The decomposition of N2O to NO2 is carried out at 280-C in chloroform- When equilibrium is reached- 0-2 mol of N2O4 and 2- 10-3 mol of NO2 are present in a 2 L solution- the equilibrium constant for the reaction N2O4- 2NO2 is-

The correct option is D 1× 10^ 5 N_2O_4 ⇌ 2NO_2 We know that k = [NO_2]^2[N_2O_4] ⇒ k = (2× 10^ 32)^2(0.22) = 10^ 60.2× 2 = 10^ 610^ 1 = 10 ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The decomposi...

Question

The decomposition of $N_{2} O$ to $N O_{2}$ is carried out at $280^{\circ} C$ in chloroform. When equilibrium is reached, $0.2 m o l$ of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mol of $N O_{2}$ are present in a $2 L$ solution, the equilibrium constant for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ is:

1 \times 10^{- 2}

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2 \times 10^{- 3}

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1 \times 10^{- 5}

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2 \times 10^{- 5}

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Solution

The correct option is D $1 \times 10^{- 5}$
$N_{2} O_{4} ⇌ 2 N O_{2}$
We know that $k = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
$\Rightarrow k = \frac{(\frac{2 \times 10^{- 3}}{2})^{2}}{(\frac{0.2}{2})} = \frac{10^{- 6}}{0.2} \times 2 = \frac{10^{- 6}}{10^{- 1}} = 10^{- 5}$
Hence, the value of k is $1 \times 10^{- 5} .$

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Similar questions

Q. The decomposition of

N_{2} O_{4}

{N O}_{2}

is carried at

280 K

in chloroform. When equilirbrium has been established,

0.2

mole of

N_{2} O_{4}

and

2 \times 10^{- 3}

mole of

{N O}_{2}

are present in

2

litre solution. The equilibrium constant for the reaction,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

is:

Q. A vessel contains

N_{2} O_{4}

N O_{2}

in 2:3 molar ratio at 10 atm under equilibrium. now

K_{p}

for

N_{2} O_{4} ⇌ 2 N O_{2}

is:

Q. For the reaction equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

the concentrations of

N_{2} O_{4}

and

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

and

1.2 \times 10^{- 2} m o l L^{- 1}

respectively. The value of

K_{c}

for the reaction is

Q. For the reaction at equilibrium:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, the concentration of

N_{2} O_{4}

and

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

and

1.2 \times 10^{- 2}

mol/L respectively. The value of

K_{c}

for the reaction is:

N_{2} O_{4}

dissociates as

N_{2} O_{4} ⇌ 2 N O_{2}

. At

55^{\circ} C

and one atmosphere,

%

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N_{2} O_{4}

50.3 %

. At what

P

and same temperature, the equilibrium mixture will have the ratio of

N_{2} O_{4} : N O_{2}

1 : 8

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The decomposition of N2O to NO2 is carried out at 280∘C in chloroform. When equilibrium is reached, 0.2 mol of N2O4 and 2×10−3 mol of NO2 are present in a 2 L solution, the equilibrium constant for the reaction N2O4⇌2NO2 is:

The correct option is D 1×10−5N2O4 ⇌2NO2We know that k=[NO2]2[N2O4]⇒k=(2×10−32)2(0.22)=10−60.2×2=10−610−1=10−5Hence, the value of k is 1×10−5.

The decomposition of $N_{2} O$ to $N O_{2}$ is carried out at $280^{\circ} C$ in chloroform. When equilibrium is reached, $0.2 m o l$ of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mol of $N O_{2}$ are present in a $2 L$ solution, the equilibrium constant for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ is: