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

Byju's Answer

Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

The compositi...

Question

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 pin hole. It is observed that the given mixture effuses 1.5 times as fast as an unknown gas 'X' effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms.[Molecular Mass of unknown gas=90]

0.4

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

0.56

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

0.8

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

0.75

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

Open in App

Solution

The correct option is A 0.56
$\frac{{R a t e}_{1}}{{R a t e}_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{1}, M_{2}$ molar masses of gases
$1.5 = \sqrt{\frac{90}{M_{1}}}$
$M_{1} = 40$
Fraction of dissociation $= \frac{40}{70.9060}$
$= 0.564$
Molar mass of ${C l}_{2} = 70.9060$

Suggest Corrections

Similar questions

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. 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

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. 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}

Q. At

800^{\circ}

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 pin hole. It is found that at

800^{\circ}

C and 1 atm mixture effuses 1.6 times as fast as

S O_{2}

effuses under the similar conditions. (Atomic weight of F =19 amu). What is the value of

K_{p}

Join BYJU'S Learning Program

The correct option is A 0.56Rate1Rate2=√M2M1M1,M2 molar masses of gases1.5=√90M1M1=40Fraction of dissociation=4070.9060=0.564Molar mass of Cl2=70.9060

The correct option is A 0.56
$\frac{{R a t e}_{1}}{{R a t e}_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{1}, M_{2}$ molar masses of gases
$1.5 = \sqrt{\frac{90}{M_{1}}}$
$M_{1} = 40$
Fraction of dissociation $= \frac{40}{70.9060}$
$= 0.564$
Molar mass of ${C l}_{2} = 70.9060$