Given that ΔH°= -311 kJ for the hydrogenation of acetylene,C₂H₂: H-C C-H(g) + 2 H₂(g) → CH₃-CH₃(g)
And the following bond dissociation energies,estimate a value for the C-to-C triple bond dissociation energy.

Question

Given that ΔH°= -311 kJ for the hydrogenation of acetylene,C₂H₂: H-C

Given that ΔH°= -311 kJ for the hydrogenation of acetylene,C<sub>2</sub>H<sub>2</sub>: H-C C-H(g) + 2 H<sub>2</sub>(g) → CH<sub>3</sub>-CH<sub>3</sub>(g) And the following bond dissociation energies,estimate a value for the C-to-C triple bond dissociation energy. A) 1050 kJ/mol B) 833 kJ/mol C) 807 kJ/mol D) 397 kJ/mol

C-H(g) + 2 H₂(g) → CH₃-CH₃(g)
And the following bond dissociation energies,estimate a value for the C-to-C triple bond dissociation energy. Given that ΔH°= -311 kJ for the hydrogenation of acetylene,C2H2: H-C C-H(g) + 2 H2(g) → CH3-CH3(g) And the following bond dissociation energies,estimate a value for the C-to-C triple bond dissociation energy. A) 1050 kJ/mol B) 833 kJ/mol C) 807 kJ/mol D) 397 kJ/mol

Quizplus · Accepted Answer

The ΔH° for the reaction is the sum of the bond dissociation energies (BDE) of the C-C triple bond and two H-H bonds, minus the BDE of the two C-H bonds formed, plus the ΔH° of formation of the product CH2=CH2. Mathematically,

ΔH°reaction = BDE(C-C) + 2BDE(H-H) - 2BDE(C-H) + ΔH°f(CH2=CH2)

Rearranging the equation and plugging in the given values,

BDE(C-C) = [2BDE(C-H) - ΔH°f(CH2=CH2) - 2×BDE(H-H)]

= [2(413 kJ/mol) - (-84 kJ/mol) - 2(435 kJ/mol)]

= 807 kJ/mol

Therefore, choice C is the best answer.

Given That ΔH°= -311 KJ for the Hydrogenation of Acetylene,C2H2

Given That ΔH°= -311 KJ for the Hydrogenation of Acetylene,C₂H₂