CameraIcon

SearchIcon

MyQuestionIcon

1

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

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

In the follow...

Question

In the following equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ $N O_{2} (g)$ is 50% of the total volume. Hence degree of dissociation and van't Hoff factor (i) respectively are:

Open in App

Solution

$N_{2} O_{4} ⇌ 2 {N O}_{2}$
$1 - x$ $2 x$
$x = 50$ % $= 0.5 =$ degree of dissociation
$i = 1 - x + 2 x / 1 = x + 1 / 1 = 1.5$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In the following equilibrium

N_{2} O_{4 (g)} ⇌ {2 N O}_{2 (g)}

{N O}_{2 (g)}

501

of the total volume. hence degree of dissociation and vant hoff factor (i) respectively are:

Q. For the following equilibrium,

N_{2} O_{4} ⇌ 2 N O_{2}

in gaseous phase,

N O_{2}

is 50% of the total volume when equilibrium is set up. Hence, percent of dissociation of

N_{2} O_{4}

Q. For the equilibrium

N_{2} O_{4} ⇌ 2 N O_{2}

in gaseous phase,

N O_{2}

50 %

of the total volume when equilibrium is set up. Hence percent of dissociation of

N_{2} O_{4}

N_{2} O_{4}

dissociates into gaseous

N O_{2}

according to the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

at 300 K and 1 atm pressure, the degree of dissociation of

N_{2} O_{4}

is 0.2. If one mole of

N_{2} O_{4}

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

Q. For the dissociation reaction,

N_{2} O_{4} ⇌ 2 N O_{2} (g)

, the equilibrium constant

K_{p}

is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of

N_{2} O_{4}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Le Chateliers Principle

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon