At -80°C, K for the reaction N₂O₄(g) 2NO₂(g)
Is 4.66 *10^-⁸. We introduce 0.034 mol of N₂O₄ into a 2.5-L vessel at -80°C and let equilibrium be established. The total pressure in the system at equilibrium will be

Question

At -80°C, K for the reaction N₂O₄(g)

At -80°C, K for the reaction N<sub>2</sub>O<sub>4</sub>(g) 2NO<sub>2</sub>(g) Is 4.66 *10<sup>-</sup><sup>8</sup>. We introduce 0.034 mol of N<sub>2</sub>O<sub>4</sub> into a 2.5-L vessel at -80°C and let equilibrium be established. The total pressure in the system at equilibrium will be A) 4.66 *10<sup>-</sup><sup>8</sup> atm. B) 0.22 atm. C) 0.09 atm. D) 0.43 atm. E) 0.23 atm.

2NO₂(g)
Is 4.66 *10^-⁸. We introduce 0.034 mol of N₂O₄ into a 2.5-L vessel at -80°C and let equilibrium be established. The total pressure in the system at equilibrium will be

Quizplus · Accepted Answer

Using the equilibrium constant (K) and the initial amount of N₂O₄, we can set up an ICE table to determine the equilibrium concentrations of each species: | | N₂O₄ | NO₂ | |-----------|----------------------------|----------------| | Initial | 0.034 mol | 0 | | Change | -x | +2x | | Equilibrium| 0.034 - x | 2x | We can then substitute these equilibrium concentrations into the equilibrium constant expression: K = [NO₂]²/[N₂O₄] = 4.66 × 10¹⁰ Plugging in the values: 4.66 × 10¹⁰ = (2x)²/(0.034 - x) Solving for x gives x = 9.17 × 10⁻³. The total pressure at equilibrium can then be determined using the ideal gas law: Ptotal = (n1 + n2)RT/V Where n1 and n2 are the moles of N₂O₄ and NO₂ respectively. Plugging in the values: n1 = 0.034 - x = 0.0248 mol n2 = 2x = 0.0183 mol R = 0.08206 L atm K⁻¹ mol⁻¹ T = -80 + 273 = 193 K V = 2.5 L Ptotal = (0.0248 + 0.0183) × 0.08206 × 193/2.5 = 0.22 atm Therefore, the correct answer is B.

At -80°C, K for the Reaction N2O4(g) 2NO2(g)

At -80°C, K for the Reaction N₂O₄(g) 2NO₂(g)