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 $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M$ and $[S O_{3}] = 1.90 M$ ?

12.229 L m o l^{- 1}

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24.5 L m o l^{- 1}

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36.0 L m o l^{- 1}

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None of the above

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Solution

The correct option is A $12.229 L m o l^{- 1}$
The formula for calculating equilibrium constant $K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$K_{c} = \frac{(1.9)^{2}}{[(0.6)^{2}] \times [0.82]}$
$K_{c} = \frac{3.61}{0.36 \times 0.82}$
$K_{c} = \frac{3.61}{0.2952}$
$∴ K_{c} = 12.229 L / m o l$

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Similar questions

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)

K_{c}

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M a n d [S O_{3}] = 1.90 M

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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