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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]

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sebassandin

Answer:

[tex]\Delta _RH=-69.165kJ[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]O_3(g)+NO(g)\rightarrow O_2(g)+NO_2(g)[/tex]

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

[tex]n_{O_3}^{available}=\frac{1.00atm*8.50L}{0.082\frac{atm*L}{molK}*298.15K}=0.348molO_3[/tex]

[tex]n_{O_3}^{consumed \ by \ NO}=\frac{1.00atm*12.00L}{0.0821\frac{atm*L}{molNOK}*298.15K}*\frac{1molO_3}{1molNO} =0.490molO_3[/tex]

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

[tex]n_{O_3}=n_{NO}=n_{O_2}=n_{NO_2}=0.348molO_3[/tex]

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

[tex]\Delta _RH=n[\Delta _fH_{products}-\Delta _fH_{reagents}]\\\Delta _RH=0.348mol*[33.85 kJ/mol+0kJ/mol-142.2 kJ/mol-90.4 kJ/mol]\\\Delta _RH=-69.165kJ[/tex]

Best regards.

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