How Do You Calculate the Enthalpy Change from Ethyne to Ethane?

[EmilyHopkins]

Calculate the standard enthalpy of ethyne

C₂H₂ + 2H₂ -------> C₂H₆

Assume the following mean bond energies

C$\equiv$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$\equiv$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)

ΔH_h - Enthalpy of hydrogenation

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

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

 

[chemisttree]

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.