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Concentration Terms

What is K c...

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 $[{S O}_{3}] = 1.90 M$ ?
$2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)$

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Solution

$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$

$K_{c} = \frac{(1.90)^{2}}{(0.60)^{2} \times 0.82} = 12.229 L / m o l$

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Q. For the reaction,

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

K_{c}

when the equilibrium concentration of

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

[S O_{3}] = 1.90 M

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Q. For the equilibrium

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

the concentration of each substance at equilibrium is:

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

[S O_{3}] = 1.90 M

_{c}

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

[S O_{3}] = 1.90 M

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

Q. For the reaction,

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

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

?

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

[S O_{3}] = 1.90 M

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

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