1.

For the cell reaction, 2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)S_(298K)^(@)=-0.32" kJ "K^(-1) What is the value of Delta_(f)H^(@)(H_(2)O)? Given: O_(2)(g)+4H^(+)(aq)+4e^(-)to2H_(2)O(l),E^(@)=1.23V

Answer»

`-189.71 " kJ "mol^(-1)`
`-285.08" kJ "mol^(-1)`
`-379.42" kJ "mol^(-1)`
`-570.16" kJ "mol^(-1)`

Solution :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`.


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