InterviewSolution
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InterviewSolution
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Calculate heat of formation of `KOH(s)` using the following equations `K(s) +H_(2)O(l) +aq rarr KOH(aq)+1//2 H_(2)O(g),DeltaH =- 48.0 kcal …(i)` `H_(2)(g) +2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.4 kcal …(ii)` `KOH(s) +(aq) rarr KOH(aq),DeltaH =- 14.0 kcal ....(iii)` |
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Answer» We have to evaluate `DeltaH` for `K(s) +1//2O_(2)(g) +1//2H_(2)(g) rarr KOH(s), DeltaH = ?` Adding equations (i) and (ii), `K(s)+H_(2)O(l) +aq rarrKOH(aq) +1//2H_(2)(g)` `DeltaH =- 48.0 kcal` `H_(2)(g)+1//2O_(2)(g) rarr H_(2)O(l)`, `DeltaH =- 68.4 kcal` `{:ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)+aqunderset(DeltaH=-116.4Kcal...(iv))(rarrKOH(aq)):}` Subtracting equation (iii) from equation (iv) `{:(KOH(s)+aqrarrunderset(DeltaH =- 102.4 kcal ..(v))(KOH(aq),)),("- - - +" ),(ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)rarrunderset(DeltaH =- 102.4 kcal)(KOH(s),))):}` |
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