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`.

Options

A. `84.55kcal mol^(-)`
B. `-84.55kcal mol^(-1)`
C. `74.55kcal mol^(-1)`
D. `-74.55kcal mol^(-1)`

Answer

B

Explanation

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`