# The rocket fuel hydrazine, "N"_2"H"_(4(g)), is formed by the reaction: 2"H"_(2(g)) + "N"_(2(g)) -> "N"_2"H"_(4(g)) The heat of reaction for this process is "95.40 kJ/mol"?

## What is the heat of formation of hydrazine? How much heat is required to create 125.0 g of hydrazine?

May 1, 2016

Here's what I got.

#### Explanation:

The standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, of a substance expresses the change in enthalpy that accompanies the formation of one mole of that substance from its constituent elements in their standard state.

The problem provides you with the following thermochemical equation

$2 {\text{H"_ (2(g)) + "N"_ (2(g)) -> "N"_ 2"H"_ (4(g))" "DeltaH_"rxn"^@ = "95.40 kJ mol}}^{- 1}$

Notice that this equation corresponds to the formation of one mole of hydrazine, which means that the enthalpy change for this reaction will be equal to the enthalpy change of formation

$\Delta {H}_{f}^{\circ} = \Delta {H}_{\text{rxn}}^{\circ}$

Since the enthalpy change for this reaction is ${\text{95.40 kJ mol}}^{- 1}$, it follows that you have

$\Delta {H}_{f}^{\circ} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} + {\text{95.40 kJ mol}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, this enthalpy change tells you that in order to produce one mole of hydrazine, you need to provide $\text{95.40 kJ}$ of energy.

Use hydrazine's molar mass to determine how many moles you have in that $\text{125.0-g}$ sample

125.0 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"H"_4)/(32.045color(red)(cancel(color(black)("g")))) = "3.901 moles N"_2"H"_4

Well, if you need $\text{95.40 kJ}$ to produce one mole of hydrazine, it follows that this many moles will require

$3.901 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles N"_2"H"_4))) * "95.40 kJ"/(1color(red)(cancel(color(black)("mole N"_2"H"_4)))) = color(green)(|bar(ul(color(white)(a/a)"372.2 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to four sig figs.