enthalpy change q

solved this using ∆rH = ∆H products - ∆H reactants but the solutions give a diff ans, was confused, am I wrong or are the solutions wrong?

iTeachChem Helper•4/13/24, 8:03 PM

@Dexter

iTeachChem Helper•4/13/24, 8:03 PM

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Comrade Rock Astley•4/13/24, 8:07 PM

Let’s say there is x amount of energy when all bonds are broken

Comrade Rock Astley•4/13/24, 8:08 PM

If H2 and Br2 are formed, then bond energy is released

Comrade Rock Astley•4/13/24, 8:08 PM

So enthalpy of system becomes x-435 -192

Comrade Rock Astley•4/13/24, 8:09 PM

Similarly, when 2HBr is formed we get x- 2(BE)

Comrade Rock Astley•4/13/24, 8:09 PM

Now try using enthalpy of products - reactants

CComrade Rock Astley Now try using enthalpy of products - reactants

Gravy BonesOP•4/13/24, 8:20 PM

used it yea, got C as the ans but the key gave D

CComrade Rock Astley So enthalpy of system becomes x-435 -192

Gravy BonesOP•4/13/24, 8:23 PM

basically I added up the ∆H of the reactants and used ∆rH = 2(∆HBr) - (∆H H2 + ∆H Br2) am I wrong?

Gravy BonesOP•4/13/24, 8:36 PM

it's just the solutions gave this ans it confused me somewhat

GGravy Bones basically I added up the ∆H of the reactants and used ∆rH = 2(∆HBr) - (∆H H2 + ∆...

Propellane•4/13/24, 9:46 PM

Kinda because the enthalpy values they've given are specifically mentioned as Bond energies (bond enthalpies). You have calculated it by taking it as "enthalpy of reaction". Difference is that the Bond enthalpy of a reaction:
(BE) = Sum of BE(reactants) - Sum of BE (products)

Propellane•4/13/24, 9:47 PM

The key point is that they're talking about bond enthalpies, a special case of enthalpy of reaction.

iTeachChem•4/14/24, 1:47 AM

voice-message.ogg384.54KB

Gravy BonesOP•4/14/24, 1:53 AM

got it thanks!!!

Gravy BonesOP•4/14/24, 1:54 AM

+solved @Propellane

iTeachChem Helper•4/14/24, 1:54 AM

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enthalpy change q

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