   Chapter 5, Problem 86GQ

Chapter
Section
Textbook Problem

Hydrazine and 1,1-dimethyIhydrazine both react spontaneously with O2 and can be used as rocket fuels.N2H4(ℓ) + O2(g) → N2(g) + 2 H2O(g) hydrazineN2H2(CH3)2(ℓ) + 4 O2(g) → 2 CO2(g) + 4 H2O(g) + N2(g) 1, 1-dimethythydrazineThe molar enthalpy of formation of N2H4(ℓ) is + 50.6 kJ/mol, and that of N2H2(CH3)2(ℓ) is + 48.9 kJ/mol. Use these values, with other ΔfH° values, to decide whether the reaction of hydrazine or 1,1-dimethylhydrazine with oxygen provides more energy per gram. 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

Explanation

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

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