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 )

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >