The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the following balanced equation..

asked by @alainf21 • over 1 year ago • Chemistry • 5 pts

4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) ΔH∘rxn = −5678 kJ

Calculate the standard enthalpy of formation (ΔH∘f) for nitroglycerin.

DeltaH of reaction=deltaH of formation for products minus deltaH of formation for reactants.

• Here we are given the DeltaH of reaction, its -5678kJ
• The products consist of 12 moles of CO2 and 10 moles of H2O. Not N2 gas or O2 gas because they are in their natural state and their deltaH of formation is zero. The problem should provide deltaH of formation for both CO2 and H2O, if not, look it up. It has to be in kJ/mole .
• deltaH of formation for H2O is -241.8 kJ/mole
• deltaH of formation for CO2 is 0393.5kJ/mole
• Nitroglycerin is our reactant and is what we need to solve for.

DeltaH=deltaHf of products - deltaHf of reactants. Now just plug in the numbers.

• -5678kJ= [(10moles * (-241.8kJ/mole))+(12moles * (-393.5kJ/mole))] - [4moles * deltaHf of nitroglycerin]
• Moles cancel out and we are left with kJ for the units.
• Notice, the reason why we multiplied deltaHf by the moles is because those values were in kJ per 1mole, and we have 12 moles of CO2 and 10 moles of H2O.
• Don't forget to add and subtract everything correctly.
• -5678kJ= -7140kJ - (4*deltaHf for nitroglycerin)
• You should get an answer of deltaHf of nitroglycerin= -365.5kJ
answered by @dashab2 • over 1 year ago