Q1: Heat of neutralisation of HCl and NaOH is - 57.46 kJ/Kg mole. The heat of ionisation of water will be __________ kJ/Kg mole.

ANS:A - 57.46

The heat of neutralization is the heat evolved or absorbed when an acid and a base react to form one mole of water. In this case, the heat of neutralization between HCl and NaOH is given as -57.46 kJ/mol. The reaction for the neutralization of HCl and NaOH is: HCl+NaOH→NaCl+H2OHCl+NaOH→NaCl+H2​O Given that the heat of neutralization is -57.46 kJ/mol, this means that when 1 mole of HCl reacts with 1 mole of NaOH, -57.46 kJ of heat is evolved. Since water (𝐻2𝑂H2​O) is formed as a product, the heat of ionization of water (𝐻2𝑂H2​O) can be determined by looking at the enthalpy change for the reverse process of water formation, which is the ionization of water (𝐻2𝑂→H++OH−H2​O→H++OH−). The reaction for the ionization of water is: H2O→H++OH−H2​O→H++OH− From Hess's law, the heat of ionization of water is the negative of the heat of neutralization: Heat of ionization of water=−(−57.46 kJ/mol)=57.46 kJ/molHeat of ionization of water=−(−57.46 kJ/mol)=57.46 kJ/mol So, the correct answer is 57.46 kJ/mol. Therefore, the answer is 57.4657.46