Question

if the bond dissociation energies of H2,Br2 and HBr are 100,50 and 90 kJ per mol respectively then the enthalpy of formation would be
a)60 kJ per mol
b)-60kJ per mol
c)-15kJ per mol
d)-30 kJ per mol​

Answer 1

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Answer 2

Answer:

Given = H2 , Br2 and HBr bond dissociation energies

            H2 = 100KJ

           Br2 = 50KJ

          HBr = 90KJ

to find =  enthalpy(ΔH)

solution  =

          ΔH = ∑bond energy of reactant - ∑bond energy in product

          ΔH= 100 + 50 - 2x90

          ΔH= 150 - 180

          ΔH= -30KJ/mol

the enthalpy of formation would be   ΔH= -30KJ/mol