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Question

At 1200∘C, the following equilibrium is established between chlorine atoms and molecule:

Cl2(g)⇌2Cl(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∘C and 1 atm pressure the mixture effuses 1.16 times as fast as krypton effuses under the same condition. The equilibrium constant Kc is:

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Solution

The correct option is **B** 6.71×10−4

The rate of effusion is inversely proportional to the molecular weight.

The molecular weight of krypton is 83.8.

1.161=√83.8M

M=62.27

Thus, the average molecular weight of the mixture of Cl2 and Cl is 62.27.

Let x be the mole fraction of Cl2. The mole fraction of Cl will be 1−x.

Hence, 71x+35.5(1−x)=62.27

x=0.754

The mole fractions of Cl2 and Cl are 0.754 and 0.246 respectively. They are also equal to the partial pressures as total pressure is 1 atm.

Kp=PCl2PCl2=0.24620.754=0.08

As Δn=1, KP=Kc(RT)

Thus, 0.08=Kc(0.082×1473)

Kc=6.71×10−4

The rate of effusion is inversely proportional to the molecular weight.

The molecular weight of krypton is 83.8.

1.161=√83.8M

M=62.27

Thus, the average molecular weight of the mixture of Cl2 and Cl is 62.27.

Let x be the mole fraction of Cl2. The mole fraction of Cl will be 1−x.

Hence, 71x+35.5(1−x)=62.27

x=0.754

The mole fractions of Cl2 and Cl are 0.754 and 0.246 respectively. They are also equal to the partial pressures as total pressure is 1 atm.

Kp=PCl2PCl2=0.24620.754=0.08

As Δn=1, KP=Kc(RT)

Thus, 0.08=Kc(0.082×1473)

Kc=6.71×10−4

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