Calculate ∆ H ° for the following reactions. (b) CH 3 Br + HCl → CH 3 Cl + HBr

Hi and welcome back. Our next problem says, calculate the delta H and then the little degree sign, meaning standard temperature and pressure or the reaction shown below. And we have CH three C two C two I plus HC is becoming CH three, CH two, CH two cl plus H I. We have a table with different types of bonds listed and their corresponding bond association enthalpy in kilojoules per mole. Our answer choices are a negative 167 kilojoules per mole B negative 251 kilojoules per mole, C 17 kilojoules per mole or D 35 kilojoules per mole. So we're given the entropy of breaking different types of bonds. That's that bond association entropy. So we are pretty simple arithmetic here of the bonds that we're breaking that entropy and then we subtract from it, the ENP due to the bonds that were forming. So let's look at our reaction and see which bonds are breaking and which performing. And it might be worth doing a quick sketch with a skeletal structure just to keep track of the attachment there because the way it's written, of course, we have the H two I, which almost makes it look like a hydrogen that iodine or collect connected, which of course they're not, but that could help us if we want. So we've got three carbons and an I A dime. So 12, three and then iodine. So the bond that we're going to break here is a carbon iodine bond. So let's write out delta H equals. So we want the bde of our bonds broken. So we'll just jot up here by our reaction in the first reactant, we had the bond broken is C I. And then we noticed that there's says for a primary carbon, which is what we have. And then our other reaction is HCL. So that would be an HCL bond. The bonds that are formed. If you look at our product, we now have the al or alkyl, it has chlorine. So our bond formed AC CL or primary carbon. And then of course, H I so back to our formula for delta H, our bonds broken C for primary carbon is 222 kilojoules per mole. And then for HCL, that is 431 kilojoules per mole. I put that in parentheses and then subtract from it and I have open parentheses. The bonds formed, we have the CCL bond for primary carbon is 339 kilojoules per mole plus, then we have H I being formed which is 297 kill jules, primo. So let's combine our terms in parentheses. So in the first set of parentheses for the bonds broken, add the vice together and we get 653 kilojoules per mole and then subtract in the second group of parentheses, 636 kilojoules per mole. And that gives me an end result of 17 kilojoules per mole. When we look at our answer choices, choice C is 17 kilojoules per mole. So that is the delta H for this particular reaction. See you in the next video.

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6. Thermodynamics and Kinetics

Enthalpy

Organic Chemistry

6. Thermodynamics and Kinetics

Enthalpy

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Problem 37b

Textbook Question

Calculate ∆H° for the following reactions.
(b) CH₃Br + HCl → CH₃Cl + HBr

Verified step by step guidance

Identify the bonds broken and formed in the reaction. In this case, the bonds broken are the C-Br bond in CH3Br and the H-Cl bond in HCl. The bonds formed are the C-Cl bond in CH3Cl and the H-Br bond in HBr.

Look up the bond dissociation energies (BDEs) for each bond involved in the reaction. These values are typically given in units of kJ/mol and can be found in a table of bond energies. For example, approximate values might be: C-Br = 276 kJ/mol, H-Cl = 431 kJ/mol, C-Cl = 338 kJ/mol, and H-Br = 366 kJ/mol.

Calculate the total energy required to break the bonds. This is the sum of the bond dissociation energies for the bonds broken: \( \text{Energy for bonds broken} = \text{BDE}_{\text{C-Br}} + \text{BDE}_{\text{H-Cl}} \).

Calculate the total energy released when the new bonds are formed. This is the sum of the bond dissociation energies for the bonds formed: \( \text{Energy for bonds formed} = \text{BDE}_{\text{C-Cl}} + \text{BDE}_{\text{H-Br}} \).

Determine \( \Delta H^\circ \) for the reaction by subtracting the energy of the bonds formed from the energy of the bonds broken: \( \Delta H^\circ = \text{Energy for bonds broken} - \text{Energy for bonds formed} \). This value will indicate whether the reaction is exothermic (negative \( \Delta H^\circ \)) or endothermic (positive \( \Delta H^\circ \)).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy Change (∆H°)

Enthalpy change (∆H°) refers to the heat content change of a system at constant pressure during a chemical reaction. It indicates whether a reaction is exothermic (releases heat, ∆H° < 0) or endothermic (absorbs heat, ∆H° > 0). Calculating ∆H° involves considering the bond energies of the reactants and products, allowing chemists to predict the energy changes associated with reactions.

Bond Energies

Bond energies are the amounts of energy required to break specific chemical bonds in a molecule. Each type of bond (e.g., C-H, C-Br, C-Cl) has a characteristic bond energy value. By summing the bond energies of the reactants and subtracting the sum of the bond energies of the products, one can calculate the overall enthalpy change for a reaction, providing insight into its energy dynamics.

Reaction Mechanism

A reaction mechanism describes the step-by-step sequence of elementary reactions that lead to the overall transformation of reactants into products. Understanding the mechanism helps in predicting the rate of reaction and the energy changes involved. In the case of the reaction CH3Br + HCl → CH3Cl + HBr, knowing the mechanism can clarify how bonds are broken and formed, influencing the calculation of ∆H°.

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