For the reaction , `2 NO (g) + Cl_(2) (g) hArr 2 NOCl (g)` and the following info is given:
`p_(NOCl)`= `0.32 atm`
`p_(NO)`= `0.22 atm`
`p_(Cl_(2))`= `0.11 atm`

then find `K_(p)`

Explanation

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

\[ 2 \text{NO} (g) + \text{Cl}_2 (g) \rightleftharpoons 2 \text{NOCl} (g) \]

we can use the formula for \( K_p \):

\[ K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Cl}_2})} \]

### Step 1: Identify the partial pressures
From the information given:
- \( P_{\text{NOCl}} = 0.32 \, \text{atm} \)
- \( P_{\text{NO}} = 0.22 \, \text{atm} \)
- \( P_{\text{Cl}_2} = 0.11 \, \text{atm} \)

### Step 2: Substitute the values into the \( K_p \) equation
Now we substitute the values into the \( K_p \) equation:

\[ K_p = \frac{(0.32)^2}{(0.22)^2 \cdot (0.11)} \]

### Step 3: Calculate the numerator
Calculate \( (0.32)^2 \):

\[ (0.32)^2 = 0.1024 \]

### Step 4: Calculate the denominator
Calculate \( (0.22)^2 \):

\[ (0.22)^2 = 0.0484 \]

Now, multiply this by \( 0.11 \):

\[ 0.0484 \cdot 0.11 = 0.005324 \]

### Step 5: Calculate \( K_p \)
Now we can calculate \( K_p \):

\[ K_p = \frac{0.1024}{0.005324} \approx 19.23 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is approximately:

\[ K_p \approx 19.23 \, \text{atm}^{-1} \]

---