1. The equilibrium pressures for reaction of decomposition of N 2 O 4 to 2 NO 2
ID: 794934 • Letter: 1
Question
1. The equilibrium pressures for reaction of decomposition of N2O4 to 2 NO2 are 0.58 bar and 1.6 bar respectively at room temperature. If the volume of the container is reduced by a factor of 2 at constant temperature, what would be the partial pressure of the gases when the equilibrium is re-established? (assume ideal behavior)
PLEASE SHOW ALL RELEVANT EQUATIONS AND WORK THE EQUATION FIRST, SO I CAN SEE HOW IT IS DONE! I have a hard time with the algebra, and would like to learn how to work that in the equation. Thank you!
The equilibrium pressures for reaction of decomposition of N2O4 to 2 NO2 are 0.58 bar and 1.6 bar respectively at room temperature. If the volume of the container is reduced by a factor of 2 at constant temperature, what would be the partial pressure of the gases when the equilibrium is re-established?Explanation / Answer
Equilibrium equation
N2O4(g) ---> 2NO2(g)
Kp = p2NO2 / pN2O4
Kp = (1.6)2 / 0.58 = 4.414
Kp = 4.414
The value of Kp will remain constant as the temperature is kept constant
The total pressure = Partial pressure of NO2 + Partial pressure of N2O4
total pressure = 1.6 +0.58 = 2.18 atm
if we halved the volume at constant temperature , the pressure will become double
So the total pressure = 2 X 2.18 = 4.36 atm
N2O4(g) ---> 2NO2(g)
Initial 0.58 1.6
Change -x +2x
Equilibr 0.58-x 1.6 +2x
Kp = 4.414 = (Partial pressure of NO2)^2 / Partial pressure of N2O4
The reaction will be [we have decreased the volume so system will move in backward direction]
N2O4(g) ---> 2NO2(g)
Initial 1.16 3.2
Change +x -2x
Equilibr 1.16+x 3.2-2x
K = 4.414 = (3.2-2x)^2 / (1.16+x)
4.414(1.16+x) = 10.24 + 4x2 - 4x
5.12 + 4.414x = 10.24 + 4x2 - 4x
x = 0.321
pressure of NO2 = 3.2-2(0.321) = 2.56 atm
Pressure of N2O4 = 1.16 + 0.321 = 1.481atm
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