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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. `-68.39+48-14.0`B. `-68.39-48.0+14.0`C. `+68.39-48.0+14.0`D. `+68.39+48.0-14.0` |
Answer» Correct Answer - B `K_((s))+(1)/(2)O_(2(g))+(1)/(2)H_(2(g))rarr KOH_((s)) , Delta H_(f) = ?` Desired equation = eq(i) + (ii) - eq (iii) `Delta H_(f)=(-68.39)+(-48)-(-14)` |
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