Ozone (O3) in the atmosphere can be converted to oxygen gas by reaction with nitric oxide (NO). (Nitrogen dioxide is also produced in the reaction.) What is the enthalpy change when 8.50 L of ozone at a pressure of 1.00 atm and 25°C reacts with 12.00 L of nitric oxide at the same initial pressure and temperature (R = 0.0821 L atm/mol K)? [ΔH°f(NO) = 90.4 kJ/mol; ΔH°f(NO2) = 33.85 kJ/mol; ΔH°f(O3) = 142.2 kJ/mol]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Thus, the moles of available ozone and its moles that are consumed by NO, considering the ideal gas equation, are:

Thus, just 0.348 moles of ozone react and produce equal amounts of oxygen and nitrogen dioxide due their 1 to 1 molar relationship:

In such a way, the enthalpy of reaction finally result:

Best regards.

thats a little if u want the answer just look it up on google thats what i do haha : )