# How would you calculate the enthalpy of formation of the reaction: 4C3H5N3O9 (l) --> 12CO2 (g) +10 H2O (g) + 6N, where delta H = -5678 J?

Oct 17, 2015

$\Delta {H}_{\text{f nitro"^@ = -"1784 kJ/mol}}$

#### Explanation:

Since your question is a little vague to begin with, I'll assume that you must determine the standard enthalpy change of formation of nitroglicerine, ${\text{C"_3"H"_5"N"_3"O}}_{9}$.

The idea here is that you need to use the standard enthalpy change of reaction, $\Delta {H}^{\circ}$, and the standard enthalpy changes of formation of the products to find the standard enthalpy change of formation of interest.

You can find the standard enthalpy changes 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 standard enthalpy change of formation of nitroglicerine, use Hess' Law, which tells you that the enthalpy change of a reaction is independent of the path taken and the number of steps needed for that reaction to take place.

This means that you can express the standard enthalpy change of reaction by using the standard enthalpy changes of formation of the reactant and of the products

$\Delta {H}_{\text{rxn"^@ = sum(n xx DeltaH_"f prod"^@) - sum(m xx DeltaH_"f react"^@)" }}$, where

$n$, $m$ - the stoichiometric coefficients of the species that take part in the reaction.

So, the standard enthalpy changes of formation for one mole of carbon dioxide, water, and nitrogen gas are

$\text{CO"_2: -"393.51 kJ/mol}$

$\text{H"_2"O": -"241.82 kJ/mol}$

$\text{N"_2: " 0 kJ/mol}$

• 12 moles of carbon dioxide
• 10 moles of water
• 6 moles of nitrogen gas

and requires

• 4 moles of nitroglicerine

Notice that the enthalpies of formation are given in kilojoules per 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 $\Delta {H}_{\text{f nitro}}^{\circ}$

-"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 to get

$4 \Delta {H}_{\text{f nitro"^@ = -"7140.32 kJ" + "5.678 kJ}}$

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