Enthalpy change of combustion per gram of methanol and isooctane is to be determined. Concept Introduction: The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state. Δ r H 0 = ΣnΔ f H 0 (products) - ΣnΔ f H 0 (reactants) The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as: Δ r H= enthalpy change number of moles

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Answer 1

Question

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

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Answer 2

Question

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

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Answer 3

Question

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

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Answer 4

Question

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

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Answer 5

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

Answer 6

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

Answer 7

Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

Answer 8

Expert Solution & Answer

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $-47.82 J/gK$ and the enthalpy change of combustion per gram of methanol is $-22.68J/gK$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

For isooctane the balanced reaction is

$C8H18+ 25/2 O2→ 8 CO2+ 9 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(8mol×-393.5kJ/mol) +(9mol×-285.83kJ/mol)]$ - $[(1mol×-259.2kJ/mol +0)]$

$ΔrH0$ = $-5461.3kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-5461.3kJ$ $/mol ×1mol114.2g$ = $-47.82 J/gK$

So, the enthalpy of isooctane is $-47.82 J/gK$

For methanol,

$2 CH3OH + 3 O2 → 2CO2 + 4 H2O$

$ΔrH0$ = $ΣnΔfH0(products) - ΣnΔfH0(reactants)$

$ΔrH0$ = $[(2mol×-393.5kJ/mol) +(4mol×-285.83kJ/mol)]$ - $[(2mol×-238.4kJ/mol +0)]$

$ΔrH0$ = $-1453.5kJ$

The change in enthalpy is $-5461.3kJ$

The change in enthalpy, $ΔH$ in kJ per mole of a given reactant for the reaction can be calculated as:

$ΔrH=enthalpy changenumber of moles$

$ΔrH0$ = $-1453.5kJ2mol×1mol32.04g$ = $-22.68J/gK$

So, the enthalpy of methanol is $-22.68J/gK$

Answer 12

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.

Answer 13

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Answer to Problem 85GQ

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