Question

Calculate the standard enthalpy change for the reaction:Cu(H2O)62+ (aq) + 4Cl-(aq) → CuCl42- (aq) + 12H2O (l)given the following standard enthalpies of formation:Cu(H2O)62+ (aq): -1846.4 kJ/molCuCl42- (aq): -3599.5 kJ/molH2O (l): -285.8 kJ/molCl- (aq): -167.2 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  [Hf products ] -  [Hf reactants ]where H is the standard enthalpy change of the reaction, and Hf represents the standard enthalpies of formation for the products and reactants.For the products:1 mol of CuCl4^2-  aq : -3599.5 kJ/mol12 mol of H2O  l : 12 *  -285.8 kJ/mol  = -3430.4 kJ/molFor the reactants:1 mol of Cu H2O 6^2+  aq : -1846.4 kJ/mol4 mol of Cl-  aq : 4 *  -167.2 kJ/mol  = -668.8 kJ/molNow, we can plug these values into the equation:H = [ -3599.5  +  -3430.4 ] - [ -1846.4  +  -668.8 ]H =  -6029.9  -  -2515.2 H = -3514.7 kJ/molThe standard enthalpy change for the reaction is -3514.7 kJ/mol.