The value of `K_p` is `1 xx 10^(-3)atm^(-1)` at `25^@C` for the reaction `2NOCI+CI_2 hArr 2NOCI`. A flask contains NO at 0.02 atm and `25^@C` . Calculate the moles of `CI_2` that must be added if 1% of NO is to be converted to NOCI at equilibrium . The volume of the flask is such that 0.2 moles of the gas produce 1atm pressure at `25^@C` (ignore the probable association of NO to `N_2O_2`)

Answer

To solve the given problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{NO} + \text{Cl}_2 \rightleftharpoons 2 \text{NOCl} \]

### Step 2: Identify initial conditions
We know:
- The initial pressure of NO, \( P_{\text{NO}} = 0.02 \, \text{atm} \)
- The initial pressure of Cl₂, \( P_{\text{Cl}_2} = P \, \text{atm} \) (unknown)
- The initial pressure of NOCl, \( P_{\text{NOCl}} = 0 \, \text{atm} \)

### Step 3: Determine equilibrium conditions
We want to convert 1% of NO to NOCl:
- 1% of 0.02 atm = \( 0.0002 \, \text{atm} \)
- At equilibrium:
- \( P_{\text{NO}} = 0.02 - 0.0002 = 0.0198 \, \text{atm} \)
- \( P_{\text{NOCl}} = 0 + 0.0002 = 0.0002 \, \text{atm} \)
- \( P_{\text{Cl}_2} = P + 0.0001 \, \text{atm} \) (since 0.0002 atm of NO reacts with 0.0001 atm of Cl₂)

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 (P_{\text{Cl}_2})} \]
Substituting the known values:
\[ 1 \times 10^{-3} = \frac{(0.0002)^2}{(0.0198)^2 (P + 0.0001)} \]

### Step 5: Solve for \( P \)
Substituting the values:
\[ 1 \times 10^{-3} = \frac{4 \times 10^{-8}}{(0.00039204)(P + 0.0001)} \]
Cross-multiplying gives:
\[ 1 \times 10^{-3} \cdot (0.00039204)(P + 0.0001) = 4 \times 10^{-8} \]
\[ 0.00039204(P + 0.0001) = 4 \times 10^{-5} \]
\[ P + 0.0001 = \frac{4 \times 10^{-5}}{0.00039204} \]
Calculating the right-hand side:
\[ P + 0.0001 \approx 0.10202 \]
Thus,
\[ P \approx 0.10202 - 0.0001 \approx 0.10201 \, \text{atm} \]

### Step 6: Calculate the volume of the flask
Using the ideal gas law:
\[ PV = nRT \]
Given that 0.2 moles produce 1 atm at 25°C (298 K):
\[ V = \frac{nRT}{P} = \frac{0.2 \times 0.0821 \times 298}{1} = 4.887 \, \text{L} \]

### Step 7: Calculate moles of Cl₂ needed
Using the ideal gas law again for the Cl₂:
\[ n = \frac{PV}{RT} \]
Substituting \( P = 0.10201 \, \text{atm} \):
\[ n = \frac{0.10201 \times 4.887}{0.0821 \times 298} \approx 0.204 \, \text{moles} \]

### Final Answer
The moles of Cl₂ that must be added is approximately:
\[ \boxed{0.204 \, \text{moles}} \]