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calculate th enthalpy change for the following reaction: H2(g) + Br2(g) -----> 2HBr(g) [given: bond enthalpies of H-H, Br-Br, H-Br, are 435kj mol, 192 kj mol, and 436 kj mol respectively]

calculate th enthalpy change for the following reaction:
 
H2(g) + Br2(g) -----> 2HBr(g)
 
[given: bond enthalpies of H-H, Br-Br, H-Br, are 435kj mol, 192 kj mol, and 436 kj mol respectively]
 

Grade:12

1 Answers

Arun
25750 Points
5 years ago
H2(g)+ Br2 (g) → 2HBr(g)

Given :
Bond enthalpy of H—H = 435 kJ mol-1
Bond enthalpy of Br—Br = 192 kJ mol-1
Bond enthalpy of H—Br = 364 kJ mol-1
 
Reaction for the given bond enthalpies are
 
H2(g)  →​2H(g)         ∆H= 435 kJmol-1  (energy required to break the bond. Hence positive. )
Br2(g) → 2Br(g)       ∆H= 192 kJmol-1
H(g)+Br(g) →​HBr(g) ∆H=  -364 kJmol-1  (energy released due to bond formation. Hence taken negative)

Multiply Third equation by two and add all three question to get to the equation
H2(g)+Br2(g)-->2HBr(g)

Hence,enthalpy change for the reaction
∆H = 435+192+2(-364)
= -101 kJ/mol-1

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