Heat evolved by the given reaction at given condition has to be determined. Concept introduction: Bond dissociation enthalpy: Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase. Δ r H = ∑ Δ H ( b o n d s b r o k e n ) − ∑ Δ H ( b o n d s f o r m e d ) Enthalpy of formation: Δ r H = ∑ Δ f H 0 ( p r o d u c t s ) − ∑ Δ f H 0 ( reactants )

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Answer 1

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

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a. The change in the Gibbs energy of a certain constant pressure process is found to fit the expression: AG-85.1 J mol −1 +36.5 J mol ¹K-1 × T A. Calculate the value of AS for the process. B. Next, use the Gibbs-Helmholtz equation: (a(AG/T)) ΔΗ - T2 to calculate the value of AH for the process.

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Answer 2

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

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Answer 3

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

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Answer 4

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

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Answer 5

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

Answer 6

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

Answer 7

Chapter 4, Problem 49Q

(a)

Interpretation Introduction

Interpretation:

Heat evolved by the given reaction at given condition has to be determined.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

Answer 8

(a)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

Bonds broken:

1 mol C≡O bond = 1(1073 kJ) = 1073 kJ3 mol H −H bond = 3(436 kJ) = 1308 kJTotal energy absorbed = 1(1073 kJ) = 2381 kJ

Bonds formed:

4 mol C-H bond = 4(416 kJ) = 1664 kJ2 mol O -H bond = 2(467 kJ) = 934 kJTotal energy released= 2598 kJ

The total energy change is given below:

(+2381 kJ) +(- 2598 kJ) = -217 kJ

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.

By combining all these terms, it is clear that as n becomes larger, more and more energy will be released.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Reason for having more or less energy is given off per mole in the formation of larger hydrocarbons has to be given without doing any calculation.

Concept introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)

Enthalpy of formation:

ΔrH=∑ΔfH0(products)−∑ΔfH0(reactants)

Answer 14

(b)

Expert Solution

Explanation of Solution

Given data:

n CO + (2n+1) H2 →CnH2n+2 + n H2O

The balanced equation when n=1 is given below:

CO + 3 H2 →CH4 + H2O

The given reaction is n CO + (2n+1) H2 →CnH2n+2 + n H2O.

From the equation it is clear that there will always be n C≡O triple bonds to break (2n+1) H −H to break. Also, the number of C-H bond forming will be (2n+2) and the number of O-H bond will be 2n and also the number of C-C bond forming will be n-1.