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What is $K_{C}$ for the following equilibriums which the equilibrium concentration of each substance is $[{SO}_{2}] = 0.60 M, [O_{2}] = 0.82 M$
$and [{SO}_{3}] = 1.90 M ?$
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$

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12.23 $M^{- 1}$

10.23 $M^{- 1}$

5.56 $M^{- 1}$

6.22 $M^{- 1}$

answer is A.

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Detailed Solution

$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$

$Given, [{SO}_{2}] = 0.60 M; [O_{2}] = 0.82 M and [{SO}_{3}] = 1.90 M$

$K_{C} = \frac{{[{SO}_{3}]}^{2}}{{[{SO}_{2}]}^{2} [O_{2}]} = \frac{(1.90 M)^{2}}{(0.60 M)^{2} (0.82 M)}$

$= 12.23 M^{- 1}$

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