One of the natural processes that destroys ozone in the atmosphere is postulated to consist of two steps: O₃(g) + UV radiation → O₂(g) + O(g)
O₃(g) + O(g) → 2 O₂(g)
The standard enthalpies of formation for ozone and for atomic oxygen are +142.7 kJ/mol and +249.2 kJ/mol, respectively. What would be the standard energy change associated with the combination of ozone and atomic oxygen to form 50.0 L of molecular oxygen, measured at -75°C and 0.12 atm?

Question

Quizplus · Accepted Answer

The reaction O₃(g) + O(g) → 2 O₂(g) has a standard enthalpy change of -106.5 kJ/mol. Using the ideal gas law, 50.0 L of O₂ at -75°C and 0.12 atm corresponds to 0.107 mol of O₂. The energy change is then -106.5 kJ/mol * 0.107 mol = -11.4 kJ. Since the reaction produces 2 moles of O₂, the energy change for 1 mole of O₂ is -72 kJ.

One of the Natural Processes That Destroys Ozone in the Atmosphere