Calculate the enthalpy of formation of water, given that the bond energies of `H-H, O=O` and `O-H` bond are `433 kJ mol^(-1), 492 kJ mol^(-1)`, and `464 kJ mol^(-1)`, respectively.

Calculate the enthalpy of formation of water, given that the bond ener

1.	Calculate the enthalpy of formation of water, given that the bond energies of `H-H, O=O` and `O-H` bond are `433 kJ mol^(-1), 492 kJ mol^(-1)`, and `464 kJ mol^(-1)`, respectively.
Answer» `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g)` `Delta_(f)H = BE` of product `-BE` of reactant `= (BE of H_(2) +(1)/(2)BE of O_(2)) - (2BE of O-H)` `= 433 +(1)/(2) xx492 -2 xx 464` `= 433 +246 - 928 =- 249 kJ`

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Calculate the enthalpy of formation of water, given that the bond energies of `H-H, O=O` and `O-H` bond are `433 kJ mol^(-1), 492 kJ mol^(-1)`, and `464 kJ mol^(-1)`, respectively.

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