Question

Calculate the standard enthalpy change for the precipitation reaction that takes place when 25.0 mL of 0.250 M barium chloride (BaCl2) and 50.0 mL of 0.100 M sodium sulfate (Na2SO4) react to form a precipitate of barium sulfate (BaSO4). The density of each solution is 1.00 g/mL and the specific heat capacity is 4.18 J/g°C. Assume a constant temperature of 25°C.

Answer 1

To calculate the standard enthalpy change for the precipitation reaction, we first need to determine the balanced chemical equation for the reaction:BaCl2 aq  + Na2SO4 aq   BaSO4 s  + 2NaCl aq Next, we need to determine the limiting reactant. To do this, we will calculate the moles of each reactant:Moles of BaCl2 =  volume in L    concentration in mol/L  =  0.025 L    0.250 mol/L  = 0.00625 molMoles of Na2SO4 =  0.050 L    0.100 mol/L  = 0.00500 molSince the stoichiometry of the reaction is 1:1, Na2SO4 is the limiting reactant.Now, we need to find the standard enthalpy change for the reaction. This can be found using the standard enthalpies of formation for each compound:H reaction  = [Hf BaSO4  + 2Hf NaCl ] - [Hf BaCl2  + Hf Na2SO4 ]The standard enthalpies of formation  Hf  for the compounds are as follows:Hf BaSO4  = -1465.6 kJ/molHf NaCl  = -411.12 kJ/molHf BaCl2  = -858.6 kJ/molHf Na2SO4  = -1387.3 kJ/molSubstitute these values into the equation:H reaction  = [ -1465.6  + 2 -411.12 ] - [ -858.6  +  -1387.3 ] = -2287.84 kJ/molSince Na2SO4 is the limiting reactant, we will use its moles to calculate the enthalpy change for the reaction:H = H reaction   moles of limiting reactant = -2287.84 kJ/mol  0.00500 mol = -11.44 kJTherefore, the standard enthalpy change for the precipitation reaction is -11.44 kJ.