Nitroglycerin is shock sensitive, and decomposes explosively: 4C3H5N3O9(l) → 12CO2(g) + 6N2(g) + O2(g) + 10H2O(g)
ΔHf° -370 -393.5 -241.8
a. What is the molar enthalpy of decomposition (detonation) of nitroglycerin?
b. Calculate the change in internal energy (ΔU) that occurs when 4 moles of nitroglycerin are detonated at 25°C.
a. To calculate the molar enthalpy of decomposition (detonation) of nitroglycerin, we need to use the Hess's Law of constant heat summation. The enthalpy change of the reaction can be determined by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
First, we need to find the enthalpy of formation (ΔHf°) of nitroglycerin using the given data. The enthalpy of formation of nitroglycerin can be calculated by summing the enthalpies of formation of all the elements in nitroglycerin:
ΔHf°(nitroglycerin) = 4ΔHf°(C3H5N3O9) - 12ΔHf°(CO2) - 6ΔHf°(N2) - ΔHf°(O2) - 10ΔHf°(H2O)
Using the given enthalpies of formation (ΔHf°) for the various substances, we can substitute these values into the equation and calculate the enthalpy of formation of nitroglycerin.
b. To calculate the change in internal energy (ΔU) when 4 moles of nitroglycerin are detonated at 25°C, we can use the equation:
ΔU = ΔH - ΔPV
Since this is an explosive reaction, the volume change (ΔV) will be significant, so we can neglect the work term (ΔPV).
Therefore, the change in internal energy (ΔU) is equal to the enthalpy change (ΔH) of the reaction. Since the given equation represents the decomposition reaction of nitroglycerin, the enthalpy change calculated in part a represents the change in internal energy as well.
So, the change in internal energy (ΔU) when 4 moles of nitroglycerin are detonated is equal to the molar enthalpy of decomposition calculated in part a.