Estimate the enthalpy change for the combustion of one mole of acetylene, c2h2, to form carbon dioxide and water vapor. be(c? h) = 456 kj/mol) be(c? c) = 962 kj/mol be(o=o) = 499 kj/mol be(c=o) = 802 kj/mol be(o? h) = 462 kj/molanswer choices: a. +653 kj/molb. ? 155 kj/molc. +1010 kj/mold. ? 1759 kj/mole. ? 1010 kj/mol

ΔH rxn =  -1010 kJ/molC₂H₂

Explanation:

To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:

C₂H₂ + 5/2O₂   ⇒   2CO₂ + H₂O

Bond Broken                                    Bonds Formed

2 C-H + 1 C≡C + 5/2 O=O                4C=O + 2 H-O

Enthalpy bonds broken:

2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol)  =  3121.5 kJ

Enthalpy bond formed:

4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ

ΔH rxn = H broken - H formed =  3121.5 kJ - 4132.0 kJ = - 1010  kJ (per mol C₂H₂ )

the answer for sodium is 0.186

the answer for rubidium is 0.251

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