JEE Main 2017 — Thermodynamics (C) Question with Solution

In order to calculate the enthalpy change for ${\mathrm{H}}_2\mathrm{O}$ at $5°\mathrm{C}$ to ice at $-5°\mathrm{C},$ we need to calculate the enthalpy change of all the transformation involved in the process.

(a) Energy change of $1 \mathrm{mol}, {\mathrm{H}}_2\mathrm{O}\left(l\right),$ at $5°\mathrm{C}\to 1 \mathrm{mol}, {\mathrm{H}}_2\mathrm{O} \left(l\right), 0°\mathrm{C}$

(b) Energy change of $1 \mathrm{mol},{\mathrm{H}}_2\mathrm{O} \left(l\right)$ at $0°\mathrm{C}\to 1 \mathrm{mol}, {\mathrm{H}}_2\mathrm{O} \left(\mathrm{s}\right) \left(\mathrm{ice}\right), 0°\mathrm{C}$

(c) Energy change of $1 \mathrm{mol}$, Ice $\left(\mathrm{s}\right)$, at $0°\mathrm{C}\to 1 \mathrm{mol},$ Ice $\left(\mathrm{s}\right), -5°\mathrm{C}$

Total $\mathrm{\Delta H}$

$={\mathrm{C}}_{\mathrm{P}}\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right]\mathrm{\Delta T}+\mathrm{\Delta H}$ freezing $+$${\mathrm{C}}_{\mathrm{P}}\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\right]\mathrm{\Delta T}$

$=\left(75.3 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}\right)(-5)\mathrm{K}+\left(-6\times {10}^3 \mathrm{J} {\mathrm{mol}}^{-1}\right)+\left(36.8 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}\right)(-5)\mathrm{K}$

$\mathrm{\Delta H}=-6.56 \mathrm{kJ} {\mathrm{mol}}^{-1}$ (exothermic process)

So, $\mathrm{\Delta H}=6.56 \mathrm{kJ} {\mathrm{mol}}^{-1}$.