The enthalpy of combustion of `H_(2)(g)` at 298 K to give `H_(2)O(g) is -249 kJ mol^(-1)`and bond enthalpies of H-H and `O=O`are `433 kJ mol^(-1)and 492 kJ mol^(-1)` respectively. The bond enthalpy of O-H is

Options

A. `464 kJ mol ^(-1)`
B. `-464 kJ mol ^(-1)`
C. `232 kJ mol ^(-1)`
D. `-232 kJ mol ^(-1)`

Answer

A

Explanation

We have

Hence `epsi _(H - H ) + (1)/(2) epsi _(O-O) - 2 epsi _(O-H) = ( 433 + (1)/(3) xx 492) kJ mol ^(-1) - 2 epsi _(O-H) =- 249 kJ mol ^(-1)`
`epsi _(O-H) = (1)/(2) [(433 + (1)/(2) xx 492) + 249] kJ mol ^(-1) = 464 kJ mol ^(-1) `