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
:?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
8.66 (a) Use bond enthalpies to estimate the enthalpy change for the reaction of hydrogen with...
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