Question

Given the enthalpy of solution of sodium hydroxide (NaOH) is -44.5 kJ/mol and the enthalpy of solution of hydrochloric acid (HCl) is -74.8 kJ/mol, calculate the enthalpy change for the precipitation reaction between NaOH and HCl, assuming all solutions are at a concentration of 1 M.

Answer 1

To calculate the enthalpy change for the precipitation reaction between NaOH and HCl, we first need to write the balanced chemical equation for the reaction:NaOH aq  + HCl aq   NaCl aq  + H2O l Now, we can use Hess's Law to determine the enthalpy change for this reaction. Hess's Law states that the enthalpy change for a reaction is the same whether it occurs in one step or several steps. In this case, we can break the reaction into two steps:1. Dissolving NaOH and HCl in water:NaOH s   NaOH aq     H1 = -44.5 kJ/molHCl g   HCl aq       H2 = -74.8 kJ/mol2. Precipitation reaction between NaOH aq  and HCl aq :NaOH aq  + HCl aq   NaCl aq  + H2O l     H3 = ?We can now use Hess's Law to find H3:H3 = H1 + H2H3 = -44.5 kJ/mol +  -74.8 kJ/mol H3 = -119.3 kJ/molThe enthalpy change for the precipitation reaction between NaOH and HCl is -119.3 kJ/mol.