The enthalpy change on freezing of 1 mol of water at 5° C to ice at -5°C is: (Given Δ textfus H = 6 kJ mol-1 at 0°C, Cp(H2O, l) = 75.3 J mol-1 K-1, Cp(H2O, s) = 36.8 J mol-1 K-1)

Question

The enthalpy change on freezing of 1 mol of water at 5° C to ice at -5°C is: (Given Δ textfus H = 6 kJ mol-1 at 0°C, Cp(H2O, l) = 75.3 J mol-1 K-1, Cp(H2O, 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

Tardigrade Admin · Accepted Answer

(1) Water at 5° C is cooled to water at 0° C The enthalpy change = nC p ( H 2 O , 1) Δ T =1 × 75.3 J / mol / K ×(5-0)° C =376.5 J /=0.3765 kJ ldots ldots(1) Here, n is the number of moles of water, C p ( H 2 O , l ) is the specific heat of liquid water and Δ T is temperature change. Also 1 kJ =1000 J (2) Liquid water at 0° C is fused at same temperature. The enthalpy change = n Δ fus H =1 × 6 kJ / mol =6 kJ ldots ldots(2) (3) Ice at 0° C is cooled to ice at -5° C The enthalpy change = nC p ( H 2 O , s ) Δ T =1 × 36.8 J / mol / K ×(0-(-5))° C =184 J /=0.184 kJ ldots ldots(3) Here, n is the number of moles of ice, C p ( H 2 O , 1) is the specific heat of ice and Δ 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

Q. The enthalpy change on freezing of 1 mol of water at $5^{\circ} C$ to ice at $- 5^{\circ} C$ is :
(Given $Δ_{fus} H = 6 k J m o l^{- 1} a t 0^{\circ} C$ ,
$C_{p} (H_{2} O, l) = 75.3 J m o l^{- 1} K^{- 1}$ ,
$C_{p} (H_{2} O, s) = 36.8 J m o l^{- 1} K^{- 1})$

$5.44 k J m o l^{- 1}$

$5.46 k J m o l^{- 1}$

$5.81 k J m o l^{- 1}$

$6.56 k J m o l^{- 1}$

$6.00 k J m o l^{- 1}$

Q. The enthalpy change on freezing of 1 mol of water at 5∘C to ice at −5∘C is : (Given Δfus​H=6kJmol−1at0∘C, Cp​(H2​O,l)=75.3Jmol−1K−1, Cp​(H2​O,s)=36.8Jmol−1K−1)

5.44kJmol−1

5.46kJmol−1

5.81kJmol−1

6.56kJmol−1