To determine the enthalpy change ($\Delta H$) for the reaction:

$\text{CH}_3\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{HCl}$

we'll use bond enthalpies for the bonds broken and formed. Here’s the general approach:

1. Identify bonds broken (reactants):

   • CH$_3$CH$_3$: Breaking one C–H bond.
   • Cl$_2$: Breaking one Cl–Cl bond.

2. Identify bonds formed (products):

   • CH$_3$CH$_2$Cl: Forming a C–Cl bond.
   • HCl: Forming an H–Cl bond.

3. Use bond enthalpies: For the calculation, you’ll need the average bond enthalpies for each bond involved. Common values are:

   • C–H bond enthalpy: ~ 413 kJ/mol
   • Cl–Cl bond enthalpy: ~ 243 kJ/mol
   • C–Cl bond enthalpy: ~ 338 kJ/mol
   • H–Cl bond enthalpy: ~ 432 kJ/mol

4. Calculate the enthalpy change: $\Delta H = \text{(Bonds Broken)} - \text{(Bonds Formed)}$ Substituting values: [ \Delta H = [(413) + (243)] - [(338) + (432)] = 656 - 770 = -114 \text{ kJ/mol} ]

Result

The enthalpy change ($\Delta H$) for this reaction is approximately -114 kJ/mol, indicating an exothermic reaction.

Would you like a breakdown of each step or have any questions?

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1. What role do bond enthalpies play in determining enthalpy change?
2. Why is this reaction exothermic?
3. How accurate are average bond enthalpies for precise calculations?
4. What would happen to the enthalpy change if Cl$_2$ was replaced with Br$_2$?
5. How does this reaction relate to the concept of bond dissociation energy?

Tip: The sign of $\Delta H$ tells us if the reaction is exothermic (negative) or endothermic (positive).