# 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"#?

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What is the heat of formation of hydrazine?

How much heat is required to create 125.0 g of hydrazine?

What is the heat of formation of hydrazine?

How much heat is required to create 125.0 g of hydrazine?

##### 1 Answer

Here's what I got.

#### Explanation:

The *standard enthalpy of formation*, **one mole** of that substance from its constituent elements in their standard state.

The problem provides you with the following **thermochemical equation**

#2"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

#DeltaH_f^@ = DeltaH_"rxn"^@#

Since the enthalpy change for this reaction is

#DeltaH_f^@ = color(green)(|bar(ul(color(white)(a/a)+"95.40 kJ mol"^(-1)color(white)(a/a)|)))#

Now, this enthalpy change tells you that in order to produce **one mole** of hydrazine, you need to provide

Use hydrazine's **molar mass** to determine how many *moles* you have in that

#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 **one mole** of hydrazine, it follows that this many moles will require

#3.901color(red)(cancel(color(black)("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"color(white)(a/a)|)))#

The answer is rounded to four **sig figs**.