Question

For the reaction: H2(g)+C2H4(g)-->C2H6(g)

Bond & Bond Enthalpy

H-H 436.4 kJ/mole

C-H 414 kJ/mol

C-C 347 kJ/mol

C=C 620 kJ/mol

Substance & delta Hf

H2 0

C2H4 52.3

C2H6 -84.7

(a) estimate the enthalpy of reaction, using the bond enthalpy values from the table in kJ/mol

(b) Calculate the enthalpy of reaction, using standard enthalpies of formation

Answer 1

Hess' Law for bond enthalpies is:

ΔH = ΣE reactant bonds broken - Σ E product bonds broken
H2(g) + C2H4(g) ---> C2H6(g)
On the reactant side, we have these bonds broken:

H2(g) + C2H4(g) --> C2H6(g)

dH rxn = dBond energies reactants - products

dH rxn = [H-H @436.4 + C=C @ 620 + 4 C-H @ 414 each] - [C-C @ 347 + 6 C-H @ 414 each]

dH rxn = [436.4 + 620 + 4 *414 ] - [ 347 + 6*414] = -118.6 kJ

(a) answer: -118.6 kJ
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(b) Calculate the enthalpy of reaction, using standard enthalpies of formation.

dH = dHf's products - reactants

dH = [C2H6 @ -84.7] - [H2 @ zero + C2H4 @ 52.3]

dH = -84.7 - 52.3 = -137.0 Kj