Chapter 5, Problem 86GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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:

â€‚Â N2H4+Â O2â†’â€‰N2+Â 2H2O

â€‚Â Î”rH0=Î£nÎ”fH0(products)â€‰-â€‰Î£nÎ”fH0(reactants)

â€‚Â Î”rH0=[(1molÃ—-393.5kJ/mol)â€‰+(0kJ/mol)]-[(1molÃ—50kJ/molâ€‰+0)]

â€‚Â Î”rH0=-446.5kJ

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

â€‚Â Î”rH=enthalpyÂ changenumberÂ ofÂ molesÂ

â€‚Â Î”rH=-446.5kJ/molâ€‰Ã—1mol50.0604Â g=-8.919Â J/gK

So, the molar enthalpy of formation of hydrazine is -8

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