Question

Calculate the enthalpy change (ΔH) of the following reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) given average bond energies of various bonds, i.e, C−H,CC,O=O,C=O,O−H as 414, 814, 499, 724 and 640 kJ/mol respectively.

Answer 1

Answer:

Given Bond enthalpies (kJmol

−1

) C−H= 414; C≡=814; O−H=640; O=O=499; C=O=724

△H

reaction

=∑△H

product

−∑△H

reactant

2C

2

H

2

(g)+5O

2

(g)→4CO

2

(g)+2H

2

O(g)

4(C−H)+2(C≡C) +5(O=O) 8(C=O)+4(O−H)

△H

rxn

=8(C=O)+4(O−H)−5(O=O)−4(C−H)−2(C≡C)

△H

rxn

=8×724+4×6405×4994×4142×814

=5792+2560−2495−1656−1628

△H

rxn

=2573 kJmol

−1

Explanation:

Hope Answer Is Correct