For a cell reaction `2H_2(g)+O_2(g)to2H_2O(l)/_\_rS_(198)^(@)=-0.32KJ//k` . What is the value of `/_\_fH_(298)^(@)(H_2O,l)` ?
Given: `O_2(g)+4H^+(aq)+4e^-)to2H_2O(l), E^(@)=1.23V`

Options

A. `-189.71 " kJ "mol^(-1)`
B. `-285.08" kJ "mol^(-1)`
C. `-379.42" kJ "mol^(-1)`
D. `-570.16" kJ "mol^(-1)`

Answer

B

Explanation

For the given cell reaction involving formation of 2 moles of `H_(2)O`,
`DeltaG^(@)=-nFE_(cell)^(@)=-4xx96500xx1.23J`
`=474780J=-474.78kJ`
(n=4 for the given cell raction)
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
orr `DeltaH^(@)=DeltaG^(@)-TDeltaS^(@)`
`=-474.78kJ+298K(-0.32kJK^(-1))`
`=-474.8-95.36kJ=-570.16 kJ`
This is enthalpy change for the formation of 2 moles of `H_(2)O`
`therefore` Enthalpy change for the formation of 1 mole of
`H_(2)O(Delta_(f)H^(@))=-(570.16)/(2)=-285.08kJ`.