Question

What is the enthalpy of precipitation for the reaction between 50 ml of 0.1 M Pb(NO3)2 solution and 75 ml of 0.1 M NaI solution? Given that the standard enthalpy of formation of PbI2 is -360 kJ/mol.

Answer 1

To calculate the enthalpy of precipitation for the reaction, we first need to determine the limiting reactant and the number of moles of the product  PbI2  formed.The balanced chemical equation for the reaction is:Pb NO3 2  aq  + 2NaI  aq   PbI2  s  + 2NaNO3  aq First, let's find the moles of each reactant:Moles of Pb NO3 2 =  0.1 mol/L  *  0.050 L  = 0.005 molMoles of NaI =  0.1 mol/L  *  0.075 L  = 0.0075 molNow, we need to determine the limiting reactant. The stoichiometry of the reaction shows that 1 mole of Pb NO3 2 reacts with 2 moles of NaI. Therefore, we can calculate the amount of NaI required to react with the given amount of Pb NO3 2:Moles of NaI required = 0.005 mol * 2 = 0.010 molSince we have only 0.0075 mol of NaI, NaI is the limiting reactant.Now, we can calculate the moles of PbI2 formed:Moles of PbI2 =  0.0075 mol NaI  *  1 mol PbI2 / 2 mol NaI  = 0.00375 molFinally, we can calculate the enthalpy of precipitation using the standard enthalpy of formation of PbI2:Enthalpy of precipitation = moles of PbI2 * standard enthalpy of formation of PbI2Enthalpy of precipitation =  0.00375 mol  *  -360 kJ/mol  = -1.35 kJSo, the enthalpy of precipitation for the reaction is -1.35 kJ.