Home
Class 12
CHEMISTRY
The ionization enthalpy of Na^(+) format...

The ionization enthalpy of `Na^(+)` formation from `Na_((g))` is `495.8kJmol^(-1)`, while the electron gain enthalpy of `Br` is `-325.0kJmol^(-1)`. Given the lattice enthalpy of `NaBr` is `-728.4kJmol^(-1)`. The energy for the formation of `NaBr` ionic solid is (-)______`xx10^(-1) kJ mol^(-1)`.

Promotional Banner

Similar Questions

Explore conceptually related problems

Ionisation of energy F^(ɵ) is 320 kJ mol^(-1) . The electronic gain enthalpy of fluorine would be

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

For an exothermic reaction ArarrB , the activation energy is 65kJmol^(-1) and enthalpy of reaction is 42kJmol^(-1) . The activation energy for the reaction BrarrA will be:

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)) .

The enthalpy of neutralisation of a strong acid by a strong base is -57.32 kJ mol^(-1) . The enthalpy of formation of water is -285.84 kJ mol^(-1) . The enthalpy of formation of hydroxyl ion is

The enthalpy of neutralisation of a strong acid by a string base is -57.32 kJ mol^(-1) . The enthalpy of formation of water is -285.84 kJ mol^(-1) . The enthalpy of formation of hydroxyl ion is

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 second electron gain enthalpies (in kJ mol^(-1) ) of oxygen and sulphur respectively are:

The second electron gain enthalpies (in kJ mol^(-1) ) of oxygen and sulphur respectively are:

The second electron gain enthalpies (in kJ mol^(-1) ) of oxygen and sulphur respectively are: