Calculate the standard enthalpy of formation for nitroglycerin.?

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Calculate the standard enthalpy of formation for nitroglycerin.

The explosive nitroglycerin (C5H5N3O3) decomposes rapidly upon ignition or sudden impact according to the following balanced equation:

4C3H5N3O9 (l) —–> 12CO2 + 10 H2O (g) + 6N2 (g)

molar enthalpy of reaction = -5678 kj

5 Answers

  • You need to look up the heat of formation of CO₂ & H2O.

    The heat of reaction is equal to the sum of the heats of formation of the products minus heat of formation of reactants. For the decomposition reaction of nitroglycerin it is:

    ΔH° = 12·ΔH°(CO₂) + 10·ΔH°(H2O) + 6·ΔH°(N₂) – 4·ΔH°(C3H5N3O9)

    ΔH° = Molar enthalpy of rxn = -5678 kJ

    ΔH°(C3H5N3O9) = the unknown we need

    The heat of formation of elements in their standard state is zero.

    ΔH°(N₂) = 0

    ΔH° = 12·ΔH°(CO₂) + 10ΔH°(H2O) – 4·ΔH°(C3H5N3O9)

    Solve for the enthalpy of formation of nitroglycerin:

    ΔH°((C3H5N3O9) = [12·ΔH°(CO₂) + 10·ΔH°(H2O) – ΔH°] / 4

    = [12·(-393.52kJ) + 10·(-241.83) – 5.678kJ] / 4

    ΔH°(C3H5N3O9) =ΔH°(nitroglycerin) = -1783.72kJ <~~Answer

  • Standard Heat Of Formation

  • The reaction, which defines the heat of formation of one mole of nitroglycerin from its constituent elements (in their standard states), is:

    3CO2(s) + 5/2H2O(g) + 3/2N2(g) + 9/2O2(g) —> C3H5N3O9(l)

    The reason for this is that elements are deemed to have zero enthalpy in their standard states, so no correction or contribution is necessary. In the equation given in the question above, the products contain compounds, water and carbon dioxide, which have a non-zero heat of formation, and allowance must be made for these. It does not specify whether the water produced in this reaction is in the gaseous or liquid state – it makes a small but significant difference. However, the evidence suggests that the heat of reaction given corresponds to the formation of water in the gaseous state.

    To make these corrections, we need to generate a Hess’s Law table, starting with the reversed form of the given equation, re-written for 1 mole of nitroglycerin and with the sign of the heat or enthalpy change also reversed:

    3CO2(s) + 5/2H2O(g) + 3/2N2(g) + 1/4O2(g) —> C3H5N3O9(l)

    ∆H = 1419.5 kJ/mol

    We add the requisite amounts to compensate for the heats of formation of water and carbon dioxide

    3C(s) + 3O2(g) —> 3CO2(g)

    ∆H = (3)(-393.5 kJ)

    5/2H2(g) + 5/4O2(g) —> 5/2H2O(g)

    ∆H = (5/2)(-241.8 kJ)

    Which when summed cancels out the water and carbon dioxide components, finishing with the first equation above. The heat of formation is the sum of the components on the RHS of each equation, giving

    ∆H (nitroglycerin) = [1419.5] + [3(-393.5)] + [(5/2)(-241.8)] = -365.5 kJ/mol

  • By definition Delta Hf is the enthalpy change when a compound is formed from its elements in their standard states. ie for the reaction 3C(s) + 3/2N2(g) + 5/2H2(g) + 9/2O2(g) ——–> C3H5(NO3)3(l) It is impossible to calculate this from the one piece of data given in the question

  • where did you get the 1419.5

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