Bond Energies

  • #1
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
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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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