Determine the standard enthalpy change for the precipitation reaction between aqueous solutions of silver nitrate (AgNO3) and potassium chloride (KCl), given the molar concentrations of AgNO3 and KCl are 0.10 M and 0.20 M, respectively. The balanced chemical equation for the reaction is:AgNO3(aq) + KCl(aq) -> AgCl(s) + KNO3(aq)The enthalpies of formation for AgNO3(aq), KCl(aq), AgCl(s), and KNO3(aq) are -365.6 kJ/mol, -436.0 kJ/mol, -127.0 kJ/mol, and -494.6 kJ/mol, respectively.

Question

Determine the standard enthalpy change for the precipitation reaction between aqueous solutions of silver nitrate (AgNO3) and potassium chloride (KCl), given the molar concentrations of AgNO3 and KCl are 0.10 M and 0.20 M, respectively. The balanced chemical equation for the reaction is:AgNO3(aq) + KCl(aq) -> AgCl(s) + KNO3(aq)The enthalpies of formation for AgNO3(aq), KCl(aq), AgCl(s), an

Answer 1

To determine the standard enthalpy change for the precipitation reaction, we can use the following equation:H_reaction =  H_f products  -  H_f reactants where H_reaction is the standard enthalpy change for the reaction, H_f products  are the enthalpies of formation for the products, and H_f reactants  are the enthalpies of formation for the reactants.For the given reaction:AgNO3 aq  + KCl aq  -> AgCl s  + KNO3 aq H_reaction = [H_f AgCl  + H_f KNO3 ] - [H_f AgNO3  + H_f KCl ]Using the given enthalpies of formation:H_reaction = [ -127.0 kJ/mol  +  -494.6 kJ/mol ] - [ -365.6 kJ/mol  +  -436.0 kJ/mol ]H_reaction =  -621.6 kJ/mol  -  -801.6 kJ/mol H_reaction = 180 kJ/molThe standard enthalpy change for the precipitation reaction between aqueous solutions of silver nitrate and potassium chloride is 180 kJ/mol.