Editing Openai/69227557-0850-8003-9c19-8f26825e337d (section)

== gpt-5-1-thinking ==
=== User: When considering the energy transfer associated to C, Cl, and H reacting to form methyl chloride, I believe that textbo… ===
When considering the energy transfer associated to C, Cl, and H reacting to form methyl chloride, I believe that textbooks say that ~1500 kJ of energy is transferred from the individual atoms to the bonds formed between the atoms. Is this accurate? How does this or does this not account for the endothermic or exothermic properties of the rection?

=== Assistant: Condensed restatement of your question: ===
Condensed restatement of your question:
Textbooks say ~1500 kJ of energy is associated with forming the bonds in CH₃Cl from C, H, and Cl atoms. Is that correct, and how does that relate (or not) to whether the overall reaction is endothermic or exothermic?

===== For gas-phase atoms: =====

C(g)+3H(g)+Cl(g)→CH3Cl(g)\text{C(g)} + 3\text{H(g)} + \text{Cl(g)} \rightarrow \text{CH}_3\text{Cl(g)}C(g)+3H(g)+Cl(g)→CH3Cl(g)
The total bond energy in CH₃Cl is roughly:
* 3 × C–H bonds ≈ 3 × 410 kJ/mol ≈ 1230 kJ/mol
* 1 × C–Cl bond ≈ 330–340 kJ/mol

Total ≈ 1.5 × 10³ kJ/mol (≈ 1500 kJ/mol).

By definition, bond enthalpy is the energy required to break a bond. So:
* To break CH₃Cl completely into separate atoms, you must put in ~1500 kJ/mol (endothermic).
* Going the other way (atoms → CH₃Cl), the system releases ~1500 kJ/mol (exothermic).

So yes: “~1500 kJ is transferred from the atoms to the bonds” is basically saying

: 

That is not yet a statement about the full reaction you do in the lab.

===== In reality, you don’t usually start with isolated atoms. You start from the elements in their standard forms, e.g.: =====

C(s, graphite)+32H2(g)+12Cl2(g)→CH3Cl(g)\text{C(s, graphite)} + \tfrac{3}{2}\text{H}_2\text{(g)} + \tfrac{1}{2}\text{Cl}_2\text{(g)}
\rightarrow \text{CH}_3\text{Cl(g)}C(s, graphite)+23H2(g)+21Cl2(g)→CH3Cl(g)
To connect this real reaction to the ~1500 kJ number, conceptually break it into steps:
# Atomize the reactants (break all their bonds, change phases as needed): - C(s, graphite) → C(g)  (sublimation/atomization; costs energy) - ½ Cl₂(g) → Cl(g)       (½ of Cl–Cl bond; costs energy) - 3/2 H₂(g) → 3 H(g)     (3/2 of H–H bond; costs energy) This is all endothermic. You’re climbing up the energy hill.
# Form CH₃Cl from atoms: - C(g) + 3H(g) + Cl(g) → CH₃Cl(g) This step is where that ≈1500 kJ/mol is released (exothermic).

The net reaction enthalpy (ΔH_rxn) is:

ΔHrxn=(energy to atomize reactants)−(energy released by forming bonds in CH3Cl)\Delta H_{\text{rxn}} = \text{(energy to atomize reactants)} - \text{(energy released by forming bonds in CH}_3\text{Cl)}ΔHrxn=(energy to atomize reactants)−(energy released by forming bonds in CH3Cl)
or equivalently (in the usual “bonds broken – bonds formed” form):

ΔHrxn≈∑E(bonds broken)−∑E(bonds formed)\Delta H_{\text{rxn}} \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})ΔHrxn≈∑E(bonds broken)−∑E(bonds formed)
So:
* The 1500 kJ is essentially the “bonds formed” term for the product.
* The reactants also have bonds and cohesive energies (C–C in graphite, H–H, Cl–Cl), which contribute to the “bonds broken” term.

Whether the overall reaction is exothermic or endothermic depends on which side “wins”:
* If bonds formed in products are stronger overall (bigger total bond energy) than those broken in reactants → exothermic (ΔH < 0).
* If bonds broken are stronger → endothermic (ΔH > 0).

===== Think of it like this: =====
* Choose a reference where separated atoms = high potential energy.
* When those atoms form CH₃Cl, the system drops by ~1500 kJ/mol in potential energy.
* That energy has to go somewhere → usually to the surroundings as heat and/or light → that’s why bond formation is exothermic.

But:
* The reaction enthalpy you see in tables (ΔH°_f for CH₃Cl) starts from standard states, not from free atoms.
* To go from standard states → atoms → CH₃Cl, you must add the atomization costs and subtract the 1500 kJ bond-formation release.
* Only that final net number tells you how exothermic or endothermic the real reaction is.

===== 1. “Is ~1500 kJ from atoms to bonds accurate?” Roughly, yes: it’s a reasonable approximate total bond energy for CH₃Cl (from isolated gas-phase atoms). It’s the energy you’d need to break the molecule into atoms, and the energy released when you form it from atoms. =====
# “How does this relate to endothermic vs exothermic?” That 1500 kJ number is only one side of the accounting (the product side, atoms → CH₃Cl). To know if the real reaction from elemental C, H₂, and Cl₂ is endothermic or exothermic, you must compare: - Energy required to break all bonds / cohesive structures in the reactants versus - Energy released when forming the CH₃Cl bonds. The difference of those two is the reaction enthalpy. The 1500 kJ is baked into the “bonds formed” term but doesn’t by itself tell you whether the overall reaction is exothermic or endothermic.

If you’d like, next step could be: we pick actual approximate bond enthalpies and walk the full numeric estimate for ΔH for
C(graphite) + 3/2 H₂ + 1/2 Cl₂ → CH₃Cl to see a concrete number.