Calculate the entropy change in melting 1 mole of ice at 273K, Δ H°f=6.025 kJ/mole-

Question

Calculate the entropy change in melting 1 mole of ice at 273K, Δ H°f=6.025 kJ/mole- (A) 11.2 JK-1 mol-1 (B) 22.1 JK-1 mol-1 (C) 15.1 JK-1 mol-1 (D) 5.1 JK-1 mol-1

Tardigrade Admin · Accepted Answer

Δ Sr=(Δ Hf/T)=(6025 J mol-1/273K)=22.1 JK-1 mol-1

Q. Calculate the entropy change in melting $1$ mole of ice at $273 K$ , $Δ H_{f}^{\circ} = 6.025 k J / m o l e -$

$11.2 J K^{- 1} m o l^{- 1}$

$22.1 J K^{- 1} m o l^{- 1}$

$15.1 J K^{- 1} m o l^{- 1}$

$5.1 J K^{- 1} m o l^{- 1}$

$Δ S_{r} = \frac{Δ H _{f}}{T} = \frac{6025 J m o l ^{- 1}}{273 K} = 22.1 J K^{- 1} m o l^{- 1}$

Q. Calculate the entropy change in melting 1 mole of ice at 273K, ΔHf∘​=6.025kJ/mole−

11.2JK−1mol−1

22.1JK−1mol−1

15.1JK−1mol−1

5.1JK−1mol−1