Since your question is a little vague to start with, I"ll assume the you must identify the standard enthalpy adjust of development of nitroglicerine, #"C"_3"H"_5"N"_3"O"_9#.

The idea below is the you should use the typical enthalpy adjust of reaction, #DeltaH^

You are watching: Standard enthalpy of formation of nitroglycerin

#, and also the standard enthalpy transforms of development of the products to uncover the standard enthalpy readjust of formation of interest.

You can find the standard enthalpy transforms of formation of carbon dioxide, water, and nitrogen gas here

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

To discover the typical enthalpy readjust of formation of nitroglicerine, usage Hess" Law, which tells you that the enthalpy readjust of a reaction is independent that the path taken and also the variety of steps necessary for the reaction to take it place.

This way that you can express the traditional enthalpy readjust of reaction by using the conventional enthalpy alters 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 that the types that take component in the reaction.

So, the standard enthalpy changes of formation 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"#

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

and calls for

4 mole of nitroglicerine

Notice the the enthalpies of development are offered in kilojoules per mole, so transform the enthalpy readjust 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 also solve because that #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")# 