The pH of pure water at `25^(@)C` and `35^(@)C` are 7 and 6 respectively. What is the heat of formation of water from `H^(+)` and `OH^(-)`?

Options

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

Answer

B

Explanation

At `25^(@)C[H^(+)]=10^(-7), :. K_(m)=10^(-14)`
At `35^(@)C[H^(+)]=10^(-6),:.K_(w)=10^(-12)`
Now using
`2.303"log"(K_(w_(2)))/(K_(w_(1)))=(DeltaH)/R((T_(2)-T_(1))/(T_(1)T_(2)))`
We get `DeltaH=84.55` kcal/mole
Thus `H_(2)OhArrH^(+)+OH^(-)`
`:.H^(+)+OH^(-)hArrH_(2)O`
`DeltaH=-84.55`kcal/mole