Calculate the standard enthalpy of formation of Cl-(aq) , given the following thermochemical data at 1 atm, and knowing that the standard enthalpies of formation of H₂(g) and H⁺(aq) are both zero. HCl(g) → H⁺(aq) + Cl-(aq) -75.15 kJ
H₂(g) + Cl₂(g) → 2HCl(g) -184.62 kJ

Question

Quizplus · Accepted Answer

The enthalpy change for the formation of 2 moles of HCl(g) from H2(g) and Cl2(g) is -184.62 kJ. For 1 mole, it is -92.31 kJ. The enthalpy change for HCl(g) dissociating into H+(aq) and Cl-(aq) is -75.15 kJ. The standard enthalpy of formation of Cl-(aq) is calculated as: -92.31 kJ + (-75.15 kJ) = -167.46 kJ/mol.

Calculate the Standard Enthalpy of Formation of Cl-(Aq), Given the Following