Chemical EquilibriumHard

Question

At 127°C and 1 atm pressure, PCl5(g) is partially dissociated into PCl2(g) and Cl2(g) as PCl5(g) $\rightleftharpoons$PCl3(g) + Cl2(g). The density of the equilibrium mixture is 3.5 g/L. Percentage dissociation of PCl5 is (R = 0.08 L-atm/K-mol,

P = 31, Cl = 35.5)

Options

A.46.28
B.86.16
C.15.86
D.53.72

Solution

$M_{mix} = \frac{dRT}{P} = \frac{3.5 \times 0.08 \times 400}{1} = 112$

Now, $\alpha = \frac{M_{PCl_{5}} - M_{mix}}{(n - 1).M_{mix}} = \frac{208.5 - 112}{(2 - 1) \times 112} = 0.8616$

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