410

You visited us 410 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Le Chatelier's Principle for Del N Greater Than Zero

9.2g of N2O4g...

Question

$9.2 g$ 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 $N_{2} O_{4} is 92$ ]

0.1

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

0.4

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

2

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

Open in App

Solution

The correct option is B $0.2$
$9.2 g$ of $N_{2} O_{4} (g)$ corresponds to $\frac{9.2}{92} = 0.1 moles$
Out of these $0.05 moles$ ( $50 % of 0.1 moles)$ dissociate to form $2 \times 0.05 = 0.1 moles of N O_{2} (g)$
The equilibrium constant is $K_{c} = \frac{[N O_{2}]^{2}}{N_{2} O_{4}} = \frac{(0.1)^{2}}{0.05} = 0.2$
Hence the correct option is B.

Suggest Corrections

Similar questions

9.2 g r a m s

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

N_{2} O_{4} (g)

is dissociated. What is the equilibrium constant? (in mol litre

^{- 1}

)

[Molecular weight of

N_{2} O_{4} = 92

]

Q. 9.2 grams 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%

N_{2} O_{4} (g)

is dissociated. What is the equilibrium constant? (in

M

) (molecular weight of

N_{2} O_{4} = 92

)

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

)

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?

Join BYJU'S Learning Program

9.2g of N2O4(g) is taken in a closed one litre vessel and heated till the following equilibrium is reached N2O4(g)⇌2NO2(g) At equilibrium 50% of N2O4(g) is dissociated. What is the equilibrium constant? [in mole lit−1] [M.wt. of N2O4 is 92]

The correct option is B 0.29.2 g of N2O4(g) corresponds to 9.292=0.1 moles Out of these 0.05 moles (50% of 0.1 moles) dissociate to form 2×0.05=0.1 moles of NO2(g) The equilibrium constant is Kc=[NO2]2N2O4=(0.1)20.05=0.2 Hence the correct option is B.