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Image for 8.66 (a) Use bond enthalpies to estimate the enthalpy change for the reaction of hydrogen with ethene: H2(g) +

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Answer #1

H2(g) + C2H4(g) -------> C2H6(g)
The bonds can be represented as
H - H + H2C=CH2 -------> H3C - CH3

From this equation we understand that 1 H - H bond and 1 C = C bond are broken and 2 C - H bonds and one C - C bond are formed.

To calculate the enthalpy change (dH) of the reaction, the bond energies are ( refered)
H - H : 432 kJ/mol
C = C : 614 kJ/mol
C - H : 413 kJ/mol
C - C : 347 kJ/mol

change in enthalpy = sum of the energies required to break old bonds (positive signs) plus the sum of the energies released in the formation of new bonds (negative signs).

Bond breaking energy:(+)
432 + 614 = 1046 kJ/mol
Bond formation energy:(-)
2(413) + 347 = 1173 kJ/mol
change in enthalpy = -1173 + 1046 = - 127 kJ/mol

2. Now chnage in enthalpy of reaction by using enthalpies of formations (again found in literature)
dHf [C2H4(g)] = + 52 kJ/mol
dHf [C2H6(g)] = - 84.7 kJ/mol
dHf [H2(g)] = 0 (dHf for the most stable form of elements is zero)

dHreaction = dHf(products) - dHf(reactants)
dHr = -84.7 - (+52) = - 136.7 kJ / mol

The difference; - 136.7 - (- 127 ) = - 9.7 kJ/mol
3. Now dH reaction by using enthalpies of formations (again found in literature)
dHf [C2H4(g)] = + 52 kJ/mol
dHf [C2H6(g)] = - 84.7 kJ/mol
dHf [H2(g)] = 0 (dHf for the most stable form of elements is zero)

dHreaction = dHf(products) - dHf(reactants)
dHr = -84.7 - (+52) = - 136.7 kJ / mol

The difference; - 136.7 - (- 127 ) = - 9.7 kJ/mol

this difference can be explained by the fact that the  bond enthalpies are average values and are not as accurate

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Answer #2

Ha(g) + C2H4(g) → C2H6(g) :?H = ?

?H = ?Hf products - ?Hf reactants

      = ?Hf C2H6(g) - ( ?Hf C2H4(g) +?Hf H2 (g) )

      = -83.85 KJ / mol - ( +52.47KJ/ mol + 0 )

      = -136.32 KJ / mol

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