Question

The enthalpy changes on freezing of 1 mol of water at ${{5}^{\circ }}C$ to ice at $-{{5}^{\circ }}C$ is:
(Given ${{\vartriangle }_{fus}}H=6KJmo{{l}^{-1}}$ at ${{0}^{\circ }}C$
${{C}_{P}}({{H}_{2}}O,l)=75.3Jmo{{l}^{-1}}{{K}^{-1}}$
${{C}_{P}}({{H}_{2}}O,S)=36.8Jmo{{l}^{-1}}{{K}^{-1}}$)
a.) $5.81KJmo{{l}^{-1}}$
b.) $6.56KJmo{{l}^{-1}}$
c.) $6.00KJmo{{l}^{-1}}$
d.) $5.44KJmo{{l}^{-1}}$

Answer 1

Hint: The enthalpy of water is changing from ${{5}^{\circ }}C$ to $-{{5}^{\circ }}C$. So, enthalpy would be changing three times. First time would be when water is cooled from to ${{0}^{\circ }}C$, second time would be fusion of liquid at ${{0}^{\circ }}C$ at the same temperature and third time would be by changing the temperature of ice from ${{0}^{\circ }}C$ to $-{{5}^{\circ }}C$.
The correct answer is B.

Step by Step answer:
Step 1 would be to bring down the temperature that cooling down of water from ${{5}^{\circ }}C$ to ${{0}^{\circ }}C$
So, enthalpy change would be equal to product of number of moles into ${{C}_{P}}({{H}_{2}}O,l)$ which is then multiplied by change in temperature.
Therefore,
$enthalpy\text{ }change~=n\times Cp(H_2O,l)\times \Delta T=1\times 75.3J/mol/K\times {{\left( 5-0 \right)}^{\circ }}C=367.5J=0.3765kJ......\left( 1 \right)$
Here, n represents the number of moles of water which are given as 1 , Cp​(H2​O,l) is the specific heat of liquid water which is given in question and ΔT is temperature change and temperature is brought down from ${{5}^{\circ }}C$to ${{0}^{\circ }}C$

Also 1kJ=1000J

Step 2 is Liquid water at ${{0}^{\circ }}C$ is fused at the same temperature.
So the enthalpy change would be equal to the number of moles multiplied by fusion heat.

$The\text{ }enthalpy\text{ }change~=n\Delta fusH=1\times 6kJ/mol=6kJ......\left( 2 \right)$

Step 3 would be Ice at ${{0}^{\circ }}C$ is cooled down to ice at $-{{5}^{\circ }}C$
So, the enthalpy change would be equal to the number of moles multiplied by specific heat which in turn is multiplied by change in temperature.

$The\text{ }enthalpy\text{ }change~=nCp(H_2O, s)\Delta T=1\times 36.8J/mol/K\times {{\left( 0-\left( -5 \right) \right)}^{\circ }}C=184J=0.184kJ......\left( 3 \right)$

where, n stands the number of moles of ice which is 1, Cp​($H_2​O$, l) is the specific heat of ice which is given in question and ΔT is temperature change which is equal to 5. Also, 1kJ=1000J

Add (1), (2) and (3)
Total enthalpy change would be equal to the sum of all the three changes in enthalpy happening during the process.

$The\text{ }enthalpy\text{ }change\text{ }for\text{ }entire\text{ }process~=0.3765kJ+6kJ+0.184kJ=6.56kJmo{{l}^{-1}}$

The enthalpy change for the entire process would be equal to 6.56kJ/mol which is option B.
Note: Enthalpy change is the amount of the heat released or absorbed in the reactions at constant pressure.