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Prove that the formation of NaCI(2) is h...

Prove that the formation of `NaCI_(2)` is highly endothermic and energetically unfavorable from the following data using Born-Haber Cycle
`{:("Heat of sublimation of"Na,=,109kJ//mol),(1^(st)IE of Na,=,494 kJ//mol),(1^(st)IE of Na^(+),=,4563 kJ//mol),("Bond Energy of" CI_(2),=,242 kJ//mol),("Electron gain enthalpy of"CI,=,-347 kJ//mol),("Lattice energy of" NaCI_(2),=,2560 kJ//mol):}`

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Calculate the lattice enthalpy of KCl from the following data by Born- Haber's Cycle. Enthalpy of sublimation of K = 89 kJ mol^(–1) Enthalpy of dissociation of Cl = 244 kJ mol^(–1) Ionization enthalpy of potassium = 425 kJ mol^(–1) Electron gain enthalpy of chlorine = –355 kJ mol^(–1) Enthalpy of formation of KCl = –438 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) .

Calculate the lattice energy of formation of NaCl from the following data : Na_((s)) +1//2Cl_(2(g)) to NaCl_((s)) Delta H_(f) -411.3 KJ.mol^(-1) Heat of sublimation of Na_((s)) = 108 . 7 kJ mol ^(-1) Ionisation energy of Na_((g)) = 49.5.0 kJ mol^(-1) Dissociation energy of Cl_(2(g)) = 244 kJ mol^(-1) Electron affinity of Cl_((g)) = - 349 . 0 kJ mol^(-1)

Give the following information, calculate the lattice enthalpy of the sodium chloride lattice: Delta ("atomisation Na") = +107 kJ mol^(-1) " " DeltaH (1^(st) "ionisation Na") = + 496 kJ mol^(-1) Delta H ("bond dissociation " CI_(2)) = +242 kJ mol^(-1) " " Delta (1^(st) "electron affinity CI") =-349 kJ mol^(-1) DeltaH( NaCI) =-411 kJ mol^(-1)

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)

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)

Calculate heat of solution of NaCl from the following data Hydration energy of Na^(+) = -389kJ mol^(-1) Hydration energy of Cl^(-) = -382kJ mol^(-1) Lattice energy of NaCl = +776 kJ mol^(-)