Answer:
Part. press. NO₂ 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 N₂O₄ and no value for NO₂.
During the reaction, y pressure has been released from N₂O₄. As ratio is 1:2, 2y will be the value for the pressure of NO₂. 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
