To calculate the standard enthalpy change for the precipitation reaction, we first need to determine the balanced chemical equation for the reaction:Pb NO3 2 aq + 2 NaI aq PbI2 s + 2 NaNO3 aq Next, we need to determine the moles of Pb NO3 2 in the 70.0 mL of 0.500 M solution:moles = volume in L concentration in mol/L moles = 70.0 mL 1 L / 1000 mL 0.500 mol/Lmoles = 0.035 mol Pb NO3 2Since the reaction is 1:1 between Pb NO3 2 and PbI2, we will also have 0.035 mol of PbI2 formed.Now, we need to find the standard enthalpy change for the reaction. We can use the standard enthalpies of formation for each compound involved in the reaction:H reaction = [ Hf products ] - [ Hf reactants ]The standard enthalpies of formation Hf for the compounds involved are in kJ/mol :Pb NO3 2 aq : -430.5NaI aq : -288.6PbI2 s : -278.0NaNO3 aq : -365.6Now, we can calculate the standard enthalpy change for the reaction:H reaction = [ 1 -278.0 + 2 -365.6 ] - [ 1 -430.5 + 2 -288.6 ]H reaction = -278.0 - 731.2 - -430.5 - 577.2 H reaction = -1009.2 + 1007.7H reaction = -1.5 kJ/molFinally, we can calculate the standard enthalpy change for the 0.035 mol of Pb NO3 2 reacting:H = moles of Pb NO3 2 H reaction H = 0.035 mol -1.5 kJ/mol H = -0.0525 kJTherefore, the standard enthalpy change for the precipitation reaction between 70.0 mL of 0.500 M lead II nitrate solution and excess sodium iodide solution at 25C is -0.0525 kJ.