The question in my textbook is as follows:

Hydrazine and 1,1-dimethylhydrazine both react spontaneously with O2 and can be used as rocket fuels.

$$\ce{ N2H4(l) + O2(g) -> N2(g) + 2H2O(g)}$$ $$\ce{N2H2(CH3)2(l) + 4O2(g) -> 2CO2(g) + 4H2O(g) + N2(g)}$$

The molar enthalpy of formation of N2H4(l) is +50.6 kJ/mol, and that of N2H2(CH3)2(l) is +48.9kJ/mol. Use these values, with other $\Delta fH^{\circ}$ values , to decide whether the reaction of hydrazine or 1,1-dimethylhydrazine with oxygen provides more energy per gram.

My attempt at the solution:

Hydrazine reaction:

$\Delta fH^{\circ}\ \ce{N2H4(l)} = +50.6kJ/mol$

$\Delta fH^{\circ}\ \ce{H2O(g)} = -241.83kJ/mol$

By Hess's law of summation

$$\Delta rH^{\circ}=\sum\Delta fH^{\circ} products -\sum\Delta fH^{\circ} reactants $$ $$ = [ 0 + 2(-241.83kJ)] - [(50.6) + 0] $$ $$ = -483.66 - 50.6 $$ $$ = -534.26 kJ/mol $$

1,1-dimethylhydrazine reaction:

$\Delta fH^{\circ}\ \ce{N2H2(CH3)2(l)} = +48.9kJ/mol$

$\Delta fH^{\circ}\ \ce{H2O(g)} = -241.83kJ/mol$

$\Delta fH^{\circ}\ \ce{CO2(g)} = --393.509kJ/mol$

By Hess's law of summation

$$\Delta rH^{\circ}=\sum\Delta fH^{\circ} products -\sum\Delta fH^{\circ} reactants $$ $$ = [ 2(-393.509kJ/mol) + 4(-241.83kJ/mol) + 0] - [48.9kJ/mol + 0] $$ $$ = -483.66kJ/mol - 50.6kJ/mol $$ $$ = -1803.238 kJ/mol $$

Now, my first reaction is to say that the 1,1-dimethylhydrazine provides more energy, as it evolves 1803.238 kJ/mol of energy as heat for the reaction vs the 534.26kJ/mol evolved by the Hydrazine. I have a strong suspicion that this is wrong however based on the fact that a quick google search reveals that the Hydrazine is indeed used in rockets. The questions does say " energy per gram " so i also tried working out the number of moles in 1g of both Hydrazine and 1,1-dimethylhydrazine and then working out how much energy per mole equivalent to 1g but 1,1-dimethylhydrazine still evolves more energy when i do those calculations.

I unfortunately don't have an answer to this problem so i'd really appreciate any feedback. Thanks in advance!

  • 1
    $\begingroup$ Nitpicking 1) Your second reaction is missing $\ce{2CO2}$ 2) Enthalpies of formation and enthalpies of reaction are in kJ/mol not kJ 3) Your value of the enthalpy of formation of CO2 isn't right 4) The third last line of the equation is wrong. Be careful with your copy pastes $\endgroup$ – orthocresol May 11 '16 at 21:51
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    $\begingroup$ Furthermore the fact that N2H4 is used as rocket fuel is hardly a sign that you are wrong. If the choice of fuel was solely based on the amount of energy released per gram, then all our vehicles would be running on nuclear fusion right now. By the way, N2H2Me2 is also used as a rocket fuel. $\endgroup$ – orthocresol May 11 '16 at 21:55
  • $\begingroup$ @orthocresol thank you - have corrected. That is true, i suppose there are actually many factors to consider such as cost of production, density of propellant etc. Thanks for the link, very interesting. $\endgroup$ – Blargian May 11 '16 at 22:08

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