1.

H_(2)+(1)/(2)O_(2)rarrH_(2)O, DeltaH=-68.39 kcalK+H_(2)O+"Water"rarrKOH(aq)+(1)/(2)H_(2),DeltaH=-48 kcalKOH+"water"rarrKOH(aq),DeltaH=-14 kcal The heat of formation of KOH is (in kcal)

Answer»

`-68.39 + 48-14`
`-68.39 - 48+14`
`68.39-48+14`
`68.39+48+14`

Solution :Aim: `K_((S))+(1)/(2)O_(2(g))+(1)/(2)H_(2(g))rarrKOH_((S))`
EQ. (II) + eq. (i) - eq. (iii) GIVES
`DeltaH=-48+(-68.39)-(-14)=-68.39-48+14`.


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