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) ; Δa H1 s =502 kJ mol -1 OH (g) longrightarrow H (g)+ O (g) ; Δa H2 S =427 kJ mol -1 What should be the mean bond enthalpy of O - H bonds in case of H 2 O molecule?

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) ; Δa H1 s =502 kJ mol -1 OH (g) longrightarrow H (g)+ O (g) ; Δa H2 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

Tardigrade Admin · Accepted Answer

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) ; Δa H1 s =502 kJ mol -1 OH (g) longrightarrow H (g)+ O (g) ; Δa H2 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 =(502+427/2)=464.5 kJ mol -1

$75 k J m o l^{- 1}$

$- 75 k J m o l^{- 1}$

$464.5 k J m o l^{- 1}$

$929 k J m o l^{- 1}$

Q. In case of H2​O molecule, the enthalpy needed to break the two O−H bonds is not the same, i.e. H2​O(g)⟶H(g)+OH(g);Δa​H1s​=502kJmol−1 OH(g)⟶H(g)+O(g);Δa​H2S​=427kJmol−1 What should be the mean bond enthalpy of O−H bonds in case of H2​O molecule?

75kJmol−1

−75kJmol−1

464.5kJmol−1

929kJmol−1

$75 k J m o l^{- 1}$

$- 75 k J m o l^{- 1}$

$464.5 k J m o l^{- 1}$

$929 k J m o l^{- 1}$