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Standard XII

Chemistry

Characteristics of Equilibrium Constant

At 1200oC, ...

Question

At $1200^{o} C$ , the following equilibrium is established between chlorine atoms & molecule.

$C l_{2} (g) ⇌ 2 C l (g)$

The composition of the equilibrium mixture may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at $1200^{o} C$ and 1 atm pressure the mixture effuses 1.16 times as fast as krypton effuses under the same condition.
Calculate the equilibrium constant $K_{c}$ :

3.1 \times 10^{5}

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6.3 \times 10^{4}

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12.6 \times 10^{5}

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None of these

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Solution

The correct option is B $6.3 \times 10^{4}$
$\frac{r_{m i x}}{r_{K r}} = \sqrt{\frac{M_{K r}}{M_{a v g}}} = 1.16$

Solving, M average $= \frac{83.8}{{1.16}^{2}}$

$\frac{71}{M_{a v g .}} = 1 + α$ .

$K_{P} = \frac{4 α^{2}}{1 - α^{2}}$

$\Rightarrow K_{C} = \frac{K_{P}}{(0.0821) (1473)} = 6.3 \times 10^{- 4}$

Hence, the correct option is $B$

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Similar questions

Q. At

1200^{\circ} C

, the following equilibrium is established between chlorine atoms and molecule:

C l_{2} (g) ⇌ 2 C l (g)

The composition of the equilibrium mixture may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at

1200^{\circ} C

and

1

atm pressure the mixture effuses

1.16

times as fast as krypton effuses under the same condition. The equilibrium constant

K_{c}

is:

Q. The composition of the equilibrium mixture

(C l_{2} ⇌ 2 C l)

, which is attained at

1200^{\circ} C

is determined by measuring the rate of effusion through a pinhole. It is observed that a 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms.

[atomic weight of Kr is 84].

Q. At

800^{o} C

, the following equilibrium is established as:

F_{2} (g) ⇌ 2 F (g)

The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at

800^{o} C

and

1

atm mixture effuses

1.6

times as fast as

{S O}_{2}

effuses under the similar conditions. (At.wt. of

F = 19

). What is the value of

K_{p}

(in atm)?

Q. The composition of the equilibrium mixture (

C l_{2} ⇌ 2 C l

), which is attained at

1200^{o} C

, is detrmined by measuring the rate of effusion through a pin-hole. It is observed that at

1.80 mm

Hg pressure, the mixture effuses

1.183

times as fast as krypton effuses under the same conditions. Calculate the fraction of the chlorine molecules dissociated into atoms. (Given : Relative atomic mass of

K r = 84

(1.183)^{2} = 1.4

)

Q. At

800 ° C

, the following equilibrium is established as:

F_{2} (g) ⇌ 2 F (g)

The composition of the equilibrium may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at

800 ° C

and

1 atm

the mixture effuses

1.6

times as fast as

S O_{2}

effuses under similar conditions

(A t . w t . o f F = 19)

. The value of

K_{P}

for the equilibrium is

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The correct option is B 6.3×104rmixrKr=√MKrMavg=1.16Solving, M average =83.81.16271Mavg.=1+α.KP=4α21−α2⇒KC=KP(0.0821)(1473)=6.3×10−4Hence, the correct option is B