Question

The $k_w$ of water at two different temperature is:

• $T{25}^{o}C$

$1.08\times {10}^{-14}$

• ${50}^{o}C$$k_w$

$5.474\times {10}^{-14}$

Assuming that $\Delta H$ of any reaction is independent of temperature, calculate the enthalpy of neutralization of a strong acid and strong base.(in $kJ/mole$)

Answer 1

$\ln \frac{k_{w_2}}{k_{w_1}}=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
$\ln \frac{5.474\times {10}^{-14}}{1.08\times {10}^{-14}}=\frac{\Delta H}{8.314}(\frac{1}{298}-\frac{1}{323})$
$\Delta =51952.6J=51.95kJ/mole$