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

The idea below is the you must use the standard enthalpy adjust of reaction, #DeltaH^


You are watching: Standard enthalpy of formation for nitroglycerin

#, and also the traditional enthalpy transforms of formation of the commodities to uncover the typical enthalpy readjust of formation of interest.

You can discover the standard enthalpy changes of development of carbon dioxide, water, and nitrogen gas here

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

To find the traditional enthalpy change of formation of nitroglicerine, usage Hess" Law, which tells you that the enthalpy adjust of a reaction is independent that the route taken and the variety of steps required for that reaction to take place.

This way that you can express the conventional enthalpy adjust of reaction by making use of the standard enthalpy transforms 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 the the types that take part in the reaction.

So, the conventional enthalpy alters of formation for one mole the 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 moles of carbon dioxide10 moles of water6 moles of nitrogen gas

and needs

4 mole of nitroglicerine

Notice the the enthalpies of development are offered in kilojoules every mole, so transform the enthalpy readjust 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 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")#


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