The change in entropy for the fusion of 1 mole of ice is [melting point of ice = 273 K, molar enthalpy of fusion for ice =6.0 kJ mol-1 ]:

Question

The change in entropy for the fusion of 1 mole of ice is [melting point of ice = 273 K, molar enthalpy of fusion for ice =6.0 kJ mol-1 ]: (A)  11.73 JK-1 mol-1  (B)  18.84 JK-1mol-1  (C)  21.97JK-1mol-1  (D)  24.47 JK-1 mol-1

Tardigrade Admin · Accepted Answer

Entropy change of fusion Δ Sf Δ Sf=(Δ Hf/T) Δ Hf=6.0 × 103 J, T=273 K Δ Sf=(6000/273) =21.97 J / K mol

Q. The change in entropy for the fusion of $1$ mole of ice is [melting point of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 k J m o l^{- 1}$ ]:

$11.73 J K^{- 1} m o l^{- 1}$

$18.84 J K^{- 1} m o l^{- 1}$

$21.97 J K^{- 1} m o l^{- 1}$

$24.47 J K^{- 1} m o l^{- 1}$

Entropy change of fusion $Δ S_{f}$
$Δ S_{f} = \frac{Δ H _{f}}{T}$
$Δ H_{f} = 6.0 \times 1 0^{3} J$ ,
$T = 273 K$
$Δ S_{f} = \frac{6000}{273}$
$= 21.97 J / K m o l$

Q. The change in entropy for the fusion of 1 mole of ice is [melting point of ice =273K, molar enthalpy of fusion for ice =6.0kJmol−1 ]:

11.73JK−1mol−1

18.84JK−1mol−1

21.97JK−1mol−1

24.47JK−1mol−1