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Avogadro's Law & Gay Lussac's Law

A 100 mL flas...

Question

A 100 mL flask of $O_{2}$ (g) at 1.0 atm and 400 K is connected to a 300 mL flask containing NO(g) at 1.5 atm and 400 K by means of a narrow tube of negligible volume to form $N O_{2}$ . $N O_{2}$ is then partially dimerized to form $N_{2} O_{4}$ (g) according to the following equation:
$2 N O_{2} ⇌ N_{2} O_{4}$
This equilibrium mixture was then effused and the effusion rate was compared to the effusion rate of $C H_{4}$ (g) under identical conditions. It was found that the effusion rate of $C H_{4}$ gas was 1.667 times that for the equilibrium mixture.
Mark the correct options about this setup.
(Take R= $\frac{1}{12} l a t m / m o l K$ )

A

Total mass of the mixture after dimerisation is 501 mg

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B

Extent of dimerisation is 53.75%

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C

Dimerisation is not possible with these conditions

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D

Total amount of NO before dimeristion is 7.5 mmol

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Solution

The correct options are
A Total mass of the mixture after dimerisation is 501 mg
D Total amount of NO before dimeristion is 7.5 mmol
$Moles of O_{2} = \frac{P V}{R T} = \frac{1 \times 12 \times 100 \times 10^{- 3}}{1 \times 400} = 3 m m o l Moles of N O = \frac{P V}{R T} = \frac{(1.5 \times 12 \times 300 \times 10^{- 3}}{1 \times 400} = 13.5 m m o l O_{2} + 2 N O \to 2 N O_{2}$

3 mmoles of $O_{2}$ will need $2 \times 3 = 6 m m o l$ .
Hence the limiting reagent is oxygen.
∴ Moles of $N O_{2}$ formed = 6 mmol. Composition of mixture before dimerization;
$O_{2}$ = 0
$N O_{2}$ = 6 mmol
$N O = (13.5 - 6) = 7.5 m m o l$
Total= 13.5 mmol
Let x mmoles of $N O_{2}$ form the dimerized $N_{2} O_{4}$ . Hence, composition of equilibrium mixture after dimerization:
$N O_{2} = 6 - x m m o l N O = 7.5 m m o l N_{2} O_{4} = \frac{x}{2} m m o l T o t a l = \frac{13.5 - x}{2} m m o l$
Total mass of mixture before dimerization = mass of mixture after dimerization
= 6×46 + 7.5 × 30 = 501 mg
Avg molar mass after dimerization
$M_{M I X} = \frac{Mass of eq. mix. dimer}{Total moles} \Rightarrow M_{M I X} = \frac{501}{\frac{13.5 - x}{2}}$
By Graham’s Law,
$\frac{r_{C H_{4}}}{r_{M I X}} = (\frac{M_{M I X}}{M_{C H_{4}}})^{\frac{1}{2}}$
$M_{M I X} = (\frac{r_{C H_{4}}}{r_{m i x}})^{2} \times M_{C H_{4}}$

$M_{M I X} = \frac{25}{9 \times 16} = \frac{400}{9}$

$\frac{501}{\frac{13.5 - x}{2}} = \frac{400}{9}$
$x = 4.455 m m o l$
∴ Extent of dimerization $= 4.455 / 6 = 74.25$ %

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Similar questions

100 m l

vessel containing

O_{2} (g)

1.0 a t m

400 K

is connected to a

300 m l

vessel containing

N O (g), 1.5 a t m

400 K

by means of a narrow tube of negligible volume where gases react to form

N O_{2}

. Final pressure of mixture will be:

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

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)

K_{P}

for the equilibrium is

Q. Calculate relative rate of effusion of

S O_{2}

C H_{4}

under given condition:

( i ) Under similar condition of pressure & temperature
( ii ) Through a container containing

S O_{2}

C H_{4}

in 3 : 2 mass ratio
( iii ) If the mixture obtained by effusing out a mixture

(n_{S O_{2}} / n_{C H_{4}} = 8 / 1)

for three effusing steps.

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}

:

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

1

atm pressure the mixture effuses

1.16

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

K_{c}

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