Question

Calculate the standard enthalpy change for the reaction of [Co(H2O)6]Cl2 and NaOH using the following balanced chemical equation and given standard enthalpies of formation:[Co(H2O)6]Cl2 (aq) + 2NaOH (aq) → [Co(H2O)6](OH)2 (s) + 2NaCl (aq)ΔH°f [Co(H2O)6]Cl2 (aq) = -415.32 kJ/molΔH°f [Co(H2O)6](OH)2 (s) = -1276.24 kJ/molΔH°f NaCl (aq) = -407.33 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:Hrxn =  Hf  products  -  Hf  reactants For the products, we have:1 mol of [Co H2O 6] OH 2  s  with Hf = -1276.24 kJ/mol2 mol of NaCl  aq  with Hf = -407.33 kJ/molFor the reactants, we have:1 mol of [Co H2O 6]Cl2  aq  with Hf = -415.32 kJ/mol2 mol of NaOH  aq  with Hf = -469.15 kJ/mol  not given in the problem, but this value is commonly known Now, we can plug these values into the formula:Hrxn =  1  -1276.24 + 2  -407.33  -  1  -415.32 + 2  -469.15 Hrxn =  -1276.24 - 814.66  -  -415.32 - 938.30 Hrxn =  -2090.90  -  -1353.62 Hrxn = -737.28 kJ/molThe standard enthalpy change for the reaction of [Co H2O 6]Cl2 and NaOH is -737.28 kJ/mol.