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

Byju's Answer

Standard XII

Chemistry

Raoult's Law

9.2g N2O4 i...

Question

$9.2$ g $N_{2} O_{4}$ is heated in a $1$ L vessel till equilibrium state is established $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
In equilibrium state $50 %$ $N_{2} O_{4}$ was dissociated, equilibrium constant will be.

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $0.2$
$N_{2} O_{4} ⇌ 2 N O_{2}$
For $92 g$ of $N_{2} O_{4}$ we get, $2 \times 46 g$ of $N O_{2}$
For $9.2 g$ of $N_{2} O_{4}$ we get, $2 \times 4.6 g$ of $N O_{2}$

Only 50% of $N_{2} O_{4}$ is dissociated,
For $4.6 g$ of $N_{2} O_{4}$ we get, $2 \times 2.3 g$ of $N O_{2}$
So in equilibrium state 4.6g of $N_{2} O_{4}$ and $2 \times 2.3 = 4.6 g$ of $N O_{2}$ is present.

$[N_{2} O_{4}] = (4.6 / 92) / 1 = 0.05$
$[N O_{2}] = (4.6 / 46) / 1 = 0.1$
Equilibrium constant is $\frac{Products}{reactants}$
$K_{c} = \frac{{[N O_{2}]}^{2}}{[N_{2} O_{4}]} = 0.01 / 0.05 = 0.2$

Suggest Corrections

Similar questions

9.2

N_{2} O_{4}

is heated in

1

L vessel till equilibrium state is established.

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

In equilibrium state,

50 %

N_{2} O_{4}

was dissociated, equilibrium constant will be :

[Molecular wt. of

N_{2} O_{4} = 92

]

18.4

g of

N_{2} O_{4}

is taken in a

1

L closed vessel and heated till the equilibrium is reached.

N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}

At equilibrium it is found that

50 %

N_{2} O_{4}

is dissociated. What will be the value of equilibrium constant?

Q. 9.2 gm of

N_{2} O_{4} (g)

is taken in a closed one-litre vessel and heated till the following equilibrium is reached

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

At equilibrium 50% of

N_{2} O_{4} (g)

is dissociated. What is the equilibrium constant?

[in mole lit

^{- 1}

] [M.wt. of

N_{2} O_{4}

is 92]

Q. 9.2 grams of

N_{2} O_{4}

is taken in a closed one litre vessel and heated till the following equilibrium is reached

N_{2} O - 4 (g) ⇌ 2 N O_{2} (g)

At equilibrium,

50 % N_{2} O_{4} (g)

is dissocated.What is the equilibrium constant (in mol

l i t r e^{- 1}

)? (molecular mass of

N_{2} O_{4} = 92

)

Q. In a container, equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

is attained at

25^{\circ}

C. The total equilibrium pressure in the container is

380

torr. If equilibrium constant of the above equilibrium is

0.667 a t m

, then the degree of dissociation of

N_{2} O_{4}

at this temperature will be:

Join BYJU'S Learning Program

9.2g N2O4 is heated in a 1L vessel till equilibrium state is established N2O4(g)⇌2NO2(g)In equilibrium state 50% N2O4 was dissociated, equilibrium constant will be.

$9.2$ g $N_{2} O_{4}$ is heated in a $1$ L vessel till equilibrium state is established $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
In equilibrium state $50 %$ $N_{2} O_{4}$ was dissociated, equilibrium constant will be.