Final answer:
To find the partial pressures of NO₂ and O₂ and the total pressure in the flask, stoichiometry and the Ideal Gas Law are used, with the reaction of ozone with nitric oxide yielding equal amounts of each product; the partial pressures of the products are equal to half of the total pressure in the flask.
Step-by-step explanation:
The student has been asked to calculate the partial pressures of products and the total pressure in a flask after the complete reaction of ozone (O₃) with nitric oxide (NO) to produce nitrogen dioxide (NO₂) and oxygen (O₂). Given that we have .202 mol of O₃ and .202 mol of NO in a 7.4 L vessel at 366K, we'll use the Ideal Gas Law and stoichiometry to find the pressures. The balanced chemical equation is:
O₃(g) + NO(g) → NO₂(g) + O₂(g)
We know that at equilibrium, for every mole of O₃ and NO that reacts, one mole of NO₂ and one mole of O₂ are produced. This means we will have .202 mol of NO₂ and .202 mol of O₂ at the end of the reaction. The total number of moles of gas in the vessel at the end is .404 mol (.202 mol NO₂ + .202 mol O₂).
Using the Ideal Gas Law (PV=nRT), where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin, we can solve for total pressure:
P = (nRT) / V = (.404 mol * 0.0821 L*atm/mol*K * 366K) / 7.4 L
After calculating P, we will find the total pressure in the vessel. Since both products are formed in equal amounts, they will each exert half of the total pressure. Therefore, the partial pressure for NO₂ and O₂ is half of the calculated total pressure.