For a cell reaction ` 2H_(2(g)) + O_(2(g)) to 2H_2O_((l)) DeltaS_(298)^(@) = - 0.32 kJ // K `. What is the value `Delta_(i) H_(298)^(@) (H_2O, l)` ? Given : `O_(2(g)) + 4H_((aq))^(+) + 4e^(-) to 2H_2O_((l)) , E^(@) = 1.23 V `

Options

A. `-285. 07 kJ// mol `
B. `-570.14 kJ // mol `
C. `285. 07 kJ//mol`
D. None of these

Answer

A

Explanation

`DeltaG = DeltaH -TDeltaS^(0) - nFE^(0) = DeltaH -TDeltaS - 4 xx 1.23 xx 96,500 = DeltaH - 298 xx (-0.32)`
`DeltaH_("for 2 moles") = -(4 xx 1.23 xx 96,500) - (298 xx 0.32)`
`DeltaH_(H_2O)` for 1 mole `=(1)/(2) [(-4 xx 1.23 xx 96,500) - (298 xx 0.32)] = -285.07 "KJ/mole"`