Question

In case of $H _{2} O$ molecule, the enthalpy needed to break the two $O - H$ bonds is not the same, i.e.
$H _{2} O (g) \longrightarrow H (g)+ OH (g) ; \Delta_{a} H_{1}^{ s }=502\, kJ\, mol ^{-1}$
$OH (g) \longrightarrow H (g)+ O (g) ; \Delta_{a} H_{2}^{ S }=427\, kJ\, mol ^{-1}$
What should be the mean bond enthalpy of $O - H$ bonds in case of $H _{2} O$ molecule?

• A. $75\, kJ\, mol ^{-1}$
• B. $-75\, kJ\, mol ^{-1}$
• C. $464.5\, kJ\, mol ^{-1}$
• D. $929\, kJ\, mol ^{-1}$

Answer 1

Answer: (C)

In case of $H _{2} O$ molecule, the enthalpy needed to break the two $O - H$ bonds is not the same, i.e.
$H _{2} O (g) \longrightarrow H (g)+ OH (g) ; \Delta_{a} H_{1}^{ s }=502\, kJ\, mol ^{-1}$ $OH (g) \longrightarrow H (g)+ O (g) ; \Delta_{a} H_{2}^{ s }=427\, kJ\, mol ^{-1}$
The mean or average bond enthalpy of $O - H$ bonds in case of $H _{2} O$ molecule is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule.
Average bond enthalpy
$=\frac{502+427}{2}=464.5\, kJ\, mol ^{-1}$