MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Characteristics of Equilibrium Constant

An equilibriu...

Question

An equilibrium mixture at $300 K$ contains $N_{2} O_{4}$ and ${N O}_{2}$ at $0.28$ and $1.1$ atm pressures respectively. If the volume of the container is doubled, calculate the new equilibrium pressure of the two gases.

Open in App

Solution

An equilibrium mixture at $300 K$ contains $N_{2} O_{4}$ and ${N O}_{2}$ at $0.28$ and $1.1$ atm pressures respectively.
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
The equilibrium constant
$K_{p} = \frac{{[p_{{N O}_{2}}]}^{2}}{p_{N_{2} O_{4}}} = \frac{{(1.1)}^{2}}{0.28} = 4.32 a t m$
If the volume of the container is doubled, the partial pressures of each gaseous component is reduced to half.
$P_{N_{2} O_{4}} = \frac{0.28}{2} = 0.14$ atm.
$P_{N O_{2}} = \frac{1.1}{2} = 0.55$ atm
When new equilibrium is stabilized, x atm of $N_{2} O_{4}$ have reacted to form 2x atm of $N O_{2}$ .
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
$P_{N_{2} O_{4}} = \frac{0.28}{2} = 0.14 - x$ atm.
$P_{N O_{2}} = \frac{1.1}{2} = 0.55 + 2 x$ atm
$K_{p} = \frac{{[p_{{N O}_{2}}]}^{2}}{p_{N_{2} O_{4}}}$
$K_{p} = \frac{{(0.55 + 2 x)}^{2}}{(0.14 - x)} = 4.32$
$0.30 + 1.1 x + 4 x^{2} = 0.605 - 4.32 x$
$4 x^{2} + 5.42 x - 0.302 = 0$
This is quadratic equation with solution.
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- (5.42) \pm \sqrt{(5.42)^{2} - 4 (4) (- 0.302)}}{2 (4)}$
$x = \frac{- 5.42 \pm 5.85}{8}$
$x = \frac{- 5.42 \pm 5.85}{8}$
$x = 0.045$ or $x = - 1.41$
The value -1.41 is discarded as it will lead to negative value of pressure.
$x = 0.045$
$p_{N_{2} O_{4}} = 0.14 - x = 0.14 - 0.045 = 0.095 a t m$
$p_{{N O}_{2}} = 0.55 + 2 x = (0.55 + 2 \times 0.045) = 0.64 a t m$
The new equilibrium pressure of the two gases is 0.64 atm.

Suggest Corrections

thumbs-up

0

Similar questions

Q. An equilibrium mixture at 300K contains

N_{2} O_{4}

N O_{2}

at 0.28 and 1.1 atm respectively. If the volume of the container is doubled and the new equilibrium pressures of

N O_{2}

x \times 10^{- 2}

x - 60

is: (nearest integer value)

Q. For the reaction

N_{2} O_{4} ⇌ 2 N O_{2}

, equilibrium mixture contains

N O_{2}

P = 1.1 a t m

N_{2} O_{4}

P = 0.28 a t m

at 350 K. The volume of the container is doubled. Calculate the equilibrium pressures of the two gases when the system reaches new equilibrium :

25^{o} C

and one atmospheric pressure, the partial pressures in an equilibrium mixture of

N_{2} O_{4}

{N O}_{2}

0.7

0.3

atmosphere, respectively. Calculate the partial pressures of these gases when they are in equilibrium at

25^{o} C

and at a total pressure of

10

At 298 K and 1 atm pressure, the partial pressures in an equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ are 0.7 and 0.3 atmosphere respectively. What will be the partial pressure of $N O_{2}$ when the gases are in equilibrium at 298 K and at a total (equilibrium) pressure of 10 atmospheres?

N_{2} O_{4}

dissociates into gaseous

{N O}_{2}

according to the reaction

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

300 K

1 a t m

pressure, the degree of dissociation of

N_{2} O_{4}

0.2

. If one mole of

N_{2} O_{4}

gas is contained in a vessel, then the density of the equilibrium mixture is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equilibrium Constants

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Characteristics of Equilibrium Constant

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon