Find out the units of `K_c and K_p` for the following equilibrium reactions :
` 2SO_2(g)+ O_2(g) hArr 2SO_3(g)`

Answer

To find the units of \( K_c \) and \( K_p \) for the equilibrium reaction:

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

we will follow these steps:

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

The equilibrium constant \( K_c \) is expressed in terms of the molar concentrations of the products and reactants. The general formula for \( K_c \) is:

\[ K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}} \]

For our reaction, this becomes:

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

### Step 2: Substitute the units for concentrations

The unit of concentration is molarity, which is expressed as moles per liter (mol/L). Thus, we can substitute the units into the expression for \( K_c \):

\[ K_c = \frac{(\text{mol/L})^2}{(\text{mol/L})^2 \cdot (\text{mol/L})} \]

### Step 3: Simplify the expression

Now, we simplify the expression:

\[ K_c = \frac{\text{mol}^2/\text{L}^2}{\text{mol}^2/\text{L}^2 \cdot \text{mol}/\text{L}} = \frac{\text{mol}^2/\text{L}^2}{\text{mol}^3/\text{L}^3} \]

This simplifies to:

\[ K_c = \frac{\text{L}^3}{\text{mol}^3} \cdot \frac{\text{mol}^2}{\text{L}^2} = \frac{\text{L}}{\text{mol}} \]

Thus, the unit for \( K_c \) is:

\[ \text{Unit of } K_c = \text{L mol}^{-1} \]

### Step 4: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is expressed in terms of the partial pressures of the products and reactants. The general formula for \( K_p \) is:

\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

### Step 5: Substitute the units for pressures

The unit of pressure is typically atmospheres (atm). Thus, we can substitute the units into the expression for \( K_p \):

\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

### Step 6: Simplify the expression

Now, we simplify the expression:

\[ K_p = \frac{(\text{atm})^2}{(\text{atm})^2 \cdot (\text{atm})} = \frac{\text{atm}^2}{\text{atm}^3} \]

This simplifies to:

\[ K_p = \frac{1}{\text{atm}} = \text{atm}^{-1} \]

Thus, the unit for \( K_p \) is:

\[ \text{Unit of } K_p = \text{atm}^{-1} \]

### Final Answer

- The unit of \( K_c \) is \( \text{L mol}^{-1} \)
- The unit of \( K_p \) is \( \text{atm}^{-1} \)