To calculate the standard enthalpy change of the precipitation reaction, we need to find the enthalpy of formation of the products and subtract the enthalpy of formation of the reactants. First, we need to find the moles of Pb NO3 2 and NaCl in the reaction.Moles of Pb NO3 2 = 0.100 mol/L * 0.050 L = 0.005 molSince there is an excess of NaCl, we don't need to calculate its moles. Now, we need to find the enthalpy of formation of PbCl2 s . We can do this using the given enthalpy of formation values for NaNO3 aq and NaCl aq and the balanced equation:Pb NO3 2 aq + 2NaCl aq PbCl2 s + 2NaNO3 aq H = Hf products - Hf reactants We know the enthalpy of formation of NaNO3 aq and NaCl aq , but we need to find the enthalpy of formation of Pb NO3 2 aq and PbCl2 s . We can use the following equations:1 Pb NO3 2 s Pb2+ aq + 2NO3- aq ; Hf12 PbCl2 s Pb2+ aq + 2Cl- aq ; Hf23 NaCl aq Na+ aq + Cl- aq ; Hf3 = -411.2 kJ/mol4 NaNO3 aq Na+ aq + NO3- aq ; Hf4 = -467.7 kJ/molNow, we can write the balanced equation using these equations:Pb NO3 2 s + 2NaCl aq PbCl2 s + 2NaNO3 aq H = Hf2 + 2Hf4 - Hf1 + 2Hf3 We have the values for Hf3 and Hf4, but we need to find Hf1 and Hf2. We can use the following relationships:Hf1 = Hf Pb2+ aq + 2Hf NO3- aq - Hf Pb NO3 2 s Hf2 = Hf Pb2+ aq + 2Hf Cl- aq - Hf PbCl2 s We can eliminate Hf Pb2+ aq from these equations:Hf2 - Hf1 = 2Hf Cl- aq - 2Hf NO3- aq - Hf PbCl2 s + Hf Pb NO3 2 s Now, we can substitute the known values into the equation:H = Hf2 + 2 -467.7 kJ/mol - Hf1 + 2 -411.2 kJ/mol H = Hf2 - Hf1 - 2 467.7 - 411.2 kJ/molH = Hf2 - Hf1 - 113.0 kJ/molSince we don't have the values for Hf1 and Hf2, we cannot calculate the exact standard enthalpy change for this reaction. However, if you can find the enthalpy of formation values for Pb NO3 2 s and PbCl2 s , you can use the above equation to calculate the standard enthalpy change.