Question

Calculate the standard enthalpy change for the reaction: [Co(H2O)6]2+(aq) + 4Cl-(aq) → [CoCl4]2-(aq) + 6H2O(l) Given the following information:• ΔH°f of [Co(H2O)6]2+ is 32.3 kJ/mol • ΔH°f of [CoCl4]2- is -341.8 kJ/mol • ΔH°f of H2O(l) is -285.8 kJ/mol • The ΔH°f of Cl- is taken as zero

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the formula:H =  Hf products  -  Hf reactants For the products, we have:1 mol of [CoCl4]2- and 6 mol of H2O l For the reactants, we have:1 mol of [Co H2O 6]2+ and 4 mol of Cl-Now, we can plug in the given values:H =  1 mol  -341.8 kJ/mol + 6 mol  -285.8 kJ/mol  -  1 mol  32.3 kJ/mol + 4 mol  0 kJ/mol H =  -341.8 kJ + 6  -285.8 kJ  -  32.3 kJ + 0 kJ H =  -341.8 kJ - 1714.8 kJ  - 32.3 kJH = -2056.6 kJ - 32.3 kJH = -2088.9 kJThe standard enthalpy change for the reaction is -2088.9 kJ.