Chlorine molecules are 10% dissociated at 975 K at a pressure of 1-0 atm (1.0% of the pressure is due to Cl atoms).
`Cl_(2) (g) Leftrightarrow 2Cl`
Calculate `K_p and K_c.`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction and Initial Conditions
The reaction given is:
\[ \text{Cl}_2 (g) \leftrightarrow 2\text{Cl} (g) \]

At 975 K and a total pressure of 1.0 atm, we know that 10% of the chlorine molecules are dissociated. This means that 10% of the initial moles of Cl₂ have converted into Cl atoms.

### Step 2: Calculate the Initial and Final Moles
Let's assume we start with 1 mole of Cl₂.

- Initial moles of Cl₂ = 1 mole
- Moles of Cl₂ dissociated = 10% of 1 mole = 0.1 moles
- Moles of Cl₂ remaining = 1 - 0.1 = 0.9 moles
- Moles of Cl formed = 2 * 0.1 = 0.2 moles (since each mole of Cl₂ produces 2 moles of Cl)

### Step 3: Calculate the Partial Pressures
The total pressure is given as 1.0 atm. The partial pressures can be calculated as follows:

- Partial pressure of Cl₂:
\[ P_{\text{Cl}_2} = \frac{\text{moles of Cl}_2}{\text{total moles}} \times \text{total pressure} \]
\[ P_{\text{Cl}_2} = \frac{0.9}{0.9 + 0.2} \times 1.0 = \frac{0.9}{1.1} \times 1.0 \approx 0.818 \text{ atm} \]

- Partial pressure of Cl:
\[ P_{\text{Cl}} = \frac{\text{moles of Cl}}{\text{total moles}} \times \text{total pressure} \]
\[ P_{\text{Cl}} = \frac{0.2}{1.1} \times 1.0 \approx 0.182 \text{ atm} \]

### Step 4: Write the Expression for Kp
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{Cl}})^2}{P_{\text{Cl}_2}} \]

Substituting the values we calculated:
\[ K_p = \frac{(0.182)^2}{0.818} \]

### Step 5: Calculate Kp
Calculating \( K_p \):
\[ K_p = \frac{0.033124}{0.818} \approx 0.0405 \]

### Step 6: Calculate Kc
To find \( K_c \), we can use the relationship between \( K_p \) and \( K_c \):
\[ K_p = K_c (RT)^{\Delta n} \]
Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 975 \, \text{K} \)
- \( \Delta n = 2 - 1 = 1 \)

Rearranging gives us:
\[ K_c = \frac{K_p}{RT} \]

Substituting the values:
\[ K_c = \frac{0.0405}{(0.0821)(975)} \]

### Step 7: Calculate Kc
Calculating \( K_c \):
\[ K_c = \frac{0.0405}{80.29575} \approx 0.000504 \]

### Final Results
- \( K_p \approx 0.0405 \)
- \( K_c \approx 0.000504 \)