Final answer:
To calculate the partial pressure of each product and the total pressure in the 12.0 L vessel after ozone (O₃) reacts completely with nitrogen monoxide (NO), producing nitrogen dioxide (NO₂) and oxygen (O₂), we can use the given initial conditions and stoichiometry of the reaction. The partial pressure of NO at equilibrium is 0.52 atm, and the total pressure in the vessel is 1.51 atm.
Step-by-step explanation:
To calculate the partial pressure of each product and the total pressure in the vessel, we can use the given initial conditions and the stoichiometry of the reaction. Let's define the change in the partial pressure of NO as 2x, then the change in the partial pressure of O₂ and N₂ is -x because 1 mol each of N₂ and O₂ is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
Using the given initial pressures of PN₂ = 0.78 atm and Po₂ = 0.21 atm, we can calculate the partial pressure of NO at equilibrium. The equation N₂(g) + O₂(g) = 2 NO(g) tells us that 1 mol of N₂ and 1 mol of O₂ react to produce 2 mol of NO. So, the change in the partial pressure of NO is 2x, and the change in the partial pressure of N₂ and O₂ is -x. We can set up an equation: 0.78 - x + 0.21 - x = 2x. Solving this equation, we find that x = 0.26 atm. Therefore, the partial pressure of NO at equilibrium is 2x = 2 * 0.26 atm = 0.52 atm.
Since we've calculated the partial pressure of NO, we can now find the total pressure in the vessel at equilibrium. The total pressure is equal to the sum of the partial pressures of all gases involved. In this case, the total pressure is PN₂ + Po₂ + PNO = 0.78 atm + 0.21 atm + 0.52 atm = 1.51 atm.