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)`

Options

A. 12.229
B. 21.229
C. 12.922
D. 21.046

Answer

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

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

we will use the equilibrium concentrations provided:

- \([SO_2] = 0.60 \, M\)
- \([O_2] = 0.82 \, M\)
- \([SO_3] = 1.90 \, M\)

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]

For our reaction, the expression becomes:

\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]} \]

### Step 2: Substitute the equilibrium concentrations into the expression

Now, we will substitute the given concentrations into the \( K_c \) expression:

\[ K_c = \frac{(1.90)^2}{(0.60)^2 \cdot (0.82)} \]

### Step 3: Calculate the values

First, calculate the numerator:

\[ (1.90)^2 = 3.61 \]

Next, calculate the denominator:

\[ (0.60)^2 = 0.36 \]
\[ 0.36 \cdot 0.82 = 0.2952 \]

Now, substitute these values back into the equation for \( K_c \):

\[ K_c = \frac{3.61}{0.2952} \]

### Step 4: Perform the final calculation

Now, divide the numerator by the denominator:

\[ K_c \approx 12.229 \]

### Conclusion

Thus, the equilibrium constant \( K_c \) for the reaction is approximately:

\[ K_c \approx 12.229 \, M^{-1} \]