For the reaction `2SO_2 + O_2 to 2SO_3`, the unit of equilibrium constant is :

Options

A. `L mol^(_1)`
B. `J mol^(-1)`
C. `mol L^(-1)`
D. `[L mol^(-1)]^2`

Explanation

To determine the unit of the equilibrium constant for the reaction \(2SO_2 + O_2 \rightleftharpoons 2SO_3\), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (K)
The equilibrium constant \(K\) for a reaction is given by the ratio of the concentrations (active masses) of the products raised to the power of their coefficients to the concentrations of the reactants raised to the power of their coefficients.

For the reaction:
\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

The expression for the equilibrium constant \(K\) can be written as:
\[ K = \frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]} \]

### Step 2: Identify the units of concentration
The concentration of a substance in a solution is expressed in moles per liter (mol/L). Therefore, the units for the concentrations in the equilibrium expression are:
- \([SO_3]\) has units of mol/L
- \([SO_2]\) has units of mol/L
- \([O_2]\) has units of mol/L

### Step 3: Substitute the units into the equilibrium expression
Now, substituting the units into the equilibrium expression:
\[ K = \frac{(mol/L)^2}{(mol/L)^2 \cdot (mol/L)} \]

### Step 4: Simplify the units
Now, we simplify the expression:
\[ K = \frac{(mol^2/L^2)}{(mol^2/L^2) \cdot (mol/L)} = \frac{(mol^2/L^2)}{(mol^3/L^3)} = \frac{L^3}{mol^3} \cdot \frac{L^2}{L^2} = \frac{L}{mol} \]

### Conclusion
Thus, the unit of the equilibrium constant \(K\) for the reaction \(2SO_2 + O_2 \rightleftharpoons 2SO_3\) is:
\[ \text{Unit of } K = \frac{L}{mol} \]