If heat of neutralisation is `-13.7kcal` at `25^(@)C` and `H_(f(H_(2)O))^(@)=-68kcal`, then standard enthalpy of `OH^(-)` would be `:`

If heat of neutralisation is -13.7 k cal and H_f^(@)H_2O =-68 k cal , then enthalpy of OH^- would be:

Statement-I : The enthalpy of neutralization of the reaction between HCl and NaOH is -13.7 kCal/mol. If the enthalpy of neutralization of oxalic acid (H_(2)C_(2)O_(4)) by a strong base is -25.4 kCal/mol, then the enthalpy change (|Delta_(r)H|) of the process H_(2)C_(2)O_(4) rarr 2H^(+)+C_(2)O_(4)^(2-) is 11.7 kCal/mol. Statement-II : H_(2)C_(2)O_(4) is a weak acid.

The heat of formation of H_(2)O(l) is -68.0 kcal, the heat of formation of H_(2)O(g) can logically be

The heat of formation of H_(2)O(l) is -68.0 kcal, the heat of formation of H_(2)O(g) can logically be

heat of neutralisation of HCl against NaOH is 13.7 kcal eq^(-1) what will be the ionisation energy of CH_(3)COOH in kcal mol^(-1) if its heat of neutralisation is 11.7 kcal eq^(-1) .