Question

The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the following balanced equation:
4C3H5N3O9(l)→12CO2(g)+10H2O(g)+6N2(g)+O2(g) ΔH∘rxn=−5678kJ

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

Answer 1

Answer:

$\Delta _fH^0_{C_3H_5N_3O_9}=-365.5kJ/mol$

Explanation:

Hello,

For the given reaction, the standard enthalpy of reaction is:

$\Delta _RH^0=12\Delta _fH^0_{CO_2}+10\Delta _fH^0_{H_2O}-4\Delta _fH^0_{C_3H_5N_3O_9}$

As long as the both nitrogen's and oxygen's standard enthalpies of formation are 0, thus, by looking for water's and carbon dioxide's standard enthalpies of formation, we compute the standard enthalpy of formation for nitroglycerin as shown below:

$\Delta _fH^0_{C_3H_5N_3O_9}=\frac{1}{4} (12\Delta _fH^0_{CO_2}+10\Delta _fH^0_{H_2O}-\Delta _RH^0)\\\Delta _fH^0_{C_3H_5N_3O_9}=\frac{1}{4mol} [12(-393.5kJ)+10(-241.8kJ)-(-5678kJ)]\\\Delta _fH^0_{C_3H_5N_3O_9}=-365.5kJ/mol$

Best regards.