For `NaCl`, lattice energy ` = - 186 kcal//mol`, the solvation energy of `Na^(+) and Cl` are `-97 and -85 kcal//"mole"` respectively. Therefore for `NaCl (s)`

If heat of formation of C Cl_4 is 316 kcal mol^-1 the dissociation energy of C-Cl is :

Calculate the enthalpy of solution of sodium chloride. The lattice energy of NaCl is +186 kcal mol^(-1) . And the solution enthalpies of cations and anions are respectively -97 and -85 kcal mol^(-1)

Calculate the lattice energy of NaCI crystal from the following data by the use of Born-Haber cycle. Sublimation energy of Na = 26 kcal//g atom, dissociation energy of CI_(2) = 54 kcal//"mole" , ionisation energy of Na(g) = 117 kcal//"mol" , electron affinity for CI(g) = 84 kcal//g atom, heat of formation of NaCI =- 99 kcal//"mole" .

If heat of solution for AB(s) is -0.95xx10^(x) kcal/mol and lattice energy for AB(s) is 700 kcal/mol and hydration energy for A^(+)(g) is -1000 kcal/mol and B^(-)(g) is -650 kcal/mol then calculate value of x.