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Law of Mass Action

The equilibri...

Question

The equilibrium constant ( $K_{P}$ ) for the reaction: $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is 0.13 $a t m^{- 1}$ at $830^{\circ}$ C. In one experiment 2.00 mol $S O_{2}$ and 2.00 mol $O_{2}$ were initially present in a flask. If 0.4 mol of $S O_{3}$ is present at equilibrium, find the total pressure.

A

1 atm

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B

0.5 atm

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C

2 atm

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D

4 atm

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Solution

The correct option is A 1 atm
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
2 2 0
1.6 1.8 0.4 at eqb
total moles =3.8 mol
$∴$ , Mole fraction:
$X_{S O_{2}} = \frac{1.6}{3.8} = 0.42$
$X_{O_{2}} = \frac{1.8}{3.8} = 0.474$
$X_{S O_{3}} = \frac{0.4}{3.8} = 0.105$
$P_{S O_{3}} = X_{S O_{3}} P_{T}$
$P_{S O_{2}} = X_{S O_{2}} P_{T}$
$P_{O_{2}} = X_{O_{2}} P_{T}$
Now given,
$K_{P} = 0.13 = \frac{(P_{S O_{3}})^{2}}{P (S O_{2})^{2} . P_{O_{2}}}$ .
$\Rightarrow 0.13 = \frac{(0.105 P_{T})^{2}}{(0.42 \times P_{T})^{2} (0.479 \times P_{T})}$
$\Rightarrow P_{T} = 1 a t m$ .

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Law of Mass Action

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