At 0- C- ice and water are in equilibrium and enthalpy change for the process H 2 O s- H 2 O 1 is 6-0 kJ mol -1- The entropy change for the conversion of ice into liquid water isA- 1638 kJ K -1 mol -1B- 45-5 k J K-1 mol -1C- 21-98 kJ K -1 mol -1D- None

The correct option is C H_2O_(s)fusion⇌ H_2O_(l); Δ S_fus=6.00kJ mol^ 1=6000J mol^ 1 T_f=0^∘ C=(0+273)K=273K Δ S_fus=Δ H_fus/T_f=6.00 kJ mol^ 1/273K=(600J m ...

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Standard XII

Chemistry

Heat Capacity at Constant Pressure

At 0∘ C, ice ...

Question

At $0^{\circ} C$ , ice and water are in equilibrium and enthalpy change for the process $H_{2} O_{(s)} ⇌ H_{2} O_{(l)} i s 6.0 k J m o l^{- 1}$ . The entropy change for the conversion of ice into liquid water is

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Solution

The correct option is C
$H_{2} O_{(s)} f u s i o n ⇌ H_{2} O_{(l)}; Δ S_{f u s} = 6.00 k J m o l^{- 1} = 6000 J m o l^{- 1} T_{f} = 0^{\circ} C = (0 + 273) K = 273 K Δ S_{f u s} = \frac{Δ H_{f u s}}{T_{f}} = \frac{6.00 k J m o l^{- 1}}{273 K} = \frac{(600 J m o l^{- 1})}{(273 K)} = 21.98 k J K^{- 1} m o l^{- 1}$

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At 0∘C, ice and water are in equilibrium and enthalpy change for the process H2O(s)⇌H2O(l)is6.0kJmol−1. The entropy change for the conversion of ice into liquid water is

The correct option is C H2O(s)fusion⇌H2O(l);ΔSfus=6.00kJmol−1=6000Jmol−1Tf=0∘C=(0+273)K=273KΔSfus=ΔHfusTf=6.00kJmol−1273K=(600Jmol−1)(273K)=21.98kJK−1mol−1

At $0^{\circ} C$ , ice and water are in equilibrium and enthalpy change for the process $H_{2} O_{(s)} ⇌ H_{2} O_{(l)} i s 6.0 k J m o l^{- 1}$ . The entropy change for the conversion of ice into liquid water is