Question

Calculate the standard enthalpy change for the precipitation reaction: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)given that the standard enthalpy of formation for AgCl(s) is -127.0 kJ/mol, and for NaNO3(aq) is -467.14 kJ/mol. The molar enthalpy of solution of NaCl and AgNO3 are -3.9 kJ/mol and -19.9 kJ/mol respectively.

Answer 1

To calculate the standard enthalpy change for the precipitation reaction, we can use the following equation:H_reaction =  H_f products  -  H_f reactants First, let's find the enthalpy of formation for the reactants:H_f AgNO3 aq   = H_f AgNO3 s   + H_sol AgNO3 H_f NaCl aq   = H_f NaCl s   + H_sol NaCl Since the enthalpy of formation for AgNO3 s  and NaCl s  are not given, we can assume they are zero because they are in their standard states. Therefore:H_f AgNO3 aq   = 0 +  -19.9 kJ/mol  = -19.9 kJ/molH_f NaCl aq   = 0 +  -3.9 kJ/mol  = -3.9 kJ/molNow, we can plug these values into the equation:H_reaction = [ -127.0 kJ/mol  +  -467.14 kJ/mol ] - [ -19.9 kJ/mol  +  -3.9 kJ/mol ]H_reaction =  -594.14 kJ/mol  -  -23.8 kJ/mol H_reaction = -570.34 kJ/molThe standard enthalpy change for the precipitation reaction is -570.34 kJ/mol.