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`

Compounds with carbon-carbon double bond, such as ethylene, C_(2)H_(4) , add hydrogen in a reaction called hydrogenation. C_(2)H_(4)(g)+H_(2)(g) rarr C_(2)H_(6)(g) Calculate enthalpy change for the reaction, using the following combustion data C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta_("comb")H^(Θ) = -1401 kJ mol^(-1) C_(2)H_(6)(g) + 7//2O_(2)(g)rarr 2CO_(2) (g) + 3H_(2)O(l) , Delta_("comb")H^(Θ) = -1550kJ H_(2)(g) + 1//2O_(2)(g) rarr H_(2)O(l) , Delta_("comb")H^(Θ) = -286.0 kJ mol^(-1)

Compounds with carbon-carbon double bond, such as ethylene, C_(2)H_(4) , add hydrogen in a reaction called hydrogenation. C_(2)H_(4)(g)+H_(2)(g) rarr C_(2)H_(6)(g) Calculate enthalpy change for the reaction, using the following combustion data C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta_("comb")H^(Θ) = -1401 kJ mol^(-1) C_(2)H_(6)(g) + 7//2O_(2)(g)rarr 2CO_(2) (g) + 3H_(2)O(l) , Delta_("comb")H^(Θ) = -1550kJ H_(2)(g) + 1//2O_(2)(g) rarr H_(2)O(l) , Delta_("comb")H^(Θ) = -286.0 kJ mol^(-1)

The enthalpy changes at 25^(@)C in successive breaking of O - H bonds of water are : H_(2)O(g) rarr H(g) + OH(g) Delta H = 498 kJ mol^(-1) OH(g) rarr H(g) + O(g) Delta H = 428 kJ mol^(-1) The bond enthalpy of the O-H bond is :

On the basis of the following thermochemical data [(Delta_(f)G^(@)H^(+) (aq) = 0] H_(2)O (l) rarr H^(+) (aq), Delta H = 57.32 kJ H_(2) (g) + (1)/(2)O_(2) (g) rarr H_(2)O (l), Delta H = - 286 kJ The value of enthalpy of formation of OH^(-) at 25^(@)C is

Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ