If the melting point of ice is 0° C and latent heat of fusion is 5.46 text KJ text mol-1, the entropy change for the conversion of 18 g of ice to water at 1 atm is

Question

If the melting point of ice is 0° C and latent heat of fusion is 5.46 text KJ text mol-1, the entropy change for the conversion of 18 g of ice to water at 1 atm is (A) zero (B)  20 text JK-1  (C)  50 text JK-1  (D)  text360 JK-1

Tardigrade Admin · Accepted Answer

: Here, Δ Hf=5.46 text kJ text mol-1 Δ Sf=(Δ Hf/T)=(5.46× 1000/273) text=20 JK-1 .

Q. If the melting point of ice is $0^{\circ} C$ and latent heat of fusion is $5.46 K J m o l^{- 1},$ the entropy change for the conversion of 18 g of ice to water at 1 atm is

zero

$20 J K^{- 1}$

$50 J K^{- 1}$

$360 J K^{- 1}$

: Here, $Δ H_{f} = 5.46 k J m o l^{- 1}$ $Δ S_{f} = \frac{Δ H _{f}}{T} = \frac{5.46 \times 1000}{273} =20 J K^{- 1}$ .

Q. If the melting point of ice is 0∘C and latent heat of fusion is 5.46KJmol−1, the entropy change for the conversion of 18 g of ice to water at 1 atm is

zero

20JK−1

50JK−1

360 JK−1

: Here, ΔHf​=5.46kJmol−1 ΔSf​=TΔHf​​=2735.46×1000​=20JK−1 .

Q. If the melting point of ice is $0^{\circ} C$ and latent heat of fusion is $5.46 K J m o l^{- 1},$ the entropy change for the conversion of 18 g of ice to water at 1 atm is

$20 J K^{- 1}$

$50 J K^{- 1}$

$360 J K^{- 1}$

: Here, $Δ H_{f} = 5.46 k J m o l^{- 1}$ $Δ S_{f} = \frac{Δ H _{f}}{T} = \frac{5.46 \times 1000}{273} =20 J K^{- 1}$ .