Enthalpy change for the process, H2O(ice) leftharpoons H2O(water) is 6.01 kJ mol-1. The entropy change of 1 mole of ice at its melting point will be

Question

Enthalpy change for the process, H2O(ice) leftharpoons H2O(water) is 6.01 kJ mol-1. The entropy change of 1 mole of ice at its melting point will be (A) 12 J K-1 mol-1 (B) 22 J K-1 mol-1 (C) 100 J K-1 mol-1 (D) 30 J K-1 mol-1

Tardigrade Admin · Accepted Answer

Δ Sfusion = (Δ Hfusion/Tfusion) = (6.01 × 1000/273) = 22 J K-1 mol-1

Q. Enthalpy change for the process,
$H_{2} O$ (ice) $⇌ H_{2} O$ (water)
is $6.01 k J m o l^{- 1}$ . The entropy change of $1$ mole of ice at its melting point will be

$12 J K^{- 1} m o l^{- 1}$

$22 J K^{- 1} m o l^{- 1}$

$100 J K^{- 1} m o l^{- 1}$

$30 J K^{- 1} m o l^{- 1}$

$Δ S_{f u s i o n} = \frac{Δ H _{f u s i o n}}{T _{f u s i o n}}$
$= \frac{6.01 \times 1000}{273}$
$= 22 J K^{- 1} m o l^{- 1}$

Q. Enthalpy change for the process, H2​O(ice) ⇌H2​O(water) is 6.01kJmol−1. The entropy change of 1 mole of ice at its melting point will be

12JK−1mol−1

22JK−1mol−1

100JK−1mol−1

30JK−1mol−1