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Calculate the free energy change when one mole of sodium chloride is dissolved in water at 298 K. (Given : Lattice energy of NaCl `= - 777.8 kJ mol^(-1)`, Hydration energy of NaCl ` = 774.1 kJ mol^(-1) and DeltaS " at" 298 K = 0.043 kJ mol^(-1)`.

Text Solution

AI Generated Solution

To calculate the free energy change (ΔG) when one mole of sodium chloride (NaCl) is dissolved in water at 298 K, we will use the following thermodynamic relationship: \[ \Delta G = \Delta H - T \Delta S \] Where: - ΔG = free energy change ...
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Calculate the free enegry change when 1mol of NaCI is dissolved in water at 298K . Given: a. Lattice enegry of NaCI =- 778 kJ mol^(-1) b. Hydration energy of NaCI - 774.1 kJ mol^(-1) c. Entropy change at 298 K = 43 J mol^(-1)

Calculate the magnitude of free energy in KJ mol^(-1) when 1 mole of a an ionic salt MX (s) is dissolved in water at 27^(@)C . Given Lattice energy of MX = 780 KJ mol^(-1) Hydration energy of MX =-775.0 KJ mol^(-1) Entropy change of dissolution at 27^(@)C = 40Jmol^(-1)K^(-1)

Knowledge Check

  • The lattice energy of NaCl is 788 kJ mol^(-1) . This means that 788 kJ of energy is required

    A
    to separate one mole of solid NaCl into one mole of `Na_((g))` and one mole of `Cl_((g))` to infinite distance
    B
    to separate one mole of solid NaCl into one mole of `Na_((g))^(+)` and one mole of `Cl_((g))^(-)` to infinite distance
    C
    to convert one mole of solid NaCl into one mole of gaseous NaCl
    D
    to convert one mole of gaseous NaCl into one mole of solid NaCl.

  • The enthalpy of solution of sodium chloride is 4 kJ mol^(-1) and its enthalpy of hydration of ion is -784 kJ mol^(-1) . Then the lattice enthalpy of NaCl (in kJ mol^(-1) ) is

    A
    a. `+"780 kJ mol"^(-1)`
    B
    b. `+"394 kJ mol"^(-1)`
    C
    c. `+"788 kJ mol"^(-1)`
    D
    d. `+"398 kJ mol"^(-1)`

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    Calculated the Gibbs energy change on dissolving one mole of sodium chloride at 25^(@)C . Lattice =+ 777.0 kJ mol^(-1) Hydration of NaCI =- 774.0 kJ mol^(-1) DeltaS at 25^(@)C = 40 J K^(-1) mol^(-1)

    Lattice energy of NaCl_((s)) is -788kJ mol^(-1) and enthalpy of hydration is -784kJ mol^(-1) . Calculate the heat of solution of NaCl_((s)) .

    Calculate heat of solution of NaCI form the following data: Hydration energy of Na^(o+)= - 389 kJ mol^(-1) Hydration energy of CI^(Θ)= - 382 kJ mol^(-1) Lattic energy of NaCI= - 776 kJ mol^(-1)

    Lattice energy of NaCl(s) is -790 kJ " mol"^(-1) and enthalpy of hydration is -785 kJ " mol"^(-1) . Calculate enthalpy of solution of NaCl(s).

    Find the enthalpy of formation of hydrogen flouride on the basis of following data: Bond energy of H-H bond =434 kJ mol^(-1) Bond energy of F-F bond =158 kJ mol^(-1) Bond enegry of H -F bond =565 kJ mol^(-1)

    Calculate the heat of formation of NaCl from the following data: Heat of sublimation of sodium = 108.5 kJ mol^(-1) Dissociation energy of chlorine = 243.0 kJ mol^(-1) lonisation energy of sodium = 495.8 kJ mol^(-1) Electron gain enthalpy of chlorine = -348.8 kJ mol^(-1) Lattice energy of sodium chloride = -758.7 kJ mol^(-1) .

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