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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)`.

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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)

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).

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

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

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) .

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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