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Question: Calculate the enthalpy change on freezing of \(1.0\) mol of water at \({10.0^o}C\) to ice at \( - {1...

Calculate the enthalpy change on freezing of 1.01.0 mol of water at 10.0oC{10.0^o}C to ice at −10.0oC - {10.0^o}C.
ΔfusH=6.03kJmol−1    at 0oC{\Delta _{fus}}H = 6.03kJmo{l^{ - 1}}\;\;{\text{at }}{{\text{0}}^o}C
CP[H2O(l)]=75.3Jmol−1K−1{C_P}\left[ {{H_2}O(l)} \right] = 75.3Jmo{l^{ - 1}}{K^{ - 1}}
CP[H2O(s)]=36.8Jmol−1K−1{C_P}\left[ {{H_2}O(s)} \right] = 36.8Jmo{l^{ - 1}}{K^{ - 1}}

Explanation

Solution

In enthalpy of transformation of 1.01.0 of water from its liquid phase at 10.0oC{10.0^o}C to its solid phase at −10.0oC - {10.0^o}C can be calculated by summation of change in enthalpy involved in the transformation of one mole of water at 10.0oC{10.0^o}C to one mole of water at 0oC{0^o}C, change in enthalpy involved in the transformation of one mole of water at 0oC{0^o}C to one mole of ice at 0oC{0^o}C and change in enthalpy involved in conversion of one mole of ice at 0oC{0^o}C to one mole of ice at −10.0oC - {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 10.0oC{10.0^o}C to one mole of water at 0oC{0^o}C:
The change in enthalpy can be expressed in terms of specific heat at constant pressure and temperature as per following relation:
ΔH1=CpΔT        ...(1)\Delta {H_1} = {C_p}\Delta T\;\;\;\;...(1)
Substituting given values for H2O(l){H_2}O(l):
⇒ΔH1=75.3(0−10)\Rightarrow \Delta {H_1} = 75.3(0 - 10)
⇒ΔH1=−753Jmol−1\Rightarrow \Delta {H_1} = - 753Jmo{l^{ - 1}}
Step-2: Change in enthalpy involved in the transformation of one mole of water at 0oC{0^o}C to one mole of ice at 0oC{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 0oC{0^o}C is 6.03kJmol−16.03kJmo{l^{ - 1}}.
Thus, ΔH2=6.03×103Jmol−1\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 0oC{0^o}C to one mole of ice at −10.0oC - {10.0^o}C:
Substituting given values for H2O(s){H_2}O(s) in equation (1):
ΔH3=CpΔT\Delta {H_3} = {C_p}\Delta T
⇒ΔH3=36.8(−10−0)\Rightarrow \Delta {H_3} = 36.8( - 10 - 0)
⇒ΔH3=−368Jmol−1\Rightarrow \Delta {H_3} = - 368Jmo{l^{ - 1}}
Therefore, overall change in enthalpy of the transformation will be as follows:
ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta {H_1} + \Delta {H_2} + \Delta {H_3}
⇒ΔH=−753−6.03×103−368\Rightarrow \Delta H = - 753 - 6.03 \times {10^3} - 368
⇒ΔH=−753−6030−368\Rightarrow \Delta H = - 753 - 6030 - 368
⇒ΔH=−7151Jmol−1  or −7.151kJmol−1\Rightarrow \Delta H = - 7151Jmo{l^{ - 1}}\;{\text{or }} - 7.151kJmo{l^{ - 1}}
Hence, the change in enthalpy on freezing of 1.01.0 mol of water at 10.0oC{10.0^o}C to ice at −10.0oC - {10.0^o}C is −7.151kJmol−1 - 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.