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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

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1.25 atm

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10.25 atm

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6.88 atm

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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
$P$ $= \frac{n}{V} \times R \times T = 0.14 \times 0.082 \times 600 = 6.88 atm$

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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

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in terms of

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and the total pressure,

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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

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, which is negligible compared with unity, the expression of

K_{P}

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and initial pressure

P

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 a degree of dissociation

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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 $R = 0.082 {L atm K}^{- 1} {mol}^{- 1}$ )