Question

If the end energies of H-H, Br-Br and H-Br are 433, 1992 and 364 kJ `mol^(-1)` respectively, then `DeltaH^(@)` for the reaction, `H_(2)(g)+Br_(2)(g)to2HB r(g)` is
A. `-261kJ`
B. `+103kJ`
C. `+261kJ`
D. `-103kJ`

Answer 1

Correct Answer - D
For the reaction,
`H_(2)(g)+Br_(2)(g) to 2HBr (g),DeltaH^(ө)=?`
`DeltaH=-[2xx`bond energy of HBr -(bond enregy of `H_(2)+` Bond energy of `Cl_(2)`)]
`DeltaH=-[2xx(364)-(433+192)]kJ`
`=-[728-625]kJ=-103kJ`