Which reaction will have a negative ∆H value

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TT0
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Homework Statement


Multiple-choice question 'Which change would have a negative ΔH value?' with five reactions labelled A to E: Na(s) to Na(g); Na(g) to Na+(g) + e-; Na+Cl-(s) to Na+(g) + Cl-(g); Cl2(g) to 2Cl(g); Cl(g) + e- to Cl-(g).

Homework Equations

The Attempt at a Solution


Negative ∆H means that the product is more stable. Change from solid to gas requires energy so cancel out A and C. Reaction D doesn't make sense. The products of reaction B and E are more stable than the reactant because there is a full shell so it seems to me both would have a ∆H value. Can someone tell me what am I missing?

Cheers!
 
Reaction D does make a perfect sense.

Ionization always requires energy.
 
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I thought chlorine didn't monatomic gases. So are you saying D is the answer? If so, why does it release energy?

Thanks a lot
 
TT0 said:
I thought chlorine didn't monatomic gases.

Every diatomic molecule can be split (forced to dissociation) at temperatures high enough. It typically requires a lot of energy (hence the need for high temperatures).
 
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I see, so D has a positive ∆H value.

Would you say both B and E have negative ∆H as the products are more stable?

Cheers!
 
TT0 said:
Would you say both B and E have negative ∆H as the products are more stable?
Borek said:
Ionization always requires energy.
 
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So I guess A is the answer then? As it does not ionise. C is sublimation which requires energy. If A is the answer, isn't energy required for a particle to turn into a gas?

Thanks!
 
I see. I got confused and forgot ionisation does not mean turning into an ion. Thanks a lot