Reaction between Nitrogen and Oxygen takes place as follows:-
$2 N_{2} (g) + O_{2} (g) ⇌ 2 N_{2} O (g)$
If a mixture of $0.482 m o l$ of $N_{2}$ and $0.933 m o l$ of $O_{2}$ is placed in a reaction vessel of volume $10 L$ and allowed to form $N_{2} O$ at a temperature for which $K e = 2.0 \times 10^{- 37}$ , determine the composition of the equilibrium mixture.

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Solution

$2 N_{2} + O_{2} ⇌ 2 N_{2} O (g)$
initial concentration of $N_{2} [N_{2}] = \frac{0.482}{10}$
$= 0.0482$
$[O_{2}] = \frac{0.933}{2} = 0.0933$
$2 N_{2} + O_{2} ⇌ 2 N_{2} O$
$0.0482 - 2 x 0.0933 - x$ $2 x$
$2 \times 10^{- 37} = \frac{[2 x]^{2}}{(0.0482 - 2 x)^{2} (0.0933 - x)}$
$2 \times 10^{- 37} = \frac{4 x^{2}}{(0.0482)^{2} (0.0933)}$
$x = 0.0329 \times 10^{- 19}$
$= 3.29 \times 10^{- 21}$
$[N_{2} O] = 2 \times 3.29 \times 10^{- 21} = 6.58 \times 10^{- 21}$
$[N_{2}] ≃ 0.0482$
$[O_{2}] ≃ 0.0933$

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Similar questions

Reaction between $N_{2}$ and $O_{2}$ takes place as follows:
$2 N_{2} (g) + O_{2} (g) \leftrightarrow 2 N_{2} O (g)$
If a mixture of 0.482 mol of $N_{2}$ and 0.933 mol of $O_{2}$ is placed in a 10 L reaction vessel and allowed to form $N_{2} O$ at a temperature for which $K_{c} = 2.0 \times 10^{- 37}$ , determine the composition of equilibrium mixture.

Q. Reaction between nitrogen and oxygen takes place as follows:

{2 N}_{2} (g) + O_{2} (g) ⇌ {2 N}_{2} O (g)

If a mixture of

0.482

N_{2}

and

0.933

mol of

N_{2}

is placed in a reaction in a reaction vessel of volume

10 L

and allowed to form

N_{2} O

at a temperature for which

K_{c} = 2.0 \times 10^{- 37}

. Determine the composition of the equilibrium mixture.

Reaction between N₂ and O₂ takes place as follows:

If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 × 10^–37, determine the composition of equilibrium mixture.

Q. If a mixture of 0.482 mol

N_{2}

and 0.933 mol of

O_{2}

is placed in a 10 L reaction vessel and allowed to form

N_{2} O

at a temperature for which

K_{C} = 2.0 \times 10^{- 37}

. The composition of equilibrium mixture are :

Reaction between N₂ and O₂ takes place as follows:

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2N2+O2⇌2N2O(g)initial concentration of N2[N2]=0.48210=0.0482[O2]=0.9332=0.09332N2+O2⇌2N2O0.0482−2x0.0933−x 2x2×10−37=[2x]2(0.0482−2x)2(0.0933−x)2×10−37=4x2(0.0482)2(0.0933)x=0.0329×10−19=3.29×10−21[N2O]=2×3.29×10−21=6.58×10−21[N2]≃0.0482[O2]≃0.0933