Answer:
Part. press. NOâ‚‚ in equilibrium is 0.590 atm
Explanation:
First of all, we determine the equilibrium:
N₂O₄ (g) ⇄ 2NO₂ (g)
These is a system of two unknown values.
In the begining we have x pressure of Nâ‚‚Oâ‚„ and no value for NOâ‚‚.
During the reaction, y pressure has been released from Nâ‚‚Oâ‚„. As ratio is 1:2, 2y will be the value for the pressure of NOâ‚‚. So in the equilibrium we have:
N₂O₄ → x - y
NO₂ → 2y
Data from the excersise states that the total pressure is 1 atm so we know that the sum of partial pressures in a mixture, will be the total one. In the equilibrium, total pressure will be:
(x-y) + 2y = 1 atm
x + y = 1 atm
Let's make the expression for Kp
Kp =  [Partial pressure NO₂]² / [ Partial pressure N₂O₄]
Kp = (2y)² / (x-y)
Kp = 4y² / (x-y)
We split the x value in the first equation:
x + y = 1 atm
x = 1 atm - y
x = 1 - y → we put this in the Kp expression
0.85 = 4y² / ( 1 - y - y)
0.85 = 4y² / 1 -2y
This is a quadractic equation
0.85 - 1.7y - 4y² = 0 where (a = -4, b = -1.7 c = 0.85)
(-b +- √(b² - 4ac)) / (2a)
(1.7 +-√((-1.7)² - 4 (-4) . 0.85) / 2 .(-4) → 0.295 = y
As [ Part. press. NO₂] in equilibrium is 2y → 0.295 . 2 = 0.590 atm
Based on the total pressure at equilibrium, the partial pressure NOâ‚‚ of at equilibrium is 0.590 atm.
From the equation of the reaction, the equilibrium is determined:
At equilibrium, Nâ‚‚Oâ‚„ and NOâ‚‚ exist in the ratio 1 : 2.
Nâ‚‚Oâ‚„ at x pressure releases y pressure to form NOâ‚‚ .
Thus at equilibrium:
N₂O₄ → x - y
NO₂ → 2y
Ptotal = 1 atm
Thus:
(x-y) + 2y = 1 atm
x + y = 1 atm
Also, Kp = 0.85
From the equation of the reaction:
Kp =  [Partial pressure NO₂]² / [ Partial pressure N₂O₄]
Kp = (2y)² / (x-y)
Kp = 4y² / (x-y)
Solving for x from the first equation:
x + y = 1 atm
x = 1 atm - y
Substitute x = 1 - y in the Kp expression
0.85 = 4y² / ( 1 - y - y)
0.85 = 4y² / 1 -2y
0.85 - 1.7y - 4y² = 0
Solving the quadractic equation:
where a = -4, b = -1.7 c = 0.85
y = (-b +- √(b² - 4ac)) / (2a)
y =(1.7 +-√((-1.7)² - 4 (-4) × 0.85) / 2 × (-4)
y = 0.295 or y = -0.720
We take positive value of y only.
Since partial pressure of NOâ‚‚ in equilibrium is 2y
Partial pressure of NO₂ = 0.295 × 2
Partial pressure of NOâ‚‚ = 0.590 atm
Therefore, the partial pressure of at equilibrium is 0.590 atm.
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