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

Byju's Answer

Standard XII

Chemistry

Introduction to Alcohols

1 mole of a g...

Question

$1$ mole of a gas $A B_{3}$ present in $10 L$ container dissociates into $A B_{2} (g)$ and $B_{2} (g)$ .
$2 A B_{3} (g) ⇌ 2 A B_{2} (g) + B_{2} (g)$
If the degree of dissociation of $A B_{3}$ is $80 %$ , then final pressure at $600 K$ is:
(Given $R = 0.082 {L atm K}^{- 1} {mol}^{- 1}$ )

5.88 atm

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

1.25 atm

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

10.25 atm

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

6.88 atm

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

Open in App

Solution

The correct option is D $6.88 atm$
$\begin{matrix} 2 A B_{3} (g) & ⇌ & 2 A B_{2} (g) & + & B_{2} t=0 & 0.1 & 0 & 0 t=eq & 0.1 - 0.08 & 0.08 & 0.04 \end{matrix}$
Total number of moles per litre $(\frac{n}{V}) = 0.14 {mol L}^{- 1}$
The final pressure at $600 K$ is
$P$ $= \frac{n}{V} \times R \times T = 0.14 \times 0.082 \times 600 = 6.88 atm$

Suggest Corrections

Similar questions

{A B}_{2} (g)

is dissociates as

{A B}_{2} (g) ⇌ {A B}_{2} (g) + \frac{1}{2} B_{2} (g)

When the initial pressure of

{A B}_{3}

800

torr and the total pressure developed at equilibrium is

900

torr. What fraction of

{A B}_{2} (g)

is dissociated?

8

mole of a gas

{A B}_{3}

are introduced into a

1.0 {d m}^{3}

vessel. It dissociates as:

2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g)

at equilibrium,

2

mole of

A_{2}

are found to be present. The equilibrium constant of the reaction is:

Q. At temperature

T

, a compound

A B_{2} (g)

dissociates according to the reaction

2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)

with degree of dissociation

α

(which is very small). The expression for

K_{P}

in terms of

α

and the total pressure,

P_{T}

is:

Q. At certain temperature, compound

A B_{2} (g)

dissociates according to the reaction,

2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)

with degree of dissociation

α

, which is negligible compared with unity, the expression of

K_{P}

, in terms of

α

and initial pressure

P

Q. Eight

mole

of a gas

A B_{3}

attain equilibrium in a closed container of volume

1 {dm}^{3}

as,

2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g)

. If at equilibrium

2 mole

A_{2}

are present then equilibrium constant is:

Join BYJU'S Learning Program

1 mole of a gas AB3 present in 10 L container dissociates into AB2(g) and B2(g). 2AB3(g)⇌2AB2(g)+B2(g) If the degree of dissociation of AB3 is 80%, then final pressure at 600 K is: (Given R=0.082 L atm K−1mol−1)

The correct option is D 6.88 atm2AB3(g)⇌2AB2(g)+B2t=00.100t=eq0.1−0.080.080.04 Total number of moles per litre (nV)=0.14 mol L−1 The final pressure at 600 K is P =nV×R×T=0.14×0.082×600=6.88 atm

$1$ mole of a gas $A B_{3}$ present in $10 L$ container dissociates into $A B_{2} (g)$ and $B_{2} (g)$ .
$2 A B_{3} (g) ⇌ 2 A B_{2} (g) + B_{2} (g)$
If the degree of dissociation of $A B_{3}$ is $80 %$ , then final pressure at $600 K$ is:
(Given $R = 0.082 {L atm K}^{- 1} {mol}^{- 1}$ )