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Derivation of Kp and Kc

What is Kc fo...

Question

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M$ and $[S O_{3}] = 1.90 M$ ?
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$

12.24 M^{- 1}

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1.23 M^{- 1}

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22.23 M^{- 1}

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32.23 M^{- 1}

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Solution

The correct option is A $12.24 M^{- 1}$
The equilibrium constant $(K_{c})$ for the give reaction is:
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$= \frac{(1.90)^{2} M^{2}}{(0.60)^{2} (0.821) M^{3}}$
$= 12.239 M^{- 1}$ (approximately)
Hence, $K_{c}$ for the equilibrium is $12.239 M^{- 1}$ .

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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)

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_{e}

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

[S O_{2}] = 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}

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

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