What is `K_(c )` for the following 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) hArr 2SO_(3)(g)`

Explanation

To find the equilibrium constant \( K_c \) for the given reaction:

**Step 1: Write the balanced chemical equation.**
The balanced equation is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

**Step 2: Write the expression for \( K_c \).**
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]
For the reaction, this becomes:
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]} \]

**Step 3: Substitute the equilibrium concentrations into the expression.**
Given the equilibrium concentrations:
- \([SO_2] = 0.60 \, M\)
- \([O_2] = 0.82 \, M\)
- \([SO_3] = 1.90 \, M\)

Substituting these values into the \( K_c \) expression:
\[ K_c = \frac{(1.90)^2}{(0.60)^2 \cdot (0.82)} \]

**Step 4: Calculate the values.**
First, calculate the numerator:
\[ (1.90)^2 = 3.61 \]

Next, calculate the denominator:
\[ (0.60)^2 = 0.36 \]
Then multiply by \([O_2]\):
\[ 0.36 \cdot 0.82 = 0.2952 \]

Now, substitute these values back into the \( K_c \) expression:
\[ K_c = \frac{3.61}{0.2952} \]

**Step 5: Perform the final calculation.**
Calculating the final value:
\[ K_c \approx 12.22 \]

**Step 6: Determine the units of \( K_c \).**
Since \( K_c \) is a ratio of concentrations, the units will be:
\[ K_c = \text{M}^{-1} \quad \text{(or mole}^{-1} \text{liter)} \]

Thus, the final answer is:
\[ K_c \approx 12.22 \, \text{M}^{-1} \]

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