For the reaction, `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))` What is `K_(c)` when the equilibrium concentration of `[SO_(2)]=0.60M,[O_(2)]=0.82Mand[SO_(3)]=1.90M` ?

Options

A. (a) `12.229Lmol^(-1)`
B. (b) `24.5Lmol^(-1)`
C. (c) `36.0Lmol^(-1)`
D. (d) `2.67xx10^(3)Lmol^(-1)`

Explanation

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

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

given the equilibrium concentrations of the species, we can follow these steps:

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

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

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

where:
- \([SO_3]\) is the equilibrium concentration of sulfur trioxide,
- \([SO_2]\) is the equilibrium concentration of sulfur dioxide,
- \([O_2]\) is the equilibrium concentration of oxygen.

### Step 2: Substitute the given concentrations into the \( K_c \) expression

From the problem, we have the following equilibrium concentrations:
- \([SO_2] = 0.60 \, M\)
- \([O_2] = 0.82 \, M\)
- \([SO_3] = 1.90 \, M\)

Now, substituting these values into the \( K_c \) expression:

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

### Step 3: Calculate the values

First, calculate \((1.90)^2\):

\[ (1.90)^2 = 3.61 \]

Next, calculate \((0.60)^2\):

\[ (0.60)^2 = 0.36 \]

Now, substitute these values back into the equation:

\[ K_c = \frac{3.61}{0.36 \times 0.82} \]

Calculate \(0.36 \times 0.82\):

\[ 0.36 \times 0.82 = 0.2952 \]

Now substitute this back into the equation for \( K_c \):

\[ K_c = \frac{3.61}{0.2952} \approx 12.22 \]

### Step 4: State the final answer with units

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

\[ K_c \approx 12.22 \, \text{L/mol} \]