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Q.

The enthalpy of combustion of H2(g) at 298 K to give H2O is –298 kJ mol–1 and bond enthalpies of H–H and O=O are 433 kJ mol–1 and 492 kJ mol–1 respectively. The bond enthalpy of O–H is 

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a

–232 kJ mol–1

b

488.5 kJ mol–1

c

464 kJ mol–1 

d

232 kJ mol–1

answer is B.

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Detailed Solution

H2+ 1/2O2→H2O ,∆H = – 298 kJ mol–1

∆H =ΔHH−H+12ΔHO=O– [2∆HO–H]

−298=[433+12×492]−[2ΔHO−H]

=ΔHO−H=433+246+2982

= 488.5 kJ /mol

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