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Q.

Bond dissociation enthalpy of H2 ,  Cl2 and HCl are 434, 242 and 431 kJ mol-1 respectively. Enthalpy of formation of HCI is.

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a

−93 kJ mol−1

b

−245 kJ mol−1

c

245 kJ mol−1

d

93 kJ mol−1

answer is A.

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Detailed Solution

ΔH=ΔHReactant−ΔHProduct

=(ΔHH2+ΔHC/2)−(2×ΔHHCl)

=(434 + 242 ) – ( 2 x 431)  =  676 - 862

ΔH=-186

ΔHf=ΔHRno moles=−1862=−93KJ

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