Question

Calculate the enthalpy change on freezing of $1.0$ mol of water at ${10.0^o}C$ to ice at $ - {10.0^o}C$.
${\Delta _{fus}}H = 6.03kJmo{l^{ - 1}}\;\;{\text{at }}{{\text{0}}^o}C$
${C_P}\left[ {{H_2}O(l)} \right] = 75.3Jmo{l^{ - 1}}{K^{ - 1}}$
${C_P}\left[ {{H_2}O(s)} \right] = 36.8Jmo{l^{ - 1}}{K^{ - 1}}$

Answer 1

Hint: In enthalpy of transformation of $1.0$ of water from its liquid phase at ${10.0^o}C$ to its solid phase at $ - {10.0^o}C$ can be calculated by summation of change in enthalpy involved in the transformation of one mole of water at ${10.0^o}C$ to one mole of water at ${0^o}C$, change in enthalpy involved in the transformation of one mole of water at ${0^o}C$ to one mole of ice at ${0^o}C$ and change in enthalpy involved in conversion of one mole of ice at ${0^o}C$ to one mole of ice at $ - {10.0^o}C$.

Complete answer:
The change in energy for the three transformations involved in the process can be calculated as follows:
Step-1: Change in enthalpy involved in the transformation of one mole of water at ${10.0^o}C$ to one mole of water at ${0^o}C$:
The change in enthalpy can be expressed in terms of specific heat at constant pressure and temperature as per following relation:
$\Delta {H_1} = {C_p}\Delta T\;\;\;\;...(1)$
Substituting given values for ${H_2}O(l)$:
$ \Rightarrow \Delta {H_1} = 75.3(0 - 10)$
$ \Rightarrow \Delta {H_1} = - 753Jmo{l^{ - 1}}$
Step-2: Change in enthalpy involved in the transformation of one mole of water at ${0^o}C$ to one mole of ice at ${0^o}C$:
When one mole of a solid is transformed or converted into liquid without changing the temperature and at constant pressure is known as enthalpy of fusion. As per question the enthalpy of fusion for water at ${0^o}C$ is $6.03kJmo{l^{ - 1}}$.
Thus, $\Delta {H_2} = 6.03 \times {10^3}Jmo{l^{ - 1}}$
Step-3: change in enthalpy involved in conversion of one mole of ice at ${0^o}C$ to one mole of ice at $ - {10.0^o}C$:
Substituting given values for ${H_2}O(s)$ in equation (1):
$\Delta {H_3} = {C_p}\Delta T$
$ \Rightarrow \Delta {H_3} = 36.8( - 10 - 0)$
$ \Rightarrow \Delta {H_3} = - 368Jmo{l^{ - 1}}$
Therefore, overall change in enthalpy of the transformation will be as follows:
$\Delta H = \Delta {H_1} + \Delta {H_2} + \Delta {H_3}$
$ \Rightarrow \Delta H = - 753 - 6.03 \times {10^3} - 368$
$ \Rightarrow \Delta H = - 753 - 6030 - 368$
$ \Rightarrow \Delta H = - 7151Jmo{l^{ - 1}}\;{\text{or }} - 7.151kJmo{l^{ - 1}}$
Hence, the change in enthalpy on freezing of $1.0$ mol of water at ${10.0^o}C$ to ice at $ - {10.0^o}C$ is $ - 7.151kJmo{l^{ - 1}}$.

Note:
It is important to note that when any substance transforms from its solid phase to its liquid phase, the change in enthalpy is positive while if the substance is changing its phase from liquid state to a solid state, then the change in enthalpy is negative. Thus, in this case the value is considered with a negative sign during the summation of enthalpies.