In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). W
hat are the equilibrium partial pressures of N₂, O₂, and NO if 
= 0.0460 for the following reaction:
2 answers:
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>
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Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again