Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 5, Problem 86GQ
Interpretation Introduction

Interpretation:

The enthalpy change of combustion per gram of N2H4 and N2H2(CH3)2 has to be calculated.

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 

  ΔH=ΔrH×number of moles

Expert Solution & Answer
Check Mark

Answer to Problem 86GQ

The enthalpy change of combustion per gram of N2H4 is -8.919 J/gK and the enthalpy change of combustion per gram of N2H2(CH3)2 is -32.93J/gK

Explanation of Solution

For N2H4 the balanced equation is:

  N2H4+ O2N2+ 2H2O

  ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

  ΔrH0=[(1mol×-393.5kJ/mol)+(0kJ/mol)]-[(1mol×50kJ/mol+0)]

  ΔrH0=-446.5kJ

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 

  ΔrH=-446.5kJ/mol×1mol50.0604 g=-8.919 J/gK

So, the molar enthalpy of formation of hydrazine is -8.919 J/gK

For N2H2(CH3)2 the balanced equation is:

  N2H2(CH3)2 + 4O2 2CO2+ 4H2O+N2

  ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

ΔrH0=[(2mol×-393.5kJ/mol)+(4mol×-285.83kJ/mol)+0]-[(1mol×48.9kJ/mol+0)]

ΔrH0=-1979.2kJ

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 

  ΔH=ΔrH×number of moles

  ΔrH=-1979.2kJ1mol×1mol60.1g=-32.93J/gK

So, the molar enthalpy of formation of 1,1-dimethylhydrazine is -32.93J/gK

Conclusion

The enthalpy change of combustion per gram of N2H4 and N2H2(CH3)2 was calculated.

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Chapter 5 Solutions

Chemistry & Chemical Reactivity

Ch. 5.8 - The standard enthalpies of formation of KNO3(s)...Ch. 5.8 - Prob. 1.2ACPCh. 5.8 - The decomposition of nitroglycerin (C3H5N3O9)...Ch. 5.8 - Prob. 2.1ACPCh. 5.8 - Prob. 2.2ACPCh. 5.8 - Prob. 2.3ACPCh. 5.8 - Prob. 2.4ACPCh. 5.8 - Prob. 2.5ACPCh. 5 - Define the terms system and surroundings. What...Ch. 5 - What determines the directionality of energy...Ch. 5 - Identify whether the following processes are...Ch. 5 - Identify whether the following processes are...Ch. 5 - The molar heat capacity of mercury is 28.1 J/mol ...Ch. 5 - The specific heat capacity of benzene (C6H6) is...Ch. 5 - The specific heat capacity of copper metal is...Ch. 5 - How much energy as heat is required to raise the...Ch. 5 - The initial temperature of a 344-g sample of iron...Ch. 5 - After absorbing 1.850 kJ of energy as heat, the...Ch. 5 - A 45.5-g sample of copper at 99.8 C is dropped...Ch. 5 - One beaker contains 156 g of water at 22 C, and a...Ch. 5 - A 182-g sample of gold at some temperature was...Ch. 5 - When 108 g of water at a temperature of 22.5 C is...Ch. 5 - A 13.8-g piece of zinc is heated to 98.8 C in...Ch. 5 - A 237-g piece of molybdenum, initially at 100.0 C,...Ch. 5 - How much energy is evolved as heat when 1.0 L of...Ch. 5 - The energy required to melt 1.00 g of ice at 0 C...Ch. 5 - How much energy is required to vaporize 125 g of...Ch. 5 - Chloromethane, CH3CI, arises from microbial...Ch. 5 - The freezing point of mercury is 38.8 C. 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