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Using bond energies, find the enthalpy of formation for HBr(g)?
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- RijlLv 51 decade agoFavorite Answer
The formation of one mole of HBr from the elements is:
½H₂ + ½Br₂ --> HBr
You need to break the H-H bonds and Br-Br bonds and make the HBr bonds.
Take one-half the bond energy per mole of the H-H bond add one-half the bond energy per mole of the Br-Br bond, and subract the bond energy per mole of H-Br.
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