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If H(2)+1//2O(2)rarr H(2)O , Delta =-68....

If `H_(2)+1//2O_(2)rarr H_(2)O , Delta =-68.39` Kcal
`K+H_(2)O+ "water" rarr KOH(aq)+1//2H_(2) , Delta H=-48.0` Kcal
`KOH+"water" rarr KOH(aq)Delta H=-14.0` Kcal the heat of formation of KOH is -

A

`-68 + 48 -14`

B

`-68-48 + 14 `

C

`68-48+14`

D

`68 +48 +14`

Text Solution

Verified by Experts

The correct Answer is:
B

Aim. `K(s)+1/2O_2(g)+1/2H_2(g) rarrKOH(s)`
Eqn(ii) + Eqn.(i)-Eqn.(iii) gives
`DeltaH =-48+(-68)-(-14)`
`=- 68-48 +14`
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