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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?

Given:
- \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)
- \( \Delta H_f^\circ(\text{NO}) = 90.4 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(\text{NO}_2) = 33.85 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(\text{O}_3) = 142.2 \, \text{kJ/mol} \)

Select one:
A. –19.7 kJ
B. –69.1 kJ
C. –1690 kJ
D. –167 kJ
E. –97.6 kJ

Answer :

Final answer:

To calculate the enthalpy change when ozone reacts with nitric oxide, determine the total enthalpy change and final gas conditions using ideal gas law principles. Therefore, the correct option is E.

Explanation:

Enthalpy Change Calculation:

First, calculate the total enthalpy change for the reaction using the enthalpy of formation values provided. Then, apply the ideal gas law to find the final pressure and temperature of the gases after the reaction.

Answer: The enthalpy change when 8.50 L of ozone reacts with 12.00 L of nitric oxide at 25°C and 1 atm is -97.6 kJ.

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