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Standard XII

Chemistry

Reaction Quotient

When N 2 O ...

Question

When $N_{2} O_{5}$ is heated at certain temperature, it dissociates as
$N_{2} O_{5} (g) ⇌ N_{2} O_{3} (g) + O_{2} (g); K_{c} = 2.5$ . At the same time $N_{2} O_{3}$ also decomposes as:
$N_{2} O_{3} (g) ⇌ N_{2} O + O_{2} (g)$ . If initially $4.0$ moles of $N_{2} O_{5}$ are taken in $1.0$ litre flask and allowed to dissociation, concentration of $O_{2}$ at equilibrium is $2.5 M$ .

Equilibrium concentration of $N_{2} O_{5}$ is:

1.0 M

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1.5 M

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2.166 M

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1.834 M

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Solution

The correct option is A $1.834 M$
$N_{2} O_{3} ⇌ N_{2} O_{5} + O_{2}$
$4 - x$ $x - y$ $x + y$
$N_{2} O_{3} ⇌ N_{2} O O_{2}$
$x - y$ $N_{2} O$ $y + x$
$ℓ [O_{2}] = x + y = 2.5$ and 2.5
$= \frac{(x + y) (x - y)}{4 - x}$
$∴ x = 2.166$
$[N_{2} O_{5}] = 4 - x = 1.846$

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Similar questions

Q. When

N_{2} O_{5}

is heated at certain temperature, is dissociates as

N_{2} O_{3} (g) ⇌ N_{2} O_{3} (g) + O_{2} (g); K_{c} = 2.5

. At the same time

N_{2} O_{3}

also decomposes as:

N_{2} O_{3} ⇌ N_{2} O (g) + O_{3} (g)

. If initially

4.0

moles of

N_{2} O_{3}

are taken in

1.0

litre flask and allowed to dissociate. Concentration of

O_{2}

at equilibrium is

2.5 M

. Equilibrium concentration of

N_{2} O_{5}

is?

Q. The rate constant

k

, for the reaction

N_{2} O_{5} (g) ⟶ 2 {N O}_{2} (g) + \frac{1}{2} O_{2} (g)

1.3 \times 10^{- 2} s^{- 1}

. Which equation given below describes the change of

[N_{2} O_{5}]

with time?

{[N_{2} O_{5}]}_{0}

and

{[N_{2} O_{5}]}_{t}

correspond to concentration of

N_{2} O_{5}

initially and at time

t

Q. When

N_{2} O_{5}

is heated at a certain temperature, it dissociates as:

N_{2} O_{5} (g) ⇌ N_{2} O_{3} (g) + O_{2} (g); K_{c} = 2.5

. At the same time

N_{2} O_{3}

also decomposes as:

N_{2} O_{3} (g) ⇌ N_{2} (g) + O_{2} (g);

. If 4.0 moles of

N_{2} O_{5}

are taken initially in a 1.0 litre flask and allowed to dissociate, the concentration of

O_{2}

at equilibrium is 2.5 M. The equilibrium concentration of

N_{2} O_{5}

is :

Q. When

N_{2} O_{5}

is heated at a certain temperature, it dissociates as:

N_{2} O_{5} (g) ⇌ N_{2} O_{3} (g) + O_{2} (g); K_{c} = 2.5

. At the same time

N_{2} O_{3}

also decomposes as:

N_{2} O_{3} (g) ⇌ N_{2} (g) + O_{2} (g);

. If 4.0 moles of

N_{2} O_{5}

are taken initially in a 1.0 litre flask and allowed to dissociate, the concentration of

O_{2}

at equilibrium is 2.5 M. The equilibrium concentration of

N_{2} O_{5}

is :

Q. For

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2},

it is found that

\frac{- d}{d t} [N_{2} O_{5}] = k_{1} [N_{2} O_{5}], \frac{d}{d t} [N O_{2}] = k_{2} [N_{2} O_{5}] a n d

\frac{d}{d t} [O_{2}] = k_{3} [N_{2} O_{3}],

then:

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The correct option is A 1.834MN2O3⇌N2O5+O24−x x−y x+yN2O3⇌N2OO2x−y N2O y+xℓ[O2]=x+y=2.5 and 2.5=(x+y)(x−y)4−x∴x=2.166[N2O5]=4−x=1.846

The correct option is A $1.834 M$
$N_{2} O_{3} ⇌ N_{2} O_{5} + O_{2}$
$4 - x$ $x - y$ $x + y$
$N_{2} O_{3} ⇌ N_{2} O O_{2}$
$x - y$ $N_{2} O$ $y + x$
$ℓ [O_{2}] = x + y = 2.5$ and 2.5
$= \frac{(x + y) (x - y)}{4 - x}$
$∴ x = 2.166$
$[N_{2} O_{5}] = 4 - x = 1.846$