Question

Calculate the standard enthalpy change for the following reaction: [CoCl(NH3)5]2+ (aq) + 4H2O (l) → [Co(H2O)6]2+ (aq) + 2Cl- (aq) + 5NH3 (g)Given that the standard enthalpy of formation of [Co(H2O)6]2+ (aq) is -1518.5 kJ/mol, [CoCl(NH3)5]2+ (aq) is -426.9 kJ/mol, and the enthalpy of vaporization of water is 44.0 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the given reaction, we can use the formula:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the reactants, we have:1 mol of [CoCl NH3 5]2+  aq  with Hf = -426.9 kJ/mol4 mol of H2O  l  with Hf = -285.8 kJ/mol  standard enthalpy of formation of liquid water For the products, we have:1 mol of [Co H2O 6]2+  aq  with Hf = -1518.5 kJ/mol5 mol of NH3  g  with Hf = -45.9 kJ/mol  standard enthalpy of formation of ammonia gas Now, we can plug these values into the formula:H = [ -1518.5  + 5 -45.9 ] - [ -426.9  + 4 -285.8 ]H =  -1518.5 - 229.5  -  -426.9 - 1143.2 H =  -1748  -  -1570.1 H = 177.9 kJ/molThe standard enthalpy change for the given reaction is 177.9 kJ/mol.