InterviewSolution
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InterviewSolution
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The `pH` of pure water at `25^(@)C` and `35^(@)C` are `7` and `6`, respectively. Calculate the heat of formation of water from `H^(o+)` and `overset(Θ)OH`.A. `84.55kcal mol^(-)`B. `-84.55kcal mol^(-1)`C. `74.55kcal mol^(-1)`D. `-74.55kcal mol^(-1)` |
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Answer» Correct Answer - B At `25^(@)C, [H^(+)]=10^(-7)` `:. K_(w)=10^(-14)` At `35^(@)C, [H^(+)]=10^(-6)` `:. K_(w)=10^(-12)` Now using, `2.303log (K_(w_(2))/(K_(w_(1))))=(Delta H)/(R ) [(T_(2)-T_(1))/(T_(1)xxT_(2))]` `2.303log (10^(-12)/(10^(-14)))=(Delta H)/(2)[(10)/(298xx308)]` `:. Delta H=84551.4cal//mol=84.551 kcal//mol` Thus, `H_(2)OhArr H^(+)+OH^(-)`, `Delta H=84.551 kcal//mol` `:. H^(+)+OH^(-)hArr H_(2)O`, `Delta H= -84.551 kcal//mol` |
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