Home
Class 11
CHEMISTRY
Calcualte the enthalpy change on freezin...

Calcualte the enthalpy change on freezing of 1.0 mole of water at `10.0^(@)C` to ice at `-10^(@)`C. `Delta_(fs)H=6.03 kJ mol^(-1)` at `0^(@)C`.
`C_(p)[H_(2)O(l)] = 75.3 J mol^(-1) K^(-1), C_(P)[H_(2)O(s)] = 36.8 Jmol^(-1)K^(-1)`

Text Solution

Verified by Experts

The enthalpy change on freezing 1.0 mol of water from `10^(@)C` to `-10^(@)C` may be expressed as :
1 mol water `(10^(@)C) overset(DeltaH_(1))to 1` mol water `(0^(@)C) overset(DeltaH_(2)) to 1` mol ice `(0^(@)C) overset(DeltaH_(3)) to` 1 mol ice `(-10^(@)C)`
`DeltaH_(1) = nC_(p)[H_(2)O(l)] xx DeltaT = 1 xx 75.3 J "mol"^(-1) K^(-1) xx (0 - 10)K`
` = -753 J "mol"^(-1) = -0.753 kJ "mol"^(-1)`
`DeltaH_(2) = n(Delta_(freezing)H) = n(-Delta_(fus)H)(because Delta_(freezing)H = -Delta_(fus)H)`
`= 1 xx (-6.03) = -6.03 kJ "mol"^(-1)`
`DeltaH_(3) = nC_(p)[H_(2)O(s)] DeltaT = 1 xx 36.8 J "mol"^(-1) K^(-1) xx (-10-0)K`
` = -368 J "mol"^(-1) = -0.368 kJ "mol"^(-1)`
`DeltaH = DeltaH_(1) + DeltaH_(2) + DeltaH_(3)`
` = -0.753 kJ "mol"^(-1) - 6.03 kJ "mol"^(-1) - 0.368 kJ "mol"^(-1)`
` = -7.151 kJ "mol"^(-1)`.
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    MODERN PUBLICATION|Exercise Conceptual Questions -1|19 Videos

  • THERMODYNAMICS

    MODERN PUBLICATION|Exercise Conceptual Questions - 2|17 Videos

  • THERMODYNAMICS

    MODERN PUBLICATION|Exercise Practice Problems|65 Videos

  • STRUCTURE OF ATOM

    MODERN PUBLICATION|Exercise Unit Practice Test|13 Videos

Similar Questions

Explore conceptually related problems

The enthalpy change of freezing of 1 mol of water at 5^(@)C to ice at -5^(@)C is (Given Delta_(fus)H=6kJ"mol"^(-1) at 0^(@)C , (C_(p)(H_(2)O,s)=36.8J"mol"^(-1)K^(-1))

Calculate the enthalpy change on freezing one mode of water at 10^(@)C to ice at temperature -10^(0)C , Delta fus H=6.03KJmol^(-1) At 0^(0) , Cp H_(2)O_((l)) = 75.3Jmol^(-1)K^(-1) and Cp H_(2)O_((s)) =36.8Jmol

What is the enthalpy change when 1.0 g of water is frozen at 0^(@)C ? Delta_(fus) = H = 6.02 kJ "mol"^(-1) .

Calculate the change in enthalpy for the following process at 1 atm H_(2)O(1, 50^(@)C) rarr H_(2)O (g, 150^(@)C) given that Delta H_(v) at 100^(@)C is 40.7 kJ mol^(-1) C_(p)(H_(2)O, l) = 75.0 J mol^(-1)K^(-1) C_(p)(H_(2)O, g) = 33.3 J mol^(-1)K^(-1)

The enthalpy change when 1 g of water is frozen at 0^@C ( Delta_(fus) H^@=-1.435 kcal/mol) is __________ .

The enthalpy change for the formation of 3.6 kg water is _____ . H_(2(g))+1/2 O_(2(g)) to H_2 O_((l)), Delta H=-284.5 kJ mol^-1

Predict the standard reaction enthalpy of 2NO_(2)(g)rarrN_(2)O_(4)(g) at 100^(@)C . Delta H^(@) at 25^(@)C is -57.2 kj.mol^(-1)C_(p)(NO_(2))=37.2j.mol^(-1)K^(-1)C_(p)(N_(2)O_(4))=77.28 J. mol^(-1)k^(-1)

Knowledge Check

  • Calculate the enthalpy change on freezing of 1.0 mole of water at 10.0^(@)C to ice at -10.0^(@)C [Delta_("fus")H=6.03kJmol^(-1) at 0^(@)C] C_(p)[H_(2)O(l)]=75.3Jmol^(-1)K^(-1) C_(p)[H_(2)O(s)]=36.8Jmol^(-1)K^(-1)

    A
    `-753Jmol^(-1)`
    B
    `-368Jmol^(-1)`
    C
    `-7.151kJmol^(-1)`
    D
    `-6.03kJmol^(-1)`

  • The enthalpy change of freezing of 1 mol of water at 5^(@)C to ice at -5^(@)C is (Given Delta_(fus)H=6kJ"mol"^(-1) at 0^(@)C , (C_(p)(H_(2)O,s)=36.8J"mol"^(-1)K^(-1))

    A
    `5.44kJ"mol"^(-1)`
    B
    `5.81kJ"mol"^(-1)`
    C
    `6.56kJ"mol"^(-1)`
    D
    `6.00kJ"mol"^(-1)`

  • Calculate the change in enthalpy for the following process at 1 atm H_(2)O(1, 50^(@)C) rarr H_(2)O (g, 150^(@)C) given that Delta H_(v) at 100^(@)C is 40.7 kJ mol^(-1) C_(p)(H_(2)O, l) = 75.0 J mol^(-1)K^(-1) C_(p)(H_(2)O, g) = 33.3 J mol^(-1)K^(-1)

    A
    64.8 KJ
    B
    52.4 KJ
    C
    48.6 KJ
    D
    46.1 KJ