Question

Calculate the standard enthalpy change (∆H°) for the precipitation reaction between aqueous solutions of lead(II) nitrate [Pb(NO3)2] and potassium iodide (KI) producing solid lead(II) iodide (PbI2) and aqueous potassium nitrate (KNO3) at a temperature of 25°C.

Answer 1

To calculate the standard enthalpy change  H  for the precipitation reaction, we need to use the following equation:H =  Hf products  -  Hf reactants where Hf represents the standard enthalpy of formation for each compound.First, let's write the balanced chemical equation for the reaction:Pb NO3 2  aq  + 2 KI  aq   PbI2  s  + 2 KNO3  aq Now, we need to find the standard enthalpies of formation  Hf  for each compound involved in the reaction. These values can be found in standard chemistry reference books or online databases. Here are the values for each compound:Hf [Pb NO3 2  aq ] = -437 kJ/molHf [KI  aq ] = -327 kJ/molHf [PbI2  s ] = -278 kJ/molHf [KNO3  aq ] = -494 kJ/molNow, we can plug these values into the equation:H = [1   -278 kJ/mol  + 2   -494 kJ/mol ] - [1   -437 kJ/mol  + 2   -327 kJ/mol ]H =  -278 - 988  -  -437 - 654 H =  -1266  -  -1091 H = -175 kJ/molSo, the standard enthalpy change  H  for the precipitation reaction between aqueous solutions of lead II  nitrate and potassium iodide at 25C is -175 kJ/mol.