`SO_(2) CI_(2)` and `CI_(2)` are introduced into a 3L vessel. Partial pressure of `SO_(2) CI_(2)` and `CI_(2)` at equilibrium are I atm and 2 atm respectively. The value of `K_(p) ` is 10 for the reaction `SO_(2) Cl_(2) (g) hArr SO_(2) (g) + Cl_(2) (g). ` The total pressure in atm at equilibrium would be ____

Answer

To solve the problem step by step, we need to analyze the given reaction and the information provided:

**Given Reaction:**
\[ \text{SO}_2\text{Cl}_2 (g) \rightleftharpoons \text{SO}_2 (g) + \text{Cl}_2 (g) \]

**Given Data:**
- Volume of the vessel = 3 L
- Partial pressure of \(\text{SO}_2\) at equilibrium, \(P_{\text{SO}_2} = 1 \, \text{atm}\)
- Partial pressure of \(\text{Cl}_2\) at equilibrium, \(P_{\text{Cl}_2} = 2 \, \text{atm}\)
- Equilibrium constant \(K_p = 10\)

**Step 1: Write the expression for \(K_p\)**
The equilibrium constant \(K_p\) for the reaction is given by:
\[ K_p = \frac{P_{\text{SO}_2} \cdot P_{\text{Cl}_2}}{P_{\text{SO}_2\text{Cl}_2}} \]

**Step 2: Substitute the known values into the \(K_p\) expression**
We know \(P_{\text{SO}_2} = 1 \, \text{atm}\) and \(P_{\text{Cl}_2} = 2 \, \text{atm}\). Thus, we can substitute these values into the equation:
\[ 10 = \frac{(1)(2)}{P_{\text{SO}_2\text{Cl}_2}} \]

**Step 3: Solve for \(P_{\text{SO}_2\text{Cl}_2}\)**
Rearranging the equation to find \(P_{\text{SO}_2\text{Cl}_2}\):
\[ P_{\text{SO}_2\text{Cl}_2} = \frac{2}{10} = 0.2 \, \text{atm} \]

**Step 4: Calculate the total pressure at equilibrium**
The total pressure at equilibrium is the sum of the partial pressures of all gases present:
\[ P_{\text{total}} = P_{\text{SO}_2\text{Cl}_2} + P_{\text{SO}_2} + P_{\text{Cl}_2} \]
Substituting the values we found:
\[ P_{\text{total}} = 0.2 \, \text{atm} + 1 \, \text{atm} + 2 \, \text{atm} = 3.2 \, \text{atm} \]

**Final Answer:**
The total pressure in atm at equilibrium would be \(3.2 \, \text{atm}\).

---