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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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Willchemistry

hey there!:

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

sebassandin

Answer:

[tex]\Delta H_{rxn}^o=-128kJ[/tex]

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:

[tex]\Delta H_{rxn}^o=\Delta H_{broken}+\Delta H_{made}[/tex]

In this manner, we infer that at the reactants for ethene, [tex]C_2H_4[/tex] 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:

[tex]\Delta H_{rxn}^o=(611kJ+436kJ)+(-347kJ-2*414kJ)\\\Delta H_{rxn}^o=-128kJ[/tex]

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