ECh. 4 - Prob. 4G.10ECh. 4 - Prob. 4H.1ASTCh. 4 - Prob. 4H.1BSTCh. 4 - Prob. 4H.2ASTCh. 4 - Prob. 4H.2BSTCh. 4 - Prob. 4H.1ECh. 4 - Prob. 4H.2ECh. 4 - Prob. 4H.3ECh. 4 - Prob. 4H.4ECh. 4 - Prob. 4H.5ECh. 4 - Prob. 4H.6ECh. 4 - Prob. 4H.7ECh. 4 - Prob. 4H.8ECh. 4 - Prob. 4H.9ECh. 4 - Prob. 4H.10ECh. 4 - Prob. 4H.11ECh. 4 - Prob. 4I.1ASTCh. 4 - Prob. 4I.1BSTCh. 4 - Prob. 4I.2ASTCh. 4 - Prob. 4I.2BSTCh. 4 - Prob. 4I.3ASTCh. 4 - Prob. 4I.3BSTCh. 4 - Prob. 4I.4ASTCh. 4 - Prob. 4I.4BSTCh. 4 - Prob. 4I.1ECh. 4 - Prob. 4I.2ECh. 4 - Prob. 4I.3ECh. 4 - Prob. 4I.4ECh. 4 - Prob. 4I.5ECh. 4 - Prob. 4I.6ECh. 4 - Prob. 4I.7ECh. 4 - Prob. 4I.8ECh. 4 - Prob. 4I.9ECh. 4 - Prob. 4I.10ECh. 4 - Prob. 4I.11ECh. 4 - Prob. 4I.12ECh. 4 - Prob. 4J.1ASTCh. 4 - Prob. 4J.1BSTCh. 4 - Prob. 4J.2ASTCh. 4 - Prob. 4J.2BSTCh. 4 - Prob. 4J.3ASTCh. 4 - Prob. 4J.3BSTCh. 4 - Prob. 4J.4ASTCh. 4 - Prob. 4J.4BSTCh. 4 - Prob. 4J.5ASTCh. 4 - Prob. 4J.5BSTCh. 4 - Prob. 4J.6ASTCh. 4 - Prob. 4J.6BSTCh. 4 - Prob. 4J.1ECh. 4 - Prob. 4J.2ECh. 4 - Prob. 4J.3ECh. 4 - Prob. 4J.4ECh. 4 - Prob. 4J.5ECh. 4 - Prob. 4J.6ECh. 4 - Prob. 4J.7ECh. 4 - Prob. 4J.8ECh. 4 - Prob. 4J.9ECh. 4 - Prob. 4J.11ECh. 4 - Prob. 4J.12ECh. 4 - Prob. 4J.13ECh. 4 - Prob. 4J.14ECh. 4 - Prob. 4J.15ECh. 4 - Prob. 4J.16ECh. 4 - Prob. 4.8ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.16ECh. 4 - Prob. 4.19ECh. 4 - Prob. 4.20ECh. 4 - Prob. 4.21ECh. 4 - Prob. 4.23ECh. 4 - Prob. 4.25ECh. 4 - Prob. 4.27ECh. 4 - Prob. 4.28ECh. 4 - Prob. 4.29ECh. 4 - Prob. 4.30ECh. 4 - Prob. 4.31ECh. 4 - Prob. 4.32ECh. 4 - Prob. 4.33ECh. 4 - Prob. 4.34ECh. 4 - Prob. 4.35ECh. 4 - Prob. 4.36ECh. 4 - Prob. 4.37ECh. 4 - Prob. 4.39ECh. 4 - Prob. 4.40ECh. 4 - Prob. 4.41ECh. 4 - Prob. 4.45ECh. 4 - Prob. 4.46ECh. 4 - Prob. 4.48ECh. 4 - Prob. 4.49ECh. 4 - Prob. 4.53ECh. 4 - Prob. 4.57ECh. 4 - Prob. 4.59E
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