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 standard energy change for the reaction can be calculated using the given standard enthalpies of formation and the ideal gas law to find the amount of moles in 50.0 L of O2 under the given conditions. The reaction is: O3(g) + O(g) → 2O2(g). The enthalpy change for the formation of 1 mole of O2 from O3 and O is the difference between the enthalpy of formation of O (249.2 kJ/mol) and O3 (142.7 kJ/mol), which is 106.5 kJ/mol for 1.5 moles of O2 formed. To find the moles of O2 at -75 °C (198 K) and 0.12 atm, use PV=nRT, solving for n gives n = PV/RT. Convert volume to m^3, pressure to Pa, and use R = 8.314 J/(mol·K). The energy change for the formation of the calculated moles of O2 is then found by multiplying the moles of O2 by the energy change per mole (106.5 kJ/mol for 1.5 moles of O2). This calculation approximates to -72 kJ, considering the stoichiometry and the conditions provided.

One of the Natural Processes That Destroys Ozone in the Atmosphere