Question

Calculate the standard enthalpy change (∆H°) for the reaction of the coordination compound [Co(NH3)6]Cl3 (aq) with NaOH (aq) to produce [Co(NH3)6](OH)3 (aq) and NaCl (aq). Given, the standard enthalpy of formation (∆H°f) for [Co(NH3)6]Cl3, [Co(NH3)6](OH)3 and NaCl are -784.8 kJ/mol, -1273.6 kJ/mol and -411.2 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants The balanced chemical equation for the reaction is:[Co NH3 6]Cl3  aq  + 3 NaOH  aq   [Co NH3 6] OH 3  aq  + 3 NaCl  aq Now, we can plug in the given standard enthalpies of formation  Hf  for each compound:H =  1  -1273.6 kJ/mol + 3  -411.2 kJ/mol  -  1  -784.8 kJ/mol H =  -1273.6 - 1233.6  -  -784.8 H = -2507.2 + 784.8H = -1722.4 kJ/molThe standard enthalpy change  H  for the reaction is -1722.4 kJ/mol.