The standard enthalpy for the given reaction has to be calculated. Concept Introduction: Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms. Standard enthalpy change for the reaction can be calculated as, Δ H r x n o = ∑ Δ H f o p r o d u c t s − ∑ Δ H f o reactants

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Answer 1

Question

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

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Answer 2

Question

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

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Answer 3

Question

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

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Answer 4

Question

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

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Answer 5

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

Answer 6

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

Answer 7

Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Answer 8

(a)

Expert Solution

Explanation of Solution

The given reaction is 2NO2(g) →N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= ΔH0(N2O4)−(2×ΔH0(NO2))ΔHrxno= (9.16 kJmol−1)−(2×33.18 kJmol−1)ΔHrxno= −57.29 kJmol−1

The standard enthalpy change for the given reaction is −57.29 kJmol−1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

Answer 13

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Answer 14

(b)

Expert Solution

Explanation of Solution

The given reaction is NO2(g) →1/2N2O4(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (1/2 ΔH0(N2O4))−(ΔH0(NO2))ΔHrxno= (1/2×9.16 kJmol−1)−(×33.18 kJmol−1)ΔHrxno= −28.6 kJmol−1

The standard enthalpy change for the given reaction is −28.6 kJmol−1

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

Answer 19

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Answer 20

(c)

Expert Solution

Explanation of Solution

The given reaction is 3NO2(g)+H2O →2HNO3(aq)+NO(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(HNO3)+ΔH0(NO))−(3×ΔH0(NO2)+ΔH0(H2O))ΔHrxno= (2×−207.36kJmol−1+90.25kJmol−1)−(3×33.18 kJmol−1+(−285.83kJmol−1))ΔHrxno= −138.2 kJmol−1

The standard enthalpy change for the given reaction is −138.2 kJmol−1

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

Answer 25

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Answer 26

(d)

Expert Solution

Explanation of Solution

The given reaction is cyclpropane(g) →propene(g).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (2 ΔH0(propene))−(ΔH0(cyclopropane))ΔHrxno= (+20.42kJmol−1)−(53.30 kJmol−1)ΔHrxno= −32.88 kJmol−1

The standard enthalpy change for the given reaction is −32.88 kJmol−1

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

Answer 31

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Answer 32

(e)

Expert Solution

Explanation of Solution

Net ionic equation is

H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Cl−(aq)+Na+(aq)+H2O(l)

The given reaction is H+(aq)+OH−(aq)→H2O(aq).

Standard enthalpy change for the reaction can be calculated as,

ΔHrxno= ∑ ΔHfoproducts− ∑ ΔHforeactants

Substituting the values,

ΔHrxno= (ΔH0(H2O))−(ΔH0(H+)+ΔH0(OH−))ΔHrxno= (−285.36kJmol−1)−(0 kJmol−1+(−229.99kJmol−1))ΔHrxno= −55.84 kJmol−1

The standard enthalpy change for the given reaction is −55.84 kJmol−1

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

Ch. 2 - Prob. 2A.1ST Ch. 2 - Prob. 2A.2ST Ch. 2 - Prob. 2B.1ST Ch. 2 - Prob. 2B.2ST Ch. 2 - Prob. 2B.3ST Ch. 2 - Prob. 2B.4ST Ch. 2 - Prob. 2B.5ST Ch. 2 - Prob. 2C.1ST Ch. 2 - Prob. 2C.2ST Ch. 2 - Prob. 2D.1ST

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