Easy
What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$
(N/A) The equilibrium constant $(K_{c})$ for the given reaction is defined as:
$K_{c} = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(1.90)^{2}}{(0.60)^{2} \times (0.82)}$
$K_{c} = \frac{3.61}{0.36 \times 0.82}$
$K_{c} = \frac{3.61}{0.2952} \approx 12.23 \, M^{-1}$
Thus,the equilibrium constant $K_{c}$ is approximately $12.23 \, M^{-1}$.
$K_{c} = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(1.90)^{2}}{(0.60)^{2} \times (0.82)}$
$K_{c} = \frac{3.61}{0.36 \times 0.82}$
$K_{c} = \frac{3.61}{0.2952} \approx 12.23 \, M^{-1}$
Thus,the equilibrium constant $K_{c}$ is approximately $12.23 \, M^{-1}$.
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