Given reaction 2NH₃(g)+ 3Cl₂(g)$\to$ N₂(g)+ 6HCl(g),you react 5.0 L of NH₃ with 5.0 L of Cl₂ measured at the same conditions in a closed container.Calculate the ratio of pressures in the container (P_final/P_initial).

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 2 moles of NH₃ react with 3 moles of Cl₂ to form 10 moles of N₂ and 6 moles of HCl. Since the volumes are measured at the same conditions, we can assume that the moles of Cl₂ is the limiting reactant. Therefore, 5 L of Cl₂ contains (5/22.4) x 3 = 0.67 moles of Cl₂. According to the stoichiometry, 0.67 x (2/3) = 0.45 moles of NH₃ are required, but we only have 0.36 moles of NH₃ (5/22.4 x 2), which is the limiting reactant. Therefore, all of the Cl₂ is consumed to form 1.5 x 0.36 = 0.54 moles of HCl and 5 x (2/3) x 0.36 = 1.2 L of N₂. The initial pressure of the mixture is P1 = (0.36/5) x R x T = 0.072 RT and the final pressure of the mixture is P2 = [(0.36/5) + (1.5 x 0.036)/5] x R x T = 0.108 RT (using the ideal gas law). Therefore, the ratio of pressures is P2/P1 = 0.108/0.072 = 1.5, which is closest to choice C (1.33).

Given Reaction 2NH3(g)+ 3Cl2(g) →\to→ N2(g)+ 6HCl(g),you React 5

Given Reaction 2NH₃(g)+ 3Cl₂(g) $\to$ N₂(g)+ 6HCl(g),you React 5