Question

(a) Calculate the enthalpy change for the hypothetical reaction 1-butene $\longrightarrow 2$ -methylpropene. The heats of formation are 1-butene, $-0.30 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-0.07 \mathrm{kcal} \mathrm{mol}^{-1}\right) ; 2$ methylpropene, $-17.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-4.13 \mathrm{kcal} \mathrm{mol}^{-1}\right)$. (b) Which butene isomer in part (a) is more stable?

Name: Problem 13, Chapter 4: Organic Chemistry
Uploaded: 2022-02-18T18:47:58-08:00
Duration: 1 min 57 s
Channel: Hunza Gilgit
Description: (a) Calculate the enthalpy change for the hypothetical reaction 1-butene ⟶2 -methylpropene. The heats of formation are 1-butene, -0.30 kJ mol^-1(-0.07 kcal mol^-1) ; 2 methylpropene, -17.3 kJ mol^-1(-4.13 kcal mol^-1). (b) Which butene isomer in part (a) is more stable?

   (a) Calculate the enthalpy change for the hypothetical reaction 1-butene $\longrightarrow 2$ -methylpropene. The heats of formation are 1-butene, $-0.30 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-0.07 \mathrm{kcal} \mathrm{mol}^{-1}\right) ; 2$ methylpropene, $-17.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-4.13 \mathrm{kcal} \mathrm{mol}^{-1}\right)$.
(b) Which butene isomer in part (a) is more stable?

Organic Chemistry

Marc Loudon, Jim… 6th Edition

Chapter 4, Problem 13

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Solved by Expert Hunza Gilgit

02/18/2022

Step 1

3 kJ/mol) - (-0.30 kJ/mol) ΔH = -17.3 kJ/mol + 0.30 kJ/mol ΔH = -17.0 kJ/mol The enthalpy change for the reaction is -17.0 kJ/mol. Show more…

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(a) Calculate the enthalpy change for the hypothetical reaction 1-butene $\longrightarrow 2$ -methylpropene. The heats of formation are 1-butene, $-0.30 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-0.07 \mathrm{kcal} \mathrm{mol}^{-1}\right) ; 2$ methylpropene, $-17.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\left(-4.13 \mathrm{kcal} \mathrm{mol}^{-1}\right)$. (b) Which butene isomer in part (a) is more stable?

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Key Concepts

Stability of Isomers

The thermodynamic stability of isomers can be inferred from their heats of formation; a lower heat of formation generally indicates a more stable compound. In comparing isomers, the one with a lower intrinsic energy is typically more thermodynamically favored and therefore more stable.

Enthalpy Change Calculation

The enthalpy change for a reaction is determined by subtracting the sum of the heats of formation of the reactants from the sum of the heats of formation of the products. This calculation is a direct application of Hess's law, allowing us to understand whether a reaction is exothermic or endothermic.

Heats of Formation

Heats of formation are the enthalpy changes associated with forming one mole of a compound from its elements in their standard states. They serve as a useful reference for comparing the intrinsic energy content of different compounds and are crucial for calculating reaction enthalpies.

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