At 1990 K and 1 atm pressure, there are equal number of `Cl_2`, molecules and Cl atoms in the reaction mixture. The value `K_(rho)` for the reaction `Cl_(2(g))hArr2Cl_(g)` under the above conditions is `"x"xx10^(-1)`. The value of x is `"_____"`
(Rounded of to the nearest integer)

Answer

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ Cl_2(g) \rightleftharpoons 2Cl(g) \]

Given that at equilibrium, there are equal numbers of \( Cl_2 \) molecules and \( Cl \) atoms, we can denote the number of moles of \( Cl_2 \) as \( n \) and the number of moles of \( Cl \) as \( 2n \) (since each \( Cl_2 \) produces 2 \( Cl \) atoms).

### Step 1: Define the equilibrium concentrations
Let the number of moles of \( Cl_2 \) at equilibrium be \( n \) and the number of moles of \( Cl \) be \( 2n \).

### Step 2: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed as:

\[ K_p = \frac{(P_{Cl})^2}{P_{Cl_2}} \]

Where \( P_{Cl} \) and \( P_{Cl_2} \) are the partial pressures of \( Cl \) and \( Cl_2 \) respectively.

### Step 3: Calculate the partial pressures
Since the total pressure \( P \) is given as 1 atm, we can express the partial pressures in terms of \( n \):

- The partial pressure of \( Cl_2 \) is given by:

\[ P_{Cl_2} = \frac{nRT}{V} \]

- The partial pressure of \( Cl \) is given by:

\[ P_{Cl} = \frac{2nRT}{V} \]

### Step 4: Substitute the partial pressures into the \( K_p \) expression
Substituting the expressions for partial pressures into the \( K_p \) expression gives:

\[ K_p = \frac{\left(\frac{2nRT}{V}\right)^2}{\frac{nRT}{V}} = \frac{4n^2(RT)^2/V^2}{nRT/V} = \frac{4n(RT)}{V} \]

### Step 5: Relate \( n \) to total pressure
Since we have equal numbers of \( Cl_2 \) and \( Cl \) at equilibrium, we can express \( n \) in terms of the total pressure. The total number of moles at equilibrium is \( n + 2n = 3n \). Therefore, we have:

\[ P = \frac{3nRT}{V} \]

Given that \( P = 1 \) atm, we can solve for \( n \):

\[ n = \frac{PV}{3RT} \]

### Step 6: Substitute \( n \) back into the \( K_p \) expression
Substituting \( n \) into the \( K_p \) expression gives:

\[ K_p = \frac{4 \cdot \frac{PV}{3RT} \cdot RT}{V} = \frac{4P}{3} \]

### Step 7: Substitute the given pressure
Substituting \( P = 1 \) atm into the equation:

\[ K_p = \frac{4 \cdot 1}{3} = \frac{4}{3} \]

### Step 8: Express \( K_p \) in the required form
The problem states that \( K_p = x \times 10^{-1} \). Therefore, we can express \( \frac{4}{3} \) as:

\[ K_p = \frac{4}{3} = 1.33 \approx 1.3 \times 10^{0} \]

Thus, \( x = 4 \) when rounded to the nearest integer.

### Final Answer
The value of \( x \) is **4**.

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