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Calculate the change in entropy for the fusion of `1 mol` of ice. The melting point of ice is `300K` and molar enthalpy of fustion for ice `= 6.0 k J mol^(-1)`.

Text Solution

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For entropy change of fustion, `Delta_(f)S^(Theta) = (Delta_(f)H^(Theta))/(T)`
Given, `Delta_(f)H^(Theta) = 6 xx 10^(3) J, T = 300 K`
`:. Delta_(f)S^(Theta) = (6 xx 10^(3))/(300) = 20.0 J K^(-1) mol^(-1)`
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(i) Calculate the change in entropy for the fusion of 1 mole of ice. The melting point of ice is 273 K and molar enthalpy of fusion for ice = 6.0 kJ/mole. (ii) Calculate the standard of entropy change for the following reaction. {:(,H^(+)(aq), + OH(aq)to, H_(2)O(l)),(S^(@)(298 K)//JK^(-1)mol^(-1),0,-10.7,+70):}

Calculate the change in entropy for the fusion of 1 mol of ice. The melting point of ice is 273 K and molar enthalpy of fusion of ice = 6KJmol^(-1)

Knowledge Check

  • The change is entropy for the fusion of 1 mole of ice is [M.P of ice = 273 K molar, enthalpy of fusion for ice = 60 kJ mol^(-1) ]

    A
    `11.73 JK^(-1) mol^(-1)`
    B
    `18.84 JK^(-1) mol^(-1)`
    C
    `219.7 JK^(-1) mol^(-1)`
    D
    `24.47 JK^(-1) mol^(-1)`

  • The entropy change for the fusion of 1 mol of a solid which melts at 27^(@)C is (latent heat of fusion =600 kcal mol^(-1) )

    A
    `2 cal deg^(-1)`
    B
    `22.2 cal deg^(-1)`
    C
    `180 kcal deg`
    D
    `0.5 cal^(-1)deg`

  • How many grams of ice at 0^(@)C can be melted by the addition of 500 J of heat (The molar heat of fusion for ice is 6.02 kJ mol^(-1) )

    A
    0.0831 g
    B
    1.50 g
    C
    3.01 g
    D
    12.0 g

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