The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is :

Question

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is : (A) 5.260cal /(molK) (B) 0.526cal /(molK) (C) 10.52 cal /(molK) (D) 21.04 cal /(molK)

Tardigrade Admin · Accepted Answer

Δ S =(Δ H / T )=(1.435 × 1000/273) =5.26 ( Cal / mol × k )

Q. The enthalpy of fusion of water is $1.435 k c a l / m o l .$ The molar entropy change for the melting of ice at $0^{\circ} C$ is :

5.260cal /(molK)

0.526cal /(molK)

10.52 cal /(molK)

21.04 cal /(molK)

$Δ S = \frac{Δ H}{T} = \frac{1.435 \times 1000}{273}$

$= 5.26 \frac{C a l}{m o l \times k}$

Q. The enthalpy of fusion of water is 1.435kcal/mol. The molar entropy change for the melting of ice at 0∘C is :