Question

For the reaction N2O4(g) = 2NO2(g), if initially only 4 atm N2O4 is present and at equilibrium, total pressure of equilibrium mixture is 6 atm then the value of Kp is

(1) 2 atm
(2) 4 atm
(3) 8 atm
(4) 16 atm​

Answer 1

Answer:

PNO2=0.64atm

PN2O4=0.05atm

Explanation:

For the first equilibrium, the total pressure is 1.1+0.28=1.38atm

The value of the equilibrium constant will be  Kp=PN2O4PNO22=0.281.12=4.32

For the second equilibrium, since the volume of the container is doubled, the total pressure will be one half i.e 21.38=0.69atm

The partial pressures of N2O4 and NO2 at equilibrium will be 0.69−x atm and x atm respectively.

The value of the equilibrium constant will be  Kp=PN2O4PNO22=0.69−xx2=4.32

PNO2=x=0.64atm and PN2O4=0.69−x=0.05atm

Answer 2

The correct answer is option (3) 8 atm.

Given:

For the reaction N$_{2}$O$_{4}$ (g) ⇄  2NO$_{2}$ (g), the  initial pressure of N$_{2}$O$_{4}$ is 4 atm.

At equilibrium, the total pressure of the mixture is 6 atm.

To Find:

The value of Kp.

Solution:

From the ideal gas equation, we have

PV = nRT, and V, R, and T are constant.

So we can apply stoichiometry in terms of the pressure of given gases

N$_{2}$O$_{4}$ (g) ⇄  2NO$_{2}$ (g)

The initial pressure at t=0 of N$_{2}$O$_{4}$ is 4 atm. Let us assume that on attaining equilibrium at t=t$_{eq}$, the pressure of N$_{2}$O$_{4}$ decreases by x atm, so by stoichiometry the pressure of NO$_{2}$ becomes 2x.

At equilibrium, P$_{total}$= P$_{N_{2} O_{4} }$ +P $_{N O_{2} }$

⇒ 6 = (4-x) + 2x

⇒ x = 2 atm

∴ At equilibrium,  P$_{N_{2} O_{4} }$= (4-x)= 4-2= 2 atm and P$_{N O_{2} }$= 2x = 4 atm.

Kp = $\frac{(P_{N_{2} O_{4} })_{2}}{(P_{N O_{2} })_{1}}$ = $\frac{4^{2} }{2}$ = 8 atm.

Therefore the correct answer is option (3) 8 atm.

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