Since your concern is a small vague to start with, I"ll assume the you must identify the typical enthalpy change of development of nitroglicerine, #"C"_3"H"_5"N"_3"O"_9#.

The idea right here is that you should use the conventional enthalpy readjust of reaction, #DeltaH^

You are watching: Standard enthalpy of formation of nitroglycerin

#, and the typical enthalpy transforms of development of the products to discover the typical enthalpy readjust of formation of interest.

You can discover the conventional enthalpy alters of formation of carbon dioxide, water, and nitrogen gas here

https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29

To uncover the standard enthalpy change of formation of nitroglicerine, usage Hess" Law, which speak you that the enthalpy change of a reaction is independent that the route taken and the variety of steps needed for that reaction to take place.

This means that you have the right to express the traditional enthalpy change of reaction by making use of the standard enthalpy changes of formation of the reactant and also of the products

#DeltaH_"rxn"^
= sum(n xx DeltaH_"f prod"^
) - sum(m xx DeltaH_"f react"^
)" "#, where

#n#, #m# - the stoichiometric coefficients of the species that take component in the reaction.

So, the standard enthalpy transforms of development for one mole the carbon dioxide, water, and also nitrogen gas are

#"CO"_2: -"393.51 kJ/mol"#

#"H"_2"O": -"241.82 kJ/mol"#

#"N"_2: " 0 kJ/mol"#

So, her reaction produces

12 moles of carbon dioxide10 moles of water6 moles of nitrogen gas

and requires

4 moles of nitroglicerine

Notice the the enthalpies of formation are given in kilojoules every mole, so convert the enthalpy change of reaction to kilojoules

#-5678color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = -"5.678 kJ"#

Plug in your values and solve for #DeltaH_"f nitro"^
#

#-"5.678 kJ" = <12color(red)(cancel(color(black)("moles"))) * (-393.51"kJ"/color(red)(cancel(color(black)("mol")))) + 10color(red)(cancel(color(black)("moles"))) * (-241.82"kJ"/color(red)(cancel(color(black)("mole")))) + 6color(red)(cancel(color(black)("moles"))) * 0"kJ"/color(red)(cancel(color(black)("mole")))> - (4 * DeltaH_"f nitro"^
)#

Rearrange come get

#4DeltaH_"f nitro"^
= -"7140.32 kJ" + "5.678 kJ"#

#DeltaH_"f nitro"^
= (-"7134.642 kJ")/4 = color(green)(-"1784 kJ/mol")#