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Question: The heat of hydrogenation for 3-methylbutene and 2-pentene are -30 kcal/mol and -28 kcal/mol respect...

The heat of hydrogenation for 3-methylbutene and 2-pentene are -30 kcal/mol and -28 kcal/mol respectively. The heats of combustion of 2-methylbutane and pentane are - 784 kcal/mol and -782 kcal/mol respectively. All the values are given under standard conditions. Taking into account that combustion of both alkanes give the same products, what is Δ\DeltaH (in kcal/mol) for the following reaction under same conditions ? 3-methylbut-1-ene ⇌\rightleftharpoons 2-pentene

A

0

B

-4

C

+2

D

2

Answer

-4

Explanation

Solution

The problem asks to calculate the enthalpy change (Δ\DeltaH) for the isomerization reaction: 3-methylbutene ⇌\rightleftharpoons 2-pentene.

We are given the following data:

  1. Heat of hydrogenation of 3-methylbutene = -30 kcal/mol. This reaction is: 3-methylbut-1-ene + H2_2 →\rightarrow 2-methylbutane. ΔHhydrog(3−methylbutene)=−30\Delta H_{hydrog}(3-methylbutene) = -30 kcal/mol

  2. Heat of hydrogenation of 2-pentene = -28 kcal/mol. This reaction is: 2-pentene + H2_2 →\rightarrow n-pentane. ΔHhydrog(2−pentene)=−28\Delta H_{hydrog}(2-pentene) = -28 kcal/mol

  3. Heat of combustion of 2-methylbutane = -784 kcal/mol. ΔHcomb(2−methylbutane)=−784\Delta H_{comb}(2-methylbutane) = -784 kcal/mol

  4. Heat of combustion of n-pentane = -782 kcal/mol. ΔHcomb(n−pentane)=−782\Delta H_{comb}(n-pentane) = -782 kcal/mol

Since 2-methylbutane and n-pentane are isomers with the molecular formula C5_5H12_{12}, their complete combustion produces the same products (CO2_2 and H2_2O).

We can use Hess's Law to find the enthalpy change for the isomerization. Let H(X)H(X) denote the enthalpy of substance X.

From the combustion data, the enthalpy difference between n-pentane and 2-methylbutane can be found: ΔHcomb(2−methylbutane)=H(products)−H(2-methylbutane)\Delta H_{comb}(2-methylbutane) = H(\text{products}) - H(2\text{-methylbutane}) ΔHcomb(n-pentane)=H(products)−H(n-pentane)\Delta H_{comb}(n\text{-pentane}) = H(\text{products}) - H(n\text{-pentane})

Subtracting the second equation from the first: ΔHcomb(2-methylbutane)−ΔHcomb(n-pentane)=[H(products)−H(2-methylbutane)]−[H(products)−H(n-pentane)]\Delta H_{comb}(2\text{-methylbutane}) - \Delta H_{comb}(n\text{-pentane}) = [H(\text{products}) - H(2\text{-methylbutane})] - [H(\text{products}) - H(n\text{-pentane})] −784−(−782)=H(n-pentane)−H(2-methylbutane)-784 - (-782) = H(n\text{-pentane}) - H(2\text{-methylbutane}) −2 kcal/mol=H(n-pentane)−H(2-methylbutane)-2 \text{ kcal/mol} = H(n\text{-pentane}) - H(2\text{-methylbutane})

From the hydrogenation data: ΔHhydrog(3-methylbutene)=H(2-methylbutane)−H(3-methylbutene)\Delta H_{hydrog}(3\text{-methylbutene}) = H(2\text{-methylbutane}) - H(3\text{-methylbutene}) ΔHhydrog(2-pentene)=H(n-pentane)−H(2-pentene)\Delta H_{hydrog}(2\text{-pentene}) = H(n\text{-pentane}) - H(2\text{-pentene})

We want to find ΔH\Delta H for the reaction: 3-methylbutene →\rightarrow 2-pentene. ΔHiso=H(2-pentene)−H(3-methylbutene)\Delta H_{iso} = H(2\text{-pentene}) - H(3\text{-methylbutene})

Rearranging the hydrogenation equations: H(3-methylbutene)=H(2-methylbutane)−ΔHhydrog(3-methylbutene)H(3\text{-methylbutene}) = H(2\text{-methylbutane}) - \Delta H_{hydrog}(3\text{-methylbutene}) H(2-pentene)=H(n-pentane)−ΔHhydrog(2-pentene)H(2\text{-pentene}) = H(n\text{-pentane}) - \Delta H_{hydrog}(2\text{-pentene})

Substitute these into the expression for ΔHiso\Delta H_{iso}: ΔHiso=[H(n-pentane)−ΔHhydrog(2-pentene)]−[H(2-methylbutane)−ΔHhydrog(3-methylbutene)]\Delta H_{iso} = [H(n\text{-pentane}) - \Delta H_{hydrog}(2\text{-pentene})] - [H(2\text{-methylbutane}) - \Delta H_{hydrog}(3\text{-methylbutene})] ΔHiso=[H(n-pentane)−H(2-methylbutane)]−ΔHhydrog(2-pentene)+ΔHhydrog(3-methylbutene)\Delta H_{iso} = [H(n\text{-pentane}) - H(2\text{-methylbutane})] - \Delta H_{hydrog}(2\text{-pentene}) + \Delta H_{hydrog}(3\text{-methylbutene})

Substitute the known values: ΔHiso=(−2 kcal/mol)−(−28 kcal/mol)+(−30 kcal/mol)\Delta H_{iso} = (-2 \text{ kcal/mol}) - (-28 \text{ kcal/mol}) + (-30 \text{ kcal/mol}) ΔHiso=−2+28−30\Delta H_{iso} = -2 + 28 - 30 ΔHiso=−4 kcal/mol\Delta H_{iso} = -4 \text{ kcal/mol}