Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) both contain a triple bond, but they differ greatly in their chemical properties. (a) Write the Lewis structures for the two substances. (b) By referring to Appendix C, look up the enthalpies of formation of acetylene and nitrogen. Which compound is more stable? (c) Write balanced chemical equations for the complete oxidation of \(\mathrm{N}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) and of acetylene to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O (g) .\) (d) Calculate the enthalpy of oxidation per mole for \(\mathrm{N}_{2}\) and for $\mathrm{C}_{2} \mathrm{H}_{2}\( (the enthalpy of formation of \)\mathrm{N}_{2} \mathrm{O}_{5}(g)\( is \)11.30 \mathrm{~kJ} / \mathrm{mol}\( ). \)(\mathbf{e})$ Both \(\mathrm{N}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{2}\) possess triple bonds with quite high bond enthalpies (Table 8.3). Calculate the enthalpy of hydrogenation per mole for both compounds: acetylene plus \(\mathrm{H}_{2}\) to make methane, \(\mathrm{CH}_{4}\); nitrogen plus \(\mathrm{H}_{2}\) to make ammonia, \(\mathrm{NH}_{3}\).

Step by step solution

01

(a) Lewis Structures for Acetylene and Nitrogen

To draw the Lewis structures for Acetylene (C2H2) and Nitrogen (N2), we first count the valence electrons. Then, we connect the atoms with bonds and distribute the remaining electrons. 1. Acetylene (C2H2): Carbon has 4 valence electrons and Hydrogen has 1 valence electron. In Acetylene, there are two Carbon atoms and two Hydrogen atoms, so we have a total of 10 valence electrons. - Connect the two Carbon atoms with a triple bond (uses 6 electrons) - Connect each Hydrogen atom to a Carbon atom with a single bond (uses 4 electrons) - No remaining electrons 2. Nitrogen (N2): Nitrogen has 5 valence electrons. In N2, there are two Nitrogen atoms, so we have a total of 10 valence electrons. - Connect the two Nitrogen atoms with a triple bond (uses 6 electrons) - Distribute the remaining 4 electrons as lone pairs on the Nitrogen atoms (2 electrons each)

02

(b) Enthalpies of Formation and Stability

Referring to Appendix C or any chemistry reference, we find the enthalpies of formation: - Acetylene (C2H2): +226.7 kJ/mol - Nitrogen (N2): 0 kJ/mol Since Nitrogen (N2) has an enthalpy of formation of 0 kJ/mol (which is lower than Acetylene's enthalpy), Nitrogen is more stable.

03

(c) Balanced Chemical Equations for Complete Oxidation

For the complete oxidation of Nitrogen (N2) to form N2O5(g): N2 + 5/2 O2 -> N2O5 For the complete oxidation of Acetylene (C2H2) to form CO2(g) and H2O(g): 2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O

04

(d) Enthalpy of Oxidation per mole

To calculate the enthalpy of oxidation per mole, we can use the equation: Enthalpy of oxidation = Enthalpy of formation (products) - Enthalpy of formation (reactants) Enthalpy of formation values (from Appendix C or a chemistry reference): N2O5(g): 11.3 kJ/mol O2(g): 0 kJ/mol CO2(g): -393.5 kJ/mol H2O(g): -241.8 kJ/mol For N2: Enthalpy of oxidation = (1 * 11.3 kJ/mol) - (0 kJ/mol + 0 kJ/mol) = 11.3 kJ/mol For C2H2: Enthalpy of oxidation = (4 * -393.5 kJ/mol + 2 * -241.8 kJ/mol) - (2 * 226.7 kJ/mol + 0 kJ/mol) = -1300.8 kJ/mol

05

(e) Enthalpy of Hydrogenation per mole

To calculate the enthalpy of hydrogenation per mole, we can use the equation: Enthalpy of hydrogenation = Enthalpy of formation (products) - Enthalpy of formation (reactants) Enthalpy of formation values (from Appendix C or a chemistry reference): CH4(g): -74.8 kJ/mol NH3(g): -45.9 kJ/mol H2(g): 0 kJ/mol For Acetylene (C2H2) to form Methane (CH4): Enthalpy of hydrogenation = (2 * -74.8 kJ/mol) - (1 * 226.7 kJ/mol + 1 * 0 kJ/mol) = -76.3 kJ/mol For Nitrogen (N2) to form Ammonia (NH3): Enthalpy of hydrogenation = (2 * -45.9 kJ/mol) - (1 * 0 kJ/mol + 3 * 0 kJ/mol) = -91.8 kJ/mol

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