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#### The pH of pure water at 250C and 450C are 7 and 5.5 respectively. Calculate the heat of dissociation of H2O into H+ and OH- ions.Option: 1 65.5 kcal/molOption: 2 98.5 kcal/molOption: 3 10 kcal/molOption: 4 Data inadequate

We have given,

$\begin{array}{ll}{\text { At } 25^{\circ} \mathrm{C} ;} & {\left[\mathrm{H}^{+}\right]=10^{-7} \quad \therefore K_{w}=10^{-14}} \\ {\text { At } 45^{\circ} \mathrm{C} ;} & {\left[\mathrm{H}^{+}\right]=10^{-5.5} \quad \therefore K_{w}=10^{-11}}\end{array}$

Now using the van't Hoff's equation for the dependance of Keq on the Temperature,

$\mathrm{2.303 \log _{10} \frac{K_{w_{2}}}{K_{w_{1}}}=\frac{\Delta H}{R}\left [ \frac{1}{T_1}-\frac{1}{T_2} \right ]}$

Here, R = 2cal /mol.K

T1 = 25 0C = 298 K

T2 = 45 0C = 318 K

putting this values in the above equation.

$2.303 \log _{10} \frac{10^{-11}}{10^{-14}}=\frac{\Delta H}{2}\left[\frac{318-298}{318 \times 298}\right]$

$\Rightarrow \Delta H = 65472.45\; cal/mol$

$\Rightarrow \Delta H = 65.5\; kcal/mol$

Hence, the correct answer is Option (1)