Calculate the standard enthalpy change for the precipitation reaction that occurs when 100 mL of 0.2 M lead nitrate solution reacts with 100 mL of 0.1 M sodium iodide solution to form solid lead iodide and aqueous sodium nitrate. The balanced chemical equation for the reaction is:Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)Given: - Molar mass of Pb(NO3)2 = 331.20 g/mol - Molar mass of NaI = 149.89 g/mol - Density of 0.2 M lead nitrate solution = 1.055 g/mL - Density of 0.1 M sodium iodide solution = 1.04 g/mL - Specific heat capacity of the reaction mixture is 4.18 J/(g · °C) - The initial temperature of the reaction mixture and the surroundings is 25 °C.

Question

Calculate the standard enthalpy change for the precipitation reaction that occurs when 100 mL of 0.2 M lead nitrate solution reacts with 100 mL of 0.1 M sodium iodide solution to form solid lead iodide and aqueous sodium nitrate. The balanced chemical equation for the reaction is:Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)Given: -

Answer 1

To calculate the standard enthalpy change for the precipitation reaction, we first need to determine the limiting reactant and the amount of heat released during the reaction.1. Determine the limiting reactant:Pb NO3 2: 100 mL  0.2 mol/L = 0.02 molNaI: 100 mL  0.1 mol/L = 0.01 molSince there are 2 moles of NaI required for every mole of Pb NO3 2, the limiting reactant is NaI.2. Calculate the mass of the reaction mixture:Mass of Pb NO3 2 solution: 100 mL  1.055 g/mL = 105.5 gMass of NaI solution: 100 mL  1.04 g/mL = 104 gTotal mass of the reaction mixture: 105.5 g + 104 g = 209.5 g3. Calculate the heat released during the reaction:We need the final temperature of the reaction mixture to calculate the heat released. However, this information is not provided. Assuming the final temperature is given as Tf, we can set up an equation to calculate the heat released.q = mcTq =  209.5 g  4.18 J/gC  Tf - 25C 4. Calculate the standard enthalpy change:To calculate the standard enthalpy change, we need the heat released per mole of the limiting reactant.H = q / moles of limiting reactantH = [ 209.5 g  4.18 J/gC  Tf - 25C ] / 0.01 molWithout the final temperature  Tf  of the reaction mixture, we cannot provide an exact value for the standard enthalpy change. If the final temperature is provided, you can plug it into the equation above to calculate the standard enthalpy change for the precipitation reaction.