Question

What is the standard enthalpy change for the reaction between copper sulfate pentahydrate and potassium hydroxide to form copper hydroxide and potassium sulfate, assuming all substances are in their standard state and the reaction occurs at 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, and Hf is the standard enthalpy of formation for each substance.First, let's write the balanced chemical equation for the reaction:CuSO45H2O  s  + 2 KOH  s   Cu OH 2  s  + K2SO4  s  + 5 H2O  l Now, we need the standard enthalpies of formation  Hf  for each substance involved in the reaction. These values can be found in standard thermodynamic tables:Hf CuSO45H2O  = -1540 kJ/molHf KOH  = -425.8 kJ/molHf Cu OH 2  = -1004 kJ/molHf K2SO4  = -1438 kJ/molHf H2O  = -285.8 kJ/molNow, we can plug these values into the equation:H = [1 -1004  + 1 -1438  + 5 -285.8 ] - [1 -1540  + 2 -425.8 ]H = [-1004 - 1438 - 1429] - [-1540 - 851.6]H = [-3871] - [-2391.6]H = -1479.4 kJ/molThe standard enthalpy change for the reaction between copper sulfate pentahydrate and potassium hydroxide to form copper hydroxide and potassium sulfate at 25C is -1479.4 kJ/mol.