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Types of Equillibrium

For the react...

Question

For the reaction, $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
What is $K_{c}$ when the equilibrium concentration of $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M a n d [S O_{3}] = 1.90 M$ ?

12.229 {L mol}^{- 1}

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24.5 {L mol}^{- 1}

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36.0 {L mol}^{- 1}

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2.67 \times 10^{3} {L mol}^{- 1}

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Solution

The correct option is A $12.229 {L mol}^{- 1}$
As We Know,
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} = \frac{(1.9)^{2}}{(0.6)^{2} (0.82)} = 12.229 {L mol}^{- 1}$
Hence, Correct Option is (a)

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Similar questions

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}

K_{c}

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

[S O_{3}] = 1.90 M

What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

[S O_{3}] = 1.90 M

_{c}

K_{c}

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

[S O_{3}] = 1.90 M

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

K_{c}

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

[S O_{3}] = 1.90 M

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

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