4D.9 7th Ed.

[Summer de Vera 2C]

Hi! For this problem, the enthalpy of formation of TNT= -67 kJ/mol, its density = 1.65 g/cm^3, and the chem eqn is 4 C7H5N3O6 (s) + 21 O2(g) --> 28 CO2(g) + 10 H2O(g)+ 6N2(g). The problem asks to find the enthalpy density of TNT, but I'm not really sure where to start. Can someone please explain?

[Nathan Tran 4K]

Hi, so you know standard enthalpy formation is based off the moles of each reactant and product. Refer to table 8.4 to find the enthalpy of the whole reaction. Remember that diatomic molecules will have an enthalpy of zero, that product enthalpy will be negative and that reactant enthalpy will be positive. Once we have this, we have the enthalpy per mole of reaction. However, we want the enthalpy of TNT, and because we are told the enthalpy of formation of TNT is -67 kJ/mol, we know that C7H5N3O6 is TNT. Thus, we need to find the enthalpy of reaction for one mole of TNT, which is why we will divide the number we got from adding all the enthalpies of formation up by four.

Now we have the enthalpy of the reaction per mole of TNT. We want to get the enthalpy of reaction per Liter. Thus, we will use our answer and do dimensional analysis. We can convert the KJ/mol (our units for our enthalpy) to KJ/g by using the molar mass of TNT. Then we can convert grams to cm cubed with the given information that the density of TNT is 1.65 g/ cm^3. Lastly, we convert cm^3 to liters by knowing that 10^3 cm cubed is equal to 1 L.