Answer:
P N₂O₄ = 3,05 atm
P NO₂ = 7,04 atm
Explanation:
48,2g of N₂O₄ are:
48,2g ₓ (1mol / 92,011g) = 0,524mol of N₂O₄ / 2,08L = 0,252M
Based on the reaction
N₂O₄(g) ⇌ 2NO₂(g) Kc = 0,619 = [NO₂]² / [N₂O₄] <em>(1)</em>
Concentrations in equilibrium are:
[N₂O₄] = 0,252M - X
[NO₂] = 2X
Replacing in (1):
0,619 = [2X]² / [0,252-X]
0,156 - 0,619X - 4X² = 0
Solving for X:
X = -0,289 → <em>False answer, there is no negative concentrations</em>
X = 0,135
Replacing:
[N₂O₄] = 0,252M - 0,135
[N₂O₄] = <em>0,117M</em>
[NO₂] = 2X
[NO₂] = 2×0,135 = 0,270M
using:
P = M×R×T
Where P is pressure, M is molarity, R is gas constant (0,082atmL/molK) and T is temperature (45 + 273,15 = 318,15K). Pressure of N₂O₄ and NO₂ are:
P N₂O₄ = 0,117M×0,082atmL/molK×318,15K = <em>3,05atm</em>
P NO₂ = 0,270M×0,082atmL/molK×318,15K = <em>7,04atm</em>
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I hope it helps!
