Question

Calculate the standard enthalpy change for the reaction: [Cu(H2O)6]2+(aq) + 2NH3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)given the following information: ΔHf° [Cu(H2O)6]2+(aq) = -203.2 kJ/molΔHf° [Cu(NH3)4(H2O)2]2+(aq) = -368.5 kJ/molΔHf° H2O(l) = -285.8 kJ/molΔHf° NH3(aq) = -80.8 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants For the reactants:1 mol of [Cu H2O 6]2+ aq  has Hf = -203.2 kJ/mol2 mol of NH3 aq  has Hf = 2 * -80.8 kJ/mol = -161.6 kJ/molFor the products:1 mol of [Cu NH3 4 H2O 2]2+ aq  has Hf = -368.5 kJ/mol4 mol of H2O l  has Hf = 4 * -285.8 kJ/mol = -1143.2 kJ/molNow, we can plug these values into the equation:H =  -368.5 + -1143.2  -  -203.2 + -161.6 H =  -1511.7  -  -364.8 H = -1146.9 kJ/molThe standard enthalpy change for the reaction is -1146.9 kJ/mol.