Question

Cl2(g)2cl(g) at 400 c the composition of equilibrium may be determined by measuring the rate by effusion of the mixture through pin hole it is found that at 400 c and 1 atm pressure the mixture effuses 1.2 times as fast as so2 effuses under similar conditions. What is the value of kp (atm). (at. Weight of cl = 35.5, s = 32, o = 16)

Answer 1

Given info : the rate of effusion of the mixture is 1.2 times as fast as the rate of effusion of SO₂ under similar conditions.

We have to find the equilibrium constant, kp (atm).

Solution : from Graham's law of diffusion, rate of diffusion is inversely proportional to square root of molecular weight.

i.e., rate ∝ 1/√M ⇒r₁/r₂ = √(M₂/M₁)

molar mass of SO₂ = 64 g/mol

Let molar mass of mixture is M

So, 1.2 = √(64/M)

⇒1.2 = 8/√(M)

⇒√M = 8/1.2 = 6.67

⇒M = 44.4 g/mol

Now. Cl₂ ⇔2Cl

At t = 0, n 0

at eql n(1 - α) 2nα

so, mole fraction of Cl₂ = (1 - α)/(1 + α) and mole fraction of Cl = 2α/(1 + α)

Now , 44.4 = (1 - α)/(1 + α) × 71 + 2α/(1 + α) × 35.5

⇒44.4(1 + α) = 71 - 71α + 71α

⇒44.4 + 44.4α = 71

⇒44.4α = 26.6

⇒α = 0.6

now Kp = [pCl]²/[PCl₂]

= [2α/(1 + α)p]²/[(1 - α)/(1 + α)p]

= [2 × 0.6 × 1/(1 + 0.6]²/[(1 - 0.6) × 1/(1 + 0.6)]

= [1.2/1.6]²/(0.4/1.6)

= (9/16)/(1/4)

= 9/4 = 2.25 atm

Therefore equivalent constant, kp = 2.25 atm