The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the following balanced equation: 4 C3H5N3O9 (l) → 12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g) ΔrxnHo = −5678 kJ Calculate the standard enthalpy of formation (ΔfHo) for nitroglycerin. The enthalpy of formation of CO2 (g) is -393.5 kJ/mol. The enthalpy of formation of H2O (g) is -241.8 kJ/mol.

Enthalpy change of a reaction is defined as the difference in enthalpy of all the products and the reactants each multiplied with their respective number of moles. The equation that is used to calculate the enthalpy change of a reaction is:

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]

For the given chemical reaction:

[tex]4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+10H_2O(g)+6N_2(g)+O_2(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(12\times \Delta H_f_{(CO_2)})+(10\times \Delta H_f_{(H_2O)})+(6\times \Delta H_f_{(N_2)})+(1\times \Delta H_f_{(O_2)})]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})][/tex]

We are given:

[tex]\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(N_2)}=-0kJ/mol\\\Delta H_f_{(CO_2)}=-393.5kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-5678kJ[/tex]

Putting values in above equation, we get:

[tex]-5678=[(12\times (-393.5))+(10\times (-241.8))+(6\times (0))+(1\times (0))]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})]\\\\\Delta H_f_{(C_3H_5N_3O_9)}=-365.5kJ/mol[/tex]

Hence, the [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.

dsdrajlin dsdrajlin · Answer 2 · 2021-10-17T21:02:14+02:00

The explosive nitroglycerin (C₃H₅N₃O₉) decomposes rapidly upon ignition or sudden impact. The standard enthalpy of formation for nitroglycerin is -1462 kJ/mol.

Let's consider the following thermochemical equation.

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

Given the standard enthalpies of formation (ΔH°f), we can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

[tex]\Delta H\° _{rxn} = \Sigma \Delta H\°_f(p) \times n_p - \Sigma \Delta H\°_f(r) \times n_r[/tex]

where,

n: number of moles in the balanced equation
p: products
r: reactants

By definition, the standard enthalpy of formation of simple substances (e.g N₂, O₂) in standard state and the most standard form is zero.

Then, we get the following expression form which we can calculate the standard enthalpy of formation of nitroglycerin.

[tex]\Delta H\° _{rxn} = \Delta H\°_f(CO_2(g)) \times 12 mol + \Delta H\°_f(H_2O(g)) \times 10 mol - \Delta H\°_f(C_3H_5N_3O_9(l)) \times 1 mol\\\\\Delta H\°_f(C_3H_5N_3O_9(l)) \times 1 mol = \Delta H\°_f(CO_2(g)) \times 12 mol + \Delta H\°_f(H_2O(g)) \times 10 mol - \Delta H\° _{rxn}\\\\\Delta H\°_f(C_3H_5N_3O_9(l)) \times 1 mol = (-393.5kJ/mol) \times 12 mol + (-241.8 kJ/mol) \times 10 mol - (-5678kJ)\\\\\Delta H\°_f(C_3H_5N_3O_9(l)) = -1462 kJ/mol[/tex]

The explosive nitroglycerin (C₃H₅N₃O₉) decomposes rapidly upon ignition or sudden impact. The standard enthalpy of formation for nitroglycerin is -1462 kJ/mol.

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