Question

Ionic product of water $\left(K_w\right)$ at two different temperatures ${25}^{o}C$ and ${50}^{o}C$ are $1.08\times {10}^{-14}$ and $5.474\times {10}^{-14}$ respectively. Assuming $\Delta H$ of any reaction to be independent of temperature, calculate enthalpy of neutralisation of strong acid with strong base.

Answer 1

$T_1=273+25=298K$

$T_2=273+50=323K$

Ionic products are

$K_1=1.08\times {10}^{-14}$

$K_2=5.474\times {10}^{-14}$

$\log \frac{K_{w_1}}{K_{w_2}}=\frac{\Delta H}{2303R}(\frac{1}{T_1}-\frac{1}{T_2})$

$\log \frac{K_{w_1}}{K_{w_2}}=\frac{\Delta H}{2303R}(\frac{1}{T_1}-\frac{1}{T_2})\log \frac{1.08{\times 10}^{-14}}{5.474{\times 10}^{-14}}=\frac{\Delta H}{2303\times 8.314}(\frac{1}{298}-\frac{1}{323})\Delta H=-0.705\times 2.303\times 8.314\times \frac{1}{0.00335-0.00309}=12.5Kcal$