What is `K_(p)` for the equation `2Cl_((g)) hArr Cl_(2(g))` ? When the system contains equal numberof `Cl_((g))` atom and `Cl_(2(g))` molecules at 1 bar and 300 k

Answer

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

\[ 2 \text{Cl}_{(g)} \rightleftharpoons \text{Cl}_2_{(g)} \]

when the system contains equal numbers of Cl atoms and Cl2 molecules at 1 bar and 300 K, we can follow these steps:

### Step 1: Understand the relationship between moles and partial pressures
Since the system contains equal numbers of Cl atoms and Cl2 molecules, we can denote the number of moles of Cl as \( n \) and the number of moles of Cl2 as \( n \) as well.

### Step 2: Calculate the total number of moles
The total number of moles in the system will be:
\[ \text{Total moles} = n + n = 2n \]

### Step 3: Calculate the mole fractions
The mole fraction of Cl2 (\( x_{\text{Cl}_2} \)) is given by:
\[ x_{\text{Cl}_2} = \frac{\text{moles of Cl}_2}{\text{total moles}} = \frac{n}{2n} = \frac{1}{2} \]

Similarly, the mole fraction of Cl (\( x_{\text{Cl}} \)) is:
\[ x_{\text{Cl}} = \frac{\text{moles of Cl}}{\text{total moles}} = \frac{n}{2n} = \frac{1}{2} \]

### Step 4: Calculate the partial pressures
Using the mole fractions and the total pressure (1 bar), we can calculate the partial pressures:
\[ P_{\text{Cl}_2} = x_{\text{Cl}_2} \times P_{\text{total}} = \frac{1}{2} \times 1 \text{ bar} = 0.5 \text{ bar} \]

\[ P_{\text{Cl}} = x_{\text{Cl}} \times P_{\text{total}} = \frac{1}{2} \times 1 \text{ bar} = 0.5 \text{ bar} \]

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{P_{\text{Cl}_2}}{(P_{\text{Cl}})^2} \]

### Step 6: Substitute the partial pressures into the \( K_p \) expression
Substituting the values we calculated:
\[ K_p = \frac{0.5 \text{ bar}}{(0.5 \text{ bar})^2} = \frac{0.5}{0.25} = 2 \]

### Conclusion
Thus, the value of \( K_p \) for the reaction at the given conditions is:
\[ K_p = 2 \]

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