Equal moles of NO2g and NOg are to be placed in a container to produce N2O according to the reaction- NO2- NO- N2O- O2- Kc - 0-914How many moles of NO2 and NO be placed in the 5-0 L container to have an equilibrium concentration of N2O to be 0-05 M-

The correct option is B 0.511 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; N ...

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

Byju's Answer

Standard XII

Chemistry

Stoichiometry

Equal moles o...

Question

Equal moles of $N O_{2} (g)$ and NO(g) are to be placed in a container to produce $N_{2} O$ according to the reaction,

$N O_{2} + N O \to N_{2} O + O_{2};$ $K_{c} = 0.914$

How many moles of $N O_{2}$ and NO be placed in the 5.0 L container to have an equilibrium concentration of $N_{2} O$ to be 0.05 M?

0.511

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

0.1023

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

0.0526

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

0.2046

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

Open in App

Solution

The correct option is B 0.511
$N O_{2} + N O ⟶ N_{2} O + O_{2}$
Intially $x$ $x$ 0 0
At eq. $\frac{x - y}{5}$ $\frac{x - y}{5}$ $\frac{y}{5}$ $\frac{y}{5}$

Given : $[N_{2} O] = 0.05 M$
$\frac{y}{5} = 0.05$
$y = 0.25 m o l e s$

$K_{c} = \frac{[N_{2} O] [O_{2}]}{[N O_{2}] [N O]}$

$0.914 = \frac{0.25 \times 0.25}{(x - 0.25)^{2}}$

$x = 0.511\ moles$

Hence, Moles of $N O_{2}$ = Moles of $N O = 0.511 m o l e s$

Therefore, option A is correct.

Suggest Corrections

Similar questions

Q. At certain temperature,

K_{C} = 1.0

for reaction:

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

equal moles of

N O

and

N O_{2}

are to be placed in

5

litre container until

N_{2} O

concentration equilibrium is

0.5

M. How many mole of

(N O + N O_{2})

must be placed in the container.

Q. An equilibrium mixture

{S O}_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)

was found to contain

0.6

mole of

{S O}_{2}

0.2

mole of

{N O}_{2}

0.8

mole of

{S O}_{2}

and

0.3

mole of

N O

in a liter. vessel. How many moles of

{N O}_{2}

per it. will have to be added to the vessel in order to increase the

N O

concentration to

0.5

mole/liter.

Q. How many of the species are paramagnetic?

N_{2} O, N O, N O_{2}, O_{2}, N O_{2}^{+}, N O^{+}, C O^{+}

Q. A mixture of

3

moles of

{S O}_{2}

4

moles of

{N O}_{2}

and

4

moles of

N O

is placed in a

2.0 L

vessel.

S O_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)

At equilibrium, the vessel is found to contain

1

mole of

S O_{2}

. Calculate the value of

K_{c}

Q. When

N_{2} O_{5}

is heated at temp. T, it dissociates as

N_{2} O_{5} ⇌ N_{2} O_{3} + O_{2}, K_{c} = 2.5

. At the same time

N_{2} O_{3} ⇌ N_{2} O + O_{2}

. If initially 4.0 moles of

N_{2} O_{5}

are taken in 1.0 litre flask and allowed to attain equilibrium, concentration of

O_{2}

was formed to be 2.5 M. Equilibrium concentration of

N_{2} O

is:

Join BYJU'S Learning Program

Equal moles of NO2(g) and NO(g) are to be placed in a container to produce N2O according to the reaction, NO2+NO→N2O+O2; Kc=0.914How many moles of NO2 and NO be placed in the 5.0 L container to have an equilibrium concentration of N2O to be 0.05 M?

The correct option is B 0.511 NO2 + NO ⟶ N2O + O2Intially x x 0 0At eq. x−y5 x−y5 y5 y5Given : [N2O]=0.05 My5=0.05y=0.25 molesKc=[N2O][O2][NO2][NO]0.914=0.25×0.25(x−0.25)2$x = 0.511\ moles$Hence, Moles of NO2 = Moles of NO=0.511 molesTherefore, option A is correct.