The enthalpy needed to break the two O – H bonds in water are as follows:
`H_(2)O(g)rarrH(g)+O-H(g)" "Delta_(a)H_(1)^(0)=502kJmol^(-1)`
`O-H(g)rarrH(g)+O(g)" "Delta_(a)H_(2)^(0)=427kJmol^(-1)`
What is the average bond enthalpy of `H_(2)O`?

The enthalpy needed to break the two O – H bonds in water are as follows: H_(2)O(g)rarrH(g)+O-H(g)" "Delta_(a)H_(1)^(0)=502kJmol^(-1) O-H(g)rarrH(g)+O(g)" "Delta_(a)H_(1)^(0)=502kJmol^(-1) What is the average bond enthalpy of H_(2)O ?

The enthalpy changes at 298 kJ in successive breaking of O-H bonds of water, are H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1) OH(g) rarr H(g)+O(g), DeltaH =428 kJ mol^(-1) The bond energy of the O-H bond is

Calculate the bond enthalpy of the O - H bond using the following thermochemical equations: H_(2) O (g) rarr H (g) + OH (g) , Delta_(bond H^(@) = 502 kJ OH (g) rarr (g) + O (g), Delta_(bond) H^(@) = 427 kJ

Calculate the standard enthalpy of formation of CH_(3)OH(l) from the following data: CH_(3)OH(l)+3/2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l), …(i), Delta_(r)H_(1)^(Θ)=-726 kJ mol^(-1) C(g)+O_(2)(g) rarr CO_(2)(g), …(ii), Delta_(c )H_(2)^(Θ)=-393 kJ mol^(-1) H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), ...(iii), Delta_(f)H_(3)^(Θ)=-286 kJ mol^(-1)

The enthalpy of vapourisation of liquid water using the data : H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l) , Delta =-285.77 KJ mol^(-1) H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(g) , Delta H=-241.84 KJ mol^(-1)

H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g) DeltaH =- 242 kJ mol^(-1) Bond energy of H_(2) and O_(2) is 436 and 500 kJ mol^(-1) , respectively. What is bond energy of O-H bond?