Question

# C2N2O2Hg(s) + O2 (g) -> Hg(g) + 2CO2(g) + N2(g) Determine the theoretical enthalpy (in kJ/mol)...

C2N2O2Hg(s) + O2 (g) -> Hg(g) + 2CO2(g) + N2(g)

Determine the theoretical enthalpy (in kJ/mol) of the mercury fulminate reaction by using the enthalpy of formation for mercury fulminate (+ 386 kJ/mol) Remember, ΔH° rxn = Σn x H°f(products) - Σn x ΔH°f(reactants). Explain your work. We are assuming a constant pressure situation.

This is can also be termed the heat of explosion, but when it is termed heat of explosion, the units are traditionally kJ/kg of substance. Convert your enthalpy of reaction (kJ/mol) to heat of explosion (kJ/kg) using the molar mass of your explosive. List that value here and use it to calculate the explosive energy of 5.00 kg.

The balanced reaction

C2N2O2Hg(s) + O2 (g) = Hg(g) + 2CO2(g) + N2(g)

Enthalpy change for the reaction

ΔH° rxn = ΣH°f(products) - ΣΔH°f(reactants)

= Hf(Hg) + Hf(N2) +2*Hf(CO2) - Hf(O2) - Hf(C2N2O2Hg)

= 61.38 + 0 + 2*(-393.5) - 0 - (386)

ΔH° rxn = - 1111.62 kJ/mol

Heat of explosion = ΔH° rxn / molar mass of C2N2O2Hg

= ( - 1111.62 kJ/mol) / (284.6236 g/mol)

= - ( 3.90557 kJ/g) x (1000g/kg)

= - 3905.57 kJ/kg

For 5 kg explosive

Heat of explosion = (- 3905.57 kJ/kg) x 5 kg

= - 19528 kJ

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