Use the bond energies provided to estimate δh°rxn for the reaction below. c2h4(g) + h2(g) → c2h6(g) δh°rxn = ? bond bond energy (kj/mol) c-c 347 c-h 414 c=c 611 c≡c 837 h-h 436 a) -128 kj b) +98 kj c) +700 kj d) -102 kj e) -166 kj

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C2H4

1 c=c ->611

4 C-H -> 4*414=1656

=> Ha=2267 kj

H2 :  H-H 436

Hb = 436

C2H6

1 C-C 347

6 C-H 6*414=2484

=> Hc=2831

H=(Ha+Hb)-Hc=2267+436-2831 = -128kj

Answer A

Explanation:

Hello,

In this case, the standard enthalpy of reaction could be computed via the bond energies when both broken or made as shown below:

In this manner, we infer that at the reactants for ethene, a double bond between carbons is broken as well as a bond between hydrogens (such values turn out positive). Furthermore, a single bond between carbons and two single bonds between carbon and hydrogen are made (such values turn out negative), in such a way, we develop the aforesaid equation to obtain:

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hypostyle hall;