Question

0-H = 460 kJ mol-
7. Calculate the enthalpy change during the reaction
H2(g) + Br.(g) - 2HBr(g)
The bond energy data is given below
H-H = 435 kJ mol
Br-Br = 192 kJ mol-1
H-Br = 368 kJ mol-1
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Answer 1

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Answer 2

H (g)+ Br  (g) → 2HBr(g)

Given :

Bond enthalpy of H—H = 435 kJ mol

Bond enthalpy of Br—Br = 192 kJ mol

Bond enthalpy of H—Br = 368 kJ mol

Reaction for the given bond enthalpies are

H (g)  →2H(g)         ∆H= 435 kJmol   (energy required to break the bond. Hence positive. )

Br (g) → 2Br(g)       ∆H= 192 kJmol

H(g)+Br(g) →HBr(g) ∆H=  -368 kJmol (energy released due to bond formation. Hence taken negative)

Multiply Third equation by two and add all three question to get to the equation

H (g)+Br (g)-->2HBr(g)

Hence,enthalpy change for the reaction

∆H = 435+192+2(-368)

= -109 kJ/mo