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?
9 Replies
@Dexter
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.Let’s say there is x amount of energy when all bonds are broken
If H2 and Br2 are formed, then bond energy is released
So enthalpy of system becomes x-435 -192
Similarly, when 2HBr is formed we get x- 2(BE)
Now try using enthalpy of products - reactants
used it yea, got C as the ans but the key gave D
basically I added up the ∆H of the reactants and used ∆rH = 2(∆HBr) - (∆H H2 + ∆H Br2) am I wrong?
it's just the solutions gave this ans it confused me somewhat
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)
The key point is that they're talking about bond enthalpies, a special case of enthalpy of reaction.
got it thanks!!!
+solved @Propellane
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