Enthalpy change for the given reaction should be calculated. The given reaction is, CaCO 3 ( s ) → CaO ( s ) + CO 2 ( g ) Concept introduction: Enthalpy ( H ) : It is the total amount of heat in a particular system. The value of standard enthalpy change ΔH ο of the given reaction is calculated by the formula, ΔH r x n ∘ = ∑ n p ΔH f ο ( products ) − ∑ n r ΔH f ο ( reactants ) Where, ΔH f ο is the standard enthalpies of formation. n is the number of moles.

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Answer 1

Question

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

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Answer 2

Question

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

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Answer 3

Question

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

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Answer 4

Question

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

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Answer 5

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

Answer 6

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

Answer 7

Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

CaCO3(s) → CaO(s) + CO2(g)

Concept introduction:

Enthalpy(H): It is the total amount of heat in a particular system.

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Where,

ΔHfο is the standard enthalpies of formation.

n is the number of moles.

Answer 8

(a)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 178.32 kJ.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The balanced equation for the given reaction follows as,

CaCO3(s) → CaO(s) + CO2(g)

Standard enthalpy of formation values is given below,

ΔHfo of CaCO3(s) = −1206.92 kJ/molΔHfo of CaO(s) = −635.09 kJ/molΔHfo of CO2 (g) = −393.51 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(1 mol×−635.09 kJ/mol) + (1 mol×−393.51 kJ/mol)] − [(1 mol×−1206.92 kJ/mol) ] = 178.32 kJ

Answer 13

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

Answer 14

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

Answer 15

(b)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is −595.16 kJ.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The balanced equation for the given reaction follows as,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Standard enthalpy of formation values is given below,

ΔHfo of HI(g) = 26.48 kJ/molΔHfo of F2(g) = 0 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of I2(s) = 0 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(2 mol×−271.1 kJ/mol) + (1 mol×0 kJ/mol)] − [(2 mol×26.48 kJ/mol) + (1 mol×0 kJ/mol) ] = −595.16 kJ

Answer 20

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

Answer 21

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

Answer 22

(c)

Expert Solution

Answer to Problem 5.82QE

Enthalpy change is 44.17 kJ.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The balanced equation for the given reaction follows as,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Standard enthalpy of formation values is given below,

ΔHfo of SF6(g) =− 1209 kJ/molΔHfo of H2O(l) = −285.83 kJ/molΔHfo of HF(g) = −271.1 kJ/molΔHfo of SO3(g) = −395.72 kJ/mol

Change in enthalpy can be calculated by the equation:

ΔHrxn∘=∑npΔHfο(products) −∑nrΔHfο(reactants)

Substitute the values as follows,

ΔHrxno = [(6 mol×−271.1 kJ/mol) + (1 mol×−395.72 kJ/mol)] − [(1 mol×− 1209 kJ/mol) + (3 mol×−285.83 kJ/mol) ] = 44.17 kJ

Answer 27

Ch. 5 - Why must the physical states of all reactants and...Ch. 5 - Why is chemical energy classified as a form of...Ch. 5 - What is the difference between the enthalpy of...Ch. 5 - Classify each process as exothermic or...Ch. 5 - Explain why the specific heat of the contents of...Ch. 5 - Prob. 5.6QE Ch. 5 - Define heat. What are its units? How does it...Ch. 5 - Prob. 5.8QE Ch. 5 - Prob. 5.9QE Ch. 5 - Prob. 5.10QE

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Answer to Problem 5.82QE

Explanation of Solution

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

Answer to Problem 5.82QE

Explanation of Solution

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).

Answer to Problem 5.82QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions

Concept explainers

Videos

Answer to Problem 5.82QE

Explanation of Solution

Interpretation: Enthalpy change for the given reaction should be calculated. The given reaction is, 2 HI(g) + F2(g) → 2 HF(g) + I2(s) Concept introduction: Refer to (a).

Answer to Problem 5.82QE

Explanation of Solution

Interpretation: Enthalpy change for the given reaction should be calculated. The given reaction is, SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g) Concept introduction: Refer to (a).

Answer to Problem 5.82QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

2 HI(g) + F2(g) → 2 HF(g) + I2(s)

Concept introduction:

Refer to (a).

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

SF6(g) + 3 H2O(l) → 6 HF(g) + SO3(g)

Concept introduction:

Refer to (a).