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Types of Equillibrium

The equilibri...

Question

The equilibrium constant $K_{p}$ for the reaction, $S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)$ is 1.125 atm at temperature T. Select the correct statements for dissociation of pure $S O_{2} C l_{2}$ (g) at equilibrium state of 2 atm equilibrium pressure and temperature T:

The degree of dissociation of

S O_{2} C l_{2}

is 0.5.

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The degree of dissociation of

S O_{2} C l_{2}

is 0.6.

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Average molar mass of final equilibrium mixture will be 84.375.

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Average molar mass of final equilibrium mixture will be 75.

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Solution

The correct options are
B The degree of dissociation of $S O_{2} C l_{2}$ is 0.6.
C Average molar mass of final equilibrium mixture will be 84.375.
Let initially, 1 mole of $S O_{2} C l_{2} (g)$ is present.
Let $α$ be the degree of dissociation.
The dissociation reaction is as shown.
$S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)$
1 0 0 ..........initial mole
$1 - α$ $α$ $α$ ..........at equilibrium
Total number of moles $= 1 - α + α + α = 2 α$
The partial pressure of a component is the product of mole fraction and total pressure P.
The equilibrium constant expression is $K_{p} = \frac{P_{S O_{2}} P_{C l_{2}}}{P_{S O_{2} C l_{2}}}$
P=2 atm and $K_{P} = 1.125 a t m$
$K p = \frac{\frac{α P}{1 + α} \times \frac{α P}{1 + α}}{\frac{(1 - α) P}{(1 + α)}} = \frac{α^{2}}{(1 - α^{2})} \times 2 = 1.125$ ;
Hence, the degree of dissociation $α = 0.6$
The equilibrium number of moles of $S O_{2} C l_{2}$ , $S O_{2}$ and $C l_{2}$ are
$1 - 0.6 = 0.4$ , 0.6 and 0.6 resprectively.
The mole fractions of $S O_{2} C l_{2}$ , $S O_{2}$ and $C l_{2}$ are
$\frac{0.4}{0.4 + 0.6 + 0.6} = 0.25$ , $\frac{0.6}{0.4 + 0.6 + 0.6} = 0.375$ and $\frac{0.6}{0.4 + 0.6 + 0.6} = 0.375$ respectively.
The molar masses of $S O_{2} C l_{2}$ , $S O_{2}$ and $C l_{2}$ are
$64 + 71$ g/mol, 64 g/mol and 71 g/mol respectively.
Average molar mass of mixture $= (64 + 71) \times 0.25 + 64 \times 0.375 + 71 \times 0.375$
$= 84.375.$
Hence, the average molar mass of the mixture is 84.375 g/mol.

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