H2(g)+ Br2(g)2HBr(g)Given :Bond enthalpy of H—H = 435 kJ mol-1Bond enthalpy of Br—Br = 192 kJ mol-1Bond enthalpy of H—Br = 364 kJ mol-1
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Solution
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