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Calculate the enthalpy change for the re...

Calculate the enthalpy change for the reaction .
`H_(2)(g) + Br_(2)(g) to 2HBr (g)` Given that the bond energies of H-H, H-Br, and Br-Br are `435 kJ mol^(-1)` , `364 kJ mol^(-1)` , and `192 kJ mol^(-1)`, respectively.

Text Solution

Verified by Experts

`DeltaH_("reaction") = sumBE_(R) - sumBE_(p)`
`=[BE(H_(2))+ BE(Br)_(2)]-[2 xxBE(HBr)]`
` = 435 + 1.92 - 2 xx 364 =- 101 kJ`
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