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

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Solution

$2 S O_{2} (g) + O_{2} (g \leftrightarrow 2 S O_{3} (g)$
Given,conc.of $S O_{2} = [S O_{2}] = 0.6 M$
conc. of $O_{2} = [O_{2}] = 0.82 M$
conc. of $S O_{3} = [S O_{3}] = 1.90 M$
so, equilibrium\quad constant $(K c) = \frac{[S O_{3}]^{2}}{[O_{2}] [S O_{2}]^{2}}$
$= \frac{(1.9)^{2}}{(0.82 \times 0.6)}$
$= 12.2289 M^{- 1} \approx 12.23 / M$
hence, equilibrium constant=12.23/M

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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}

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)

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[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}

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

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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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