For equilibrium reaction at 1900K temperature and 1 atm pressure
`Cl_2(g) harr 2Cl(g)`
at equilibrium both `Cl_2` & Cl have equal number of moles, if the value of `K_p` is `a xx 10^(-1)` then value of a is

Answer

To solve the equilibrium reaction \( \text{Cl}_2(g) \rightleftharpoons 2\text{Cl}(g) \) at a temperature of 1900 K and a pressure of 1 atm, where at equilibrium both \( \text{Cl}_2 \) and \( \text{Cl} \) have equal number of moles, we can follow these steps:

### Step 1: Define the equilibrium condition
At equilibrium, let the number of moles of \( \text{Cl}_2 \) be \( n \) and the number of moles of \( \text{Cl} \) be \( n \). Therefore, the total number of moles at equilibrium will be \( n + 2n = 3n \).

### Step 2: Calculate the partial pressures
Given that the total pressure \( P \) is 1 atm, we can express the partial pressures of the gases in terms of their mole fractions:
- The partial pressure of \( \text{Cl}_2 \) is given by:
\[ P_{\text{Cl}_2} = \frac{n}{3n} \times P = \frac{1}{3} \, \text{atm} \]
- The partial pressure of \( \text{Cl} \) is given by:
\[ P_{\text{Cl}} = \frac{2n}{3n} \times P = \frac{2}{3} \, \text{atm} \]

### Step 3: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{Cl}})^2}{P_{\text{Cl}_2}} = \frac{\left(\frac{2}{3}\right)^2}{\frac{1}{3}} \]

### Step 4: Simplify the expression
Calculating \( K_p \):
\[ K_p = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{9} \times 3 = \frac{4}{3} \]

### Step 5: Express \( K_p \) in the required form
We need to express \( K_p \) in the form \( a \times 10^{-1} \):
\[ K_p = \frac{4}{3} \approx 1.33 \]
To express this in the form \( a \times 10^{-1} \):
\[ 1.33 = 0.133 \times 10^{1} \]
Thus, \( a = 4 \).

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

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