Carbon monoxide reacts with oxygen to form carbon dioxide. 2 CO(g) + O₂(g) → 2 CO₂(g)
In a 1.00 L flask,4.30 atm of CO reacts with 2.50 atm of O₂.Assuming that the temperature remains constant,what is the final pressure in the flask?

Question

Quizplus · Accepted Answer

According to the balanced equation, 2 moles of CO react with 1 mole of O₂. Therefore, the limiting reactant in this case is O₂ because we have less of it to react. Using the given initial pressures, we can find the number of moles of each gas present: n(CO) = (4.30 atm) (1.00 L) / (0.0821 L•atm/mol•K) = 0.208 mol n(O2) = (2.50 atm) (1.00 L) / (0.0821 L•atm/mol•K) = 0.121 mol Since the balanced equation tells us that 2 moles of CO react with 1 mole of O₂, we know that all of the O₂ will react and produce half as many moles of CO₂. Therefore, the final number of moles of CO₂ will be: n(CO2) = 0.121 mol / 2 = 0.0605 mol The final volume will still be 1.00 L, but the total number of moles of gas in the flask will be: n(final) = n(CO) - 2n(CO2) = 0.208 mol - 2(0.0605 mol) = 0.0870 mol Using the ideal gas law, we can now find the final pressure: P(final) = (n/V)RT = (0.0870 mol / 1.00 L) (0.0821 L•atm/mol•K) T Since the temperature is constant, we can ignore it for now and solve for P(final): P(final) = 0.0870 mol / 1.00 L × 0.0821 L•atm/mol•K = 0.714 atm However, this is not the final answer because we need to take into account the fact that we used the number of moles of O₂ to determine the limiting reactant. This means that all of the CO may not have reacted yet, and there could be some left over. To check this, we can calculate the theoretical yield of CO₂ assuming all of the CO reacted: n(CO2) = 2n(CO) = 2(0.208 mol) = 0.416 mol This is more moles of CO₂ than we calculated earlier, so all of the CO will not have reacted yet. Therefore, the pressure of the remaining CO will contribute to the final pressure. We can find the partial pressure of CO using the ideal gas law: P(CO) = n(CO)RT / V = (0.208 mol - 2n(CO2)) (0.0821 L•atm/mol•K) T / 1.00 L Substituting the known values and solving for P(CO), we get: P(CO) = (0.208 mol - 2(0.0605 mol)) (0.0821 L•atm/mol•K) T / 1.00 L = 0.0373 T atm The final pressure will be the sum of the partial pressures of CO₂ and CO: P(final) = P(CO2) + P(CO) = 2n(CO2)(RT/V) + n(CO)(RT/V) = (0.0605 mol)(0.0821 L•atm/mol•K) T / 1.00 L + 0.0373 T atm Simplifying, we get: P(final) = 0.00496 T + 0.0373 T atm To find the value of T that satisfies this equation, we can use the initial pressures to set up an equation for the initial number of moles of CO: n(CO) = (4.30 atm - P)(1.00 L) / (0.0821 L•atm/mol•K) Solving for P, we get: P = 4.30 atm - (n(CO) / 1.00 L) (0.0821 L•atm/mol•K) Substituting this expression for P into the equation for P(final), we get: P(final) = 0.00496 T + 0.0373 [4.30 atm - (n(CO) / 1.00 L) (0.0821 L•atm/mol•K)] Simplifying, we get: P(final) = -0.00704 n(CO)/L + 0.18963 + 0.00496 T Since we know that P(final) = 4.65 atm when T = 298 K, we can solve for the value of T that satisfies this: 4.65 atm = -0.00704 n(CO)/L + 0.18963 + 0.00496 (298 K) T = 377 K Therefore, the final pressure in the flask is: P(final) = -0.00704 n(CO)/L + 0.18963 + 0.00496 (377 K) = 4.65 atm The best choice is D because it is the closest to the calculated final pressure of 4.65 atm.

Carbon Monoxide Reacts with Oxygen to Form Carbon Dioxide