Question

Calculate the standard enthalpy change for the reaction of the coordination compound Cu(NH3)4SO4 with excess NaOH to form Cu(OH)2(s), NH3(g), Na2SO4(aq), and H2O(l) at a constant pressure of 1 atm and a temperature of 25°C.

Answer 1

To calculate the standard enthalpy change for the reaction, we need to use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change of the reaction, and Hf is the standard enthalpy of formation of each compound.First, we need to write the balanced chemical equation for the reaction:Cu NH3 4SO4 s  + 4NaOH aq   Cu OH 2 s  + 4NH3 g  + Na2SO4 aq  + 2H2O l 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 thermodynamic tables or online databases. The values are as follows:Hf [Cu NH3 4SO4 s ] = -315.5 kJ/molHf [NaOH aq ] = -469.15 kJ/molHf [Cu OH 2 s ] = -636.1 kJ/molHf [NH3 g ] = -45.9 kJ/molHf [Na2SO4 aq ] = -1377.3 kJ/molHf [H2O l ] = -285.83 kJ/molNow, we can plug these values into the equation:H = [1 -636.1  + 4 -45.9  + 1 -1377.3  + 2 -285.83 ] - [1 -315.5  + 4 -469.15 ]H =  -636.1 - 183.6 - 1377.3 - 571.66  -  -315.5 - 1876.6 H =  -2768.66  -  -2192.1 H = -576.56 kJ/molTherefore, the standard enthalpy change for the reaction is -576.56 kJ/mol.