Q.

The enthalpy change at 298 K for decomposition is given in following steps :

Step-1: H2O(g)H(g)+OH(g) ΔH=498kJ/mol1

Step-2: OH(g)H(g)+O(g) ΔH=428kJ/mol1

Then, value of mean bond enthalpy of O-H bond will be:

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a

498 kJ/mol

b

463 kJ/mol

c

428 kJ/mol

d

70 kJ/mol

answer is B.

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Detailed Solution

H2OH+OH;ΔH=498kJ/molOHH+O;ΔH=428KJ/ mol 

The bond enthalpy of OH bond will be 

The average of both enthalpy required in 1 and 2

ΔHOH=ΔH1,OH+ΔH2,OH2=498+42824632463KJ/mol

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