Bond Energies

  • #1
EmilyHopkins
8
0
Calculate the standard enthalpy of ethyne

C2H2 + 2H2 -------> C2H6

Assume the following mean bond energies

C[itex]\equiv[/itex]C = 813 KJ mole-1

C - C = 364 KJ mole-1

C - H = 413 KJ mole -1

H - H = 436 KJ mole -1


Attempted Solution:

Bonds broken
1 C[itex]\equiv[/itex]C = 813 KJ mole-1 ( 813 KJ mole-1)
2 H - H = 436 KJ mole -1 (2 x 436 KJ mole -1)

Bonds formed
4 C - H = 413 KJ mole -1 (4x 413 KJ mole-1)
1 C - C = 364 KJ mole-1 (364 KJ mole-1)

ΔHh - Enthalpy of hydrogenation

ΔHh = ((813 + 2x 436) - ((4x 413) + 364)) KJ mole -1
ΔHh = 1685 KJ mole-1 - 2016 KJ mole-1
ΔHh = -331 KJ mole -1

--
Is my method and the way I worked it out correct ?
 

Answers and Replies

  • #2
chemisttree
Science Advisor
Homework Helper
Gold Member
3,710
704
Looks good. You might check that by subtracting the two enthalpies of formation for ethyne and ethane. The difference should be about the same as your answer.
 

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