at-room-temperature-the-reaction-between-propene-ch2-ch-ch3-and-hydrogen-gas-h2-

Question

At room temperature the reaction between propene (CH₂=CH−CH₃) and hydrogen gas (H₂) to form propane (CH₃−CH₂−CH₃) goes to completion when a platinum catalyst is present in the reaction mixture.
Determine the signs of ΔH° and ΔS° for this addition reaction based on the information given above.
Explain these signs in terms of freedom of motion and bonding.
CH₂=CH−CH₃ + H₂ (Pt catalyst) → CH₃−CH₂−CH₃

Accepted Answer

∆Sº is negative because there is a loss of freedom of motion.
∆Hº is negative because a strong sigma bond and a weak pi bond are replaced by two strong sigma bonds.

Answer

∆Sº is negative because there is a loss of freedom of motion.
∆Hº is positive because alkanes have higher energy than the corresponding alkenes.

Answer

∆Sº is positive because alkanes have more freedom of motion.
∆Hº is negative because a strong sigma bond and a weak pi bond are replaced by two strong sigma bonds.

Answer

∆Sº is positive because alkanes have more freedom of motion.
∆Hº is positive because alkanes have higher energy than the corresponding alkenes.

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