Question

The enthalpy change on freezing of 1 mol of water at $5^{\circ} C$ to ice at $-5^{\circ}C$ is :
(Given $\Delta_{\text{fus}} H = 6 \, kJ \, mol^{-1} \, at \,0^{\circ}C$,
$C_p(H_2O, l) = 75.3 \, J \, mol^{-1} \, K^{-1}$,
$C_p(H_2O, s) = 36.8 \, J \, mol^{-1} \, K^{-1})$

• A. $5.44 \, kJ \, mol^{-1}$
• B. $5.46 \, kJ \, mol^{-1}$
• C. $5.81 \, kJ \, mol^{-1}$
• D. $6.56 \, kJ \, mol^{-1}$
• E. $6.00 \, kJ \, mol^{-1}$

Answer 1

Answer: (C)

(1) Water at $5^{\circ} C$ is cooled to water at $0^{\circ} C$
The enthalpy change $= nC _{ p }\left( H _{2} O , 1\right) \Delta_{ T }=1 \times 75.3 J / mol / K \times(5-0)^{\circ} C$
$=376.5\, J /=0.3765\, kJ \ldots \ldots(1)$
Here, $n$ is the number of moles of water, $C _{ p }\left( H _{2} O , l \right)$ is the specific heat of liquid water and $\Delta T$ is temperature change.
Also $1\, kJ =1000\, J$
(2) Liquid water at $0^{\circ} C$ is fused at same temperature.
The enthalpy change $= n \Delta_{ fus } H =1 \times 6\, kJ / mol =6\, kJ \ldots \ldots(2)$
(3) Ice at $0^{\circ} C$ is cooled to ice at $-5^{\circ} C$
The enthalpy change $= nC _{ p }\left( H _{2} O , s \right) \Delta_{ T }$
$=1 \times 36.8\, J / mol / K \times(0-(-5))^{\circ} C$
$=184\, J /=0.184\, kJ \ldots \ldots(3)$
Here, $n$ is the number of moles of ice, $C _{ p }\left( H _{2} O , 1\right)$ is the specific heat of ice and $\Delta T$ is temperature change. Also $1\, kJ =1000\, J$
Add (1), (2) and (3)
The enthalpy change for entire process
$=0.3765\, kJ +6\, kJ +0.184\, kJ =6.56\, kJ\, mol ^{-1}$