Since your inquiry is a little vague to start with, I"ll assume the you must determine the typical enthalpy readjust of formation of nitroglicerine, #"C"_3"H"_5"N"_3"O"_9#.

The idea below is that you have to use the conventional enthalpy change of reaction, #DeltaH^

You are watching: Calculate the standard enthalpy of formation (δh∘for nitroglycerin

#, and also the typical enthalpy transforms of formation of the assets to find the traditional enthalpy adjust of formation of interest.

You can find 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 find the typical enthalpy change of formation of nitroglicerine, use Hess" Law, which speak you the the enthalpy change of a reaction is independent of the route taken and also the variety of steps needed for that reaction to take place.

This means that you can express the traditional enthalpy readjust of reaction by using the standard enthalpy alters of development 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 traditional enthalpy transforms of formation for one mole that carbon dioxide, water, and 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 mole of carbon dioxide10 mole of water6 mole of nitrogen gas

and calls for

4 moles of nitroglicerine

Notice that the enthalpies of development are offered in kilojoules per mole, so convert the enthalpy adjust of reaction come kilojoules

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

Plug in her values and also 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")# 