For the reaction `F_2 Leftrightarrow 2F,` calculate the degree of dissociation and density of fluorine at 4 atm and 1000 K, when `K_p= 1.4 xx 10^(-2)" atm."`
If `K_p (760^@)=2 xx 10^(-5)` atm and `K_p (960^@)= 4 xx 10^(-2)` atm, calculate `triangleH^@` for the dissociation of fluorine.

Answer

To solve the problem step by step, we will first calculate the degree of dissociation (α) of fluorine (F₂) at the given conditions and then find the density of fluorine gas. Finally, we will calculate the enthalpy change (ΔH) for the dissociation of fluorine using the provided Kp values at different temperatures.

### Step 1: Write the equilibrium expression

For the reaction:
\[ F_2 \rightleftharpoons 2F \]

The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_F)^2}{(P_{F_2})} \]

Where:
- \( P_F \) is the partial pressure of fluorine atoms (F).
- \( P_{F_2} \) is the partial pressure of fluorine molecules (F₂).

### Step 2: Define the degree of dissociation

Let the initial pressure of F₂ be \( P_0 \) (which is 4 atm). At equilibrium, if α is the degree of dissociation, then:
- The pressure of F₂ at equilibrium will be \( P_{F_2} = P_0 (1 - \alpha) \)
- The pressure of F at equilibrium will be \( P_F = 2 P_0 \alpha \)

### Step 3: Substitute into the equilibrium expression

Substituting these into the equilibrium expression gives:
\[ K_p = \frac{(2P_0 \alpha)^2}{P_0 (1 - \alpha)} \]

This simplifies to:
\[ K_p = \frac{4P_0^2 \alpha^2}{P_0 (1 - \alpha)} \]
\[ K_p = \frac{4P_0 \alpha^2}{(1 - \alpha)} \]

### Step 4: Substitute known values

Given \( K_p = 1.4 \times 10^{-2} \) atm and \( P_0 = 4 \) atm:
\[ 1.4 \times 10^{-2} = \frac{4 \times 4 \alpha^2}{(1 - \alpha)} \]
\[ 1.4 \times 10^{-2} = \frac{16 \alpha^2}{(1 - \alpha)} \]

### Step 5: Rearranging the equation

Rearranging gives:
\[ 1.4 \times 10^{-2} (1 - \alpha) = 16 \alpha^2 \]
\[ 1.4 \times 10^{-2} - 1.4 \times 10^{-2} \alpha = 16 \alpha^2 \]
\[ 16 \alpha^2 + 1.4 \times 10^{-2} \alpha - 1.4 \times 10^{-2} = 0 \]

### Step 6: Solve the quadratic equation

Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where:
- \( a = 16 \)
- \( b = 1.4 \times 10^{-2} \)
- \( c = -1.4 \times 10^{-2} \)

Calculating the discriminant:
\[ D = (1.4 \times 10^{-2})^2 - 4 \times 16 \times (-1.4 \times 10^{-2}) \]

Calculating \( D \):
\[ D = 1.96 \times 10^{-4} + 0.896 \]
\[ D = 0.896196 \]

Now substituting back into the quadratic formula:
\[ \alpha = \frac{-1.4 \times 10^{-2} \pm \sqrt{0.896196}}{32} \]

Calculating the positive root (since α must be positive):
\[ \alpha \approx \frac{-1.4 \times 10^{-2} + 0.946}{32} \]
\[ \alpha \approx 0.0295 \]

### Step 7: Calculate the density of fluorine gas

Density (ρ) can be calculated using the ideal gas equation:
\[ \rho = \frac{P \cdot M}{R \cdot T} \]
Where:
- \( P = 4 \) atm
- \( M \) (molar mass of F₂) = 38 g/mol
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 1000 \) K

Substituting the values:
\[ \rho = \frac{4 \cdot 38}{0.0821 \cdot 1000} \]
\[ \rho \approx \frac{152}{82.1} \]
\[ \rho \approx 1.85 \text{ g/L} \]

### Step 8: Calculate ΔH using Kp values at different temperatures

Using the van 't Hoff equation:
\[ \ln \left( \frac{K_{p2}}{K_{p1}} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

Where:
- \( K_{p1} = 2 \times 10^{-5} \) atm at \( T_1 = 760 \) °C = 1033 K
- \( K_{p2} = 4 \times 10^{-2} \) atm at \( T_2 = 960 \) °C = 1233 K

Substituting the values:
\[ \ln \left( \frac{4 \times 10^{-2}}{2 \times 10^{-5}} \right) = -\frac{\Delta H^\circ}{0.0821} \left( \frac{1}{1233} - \frac{1}{1033} \right) \]

Calculating the left side:
\[ \ln(2000) \approx 7.6 \]

Calculating the right side:
\[ \frac{1}{1233} - \frac{1}{1033} \approx -0.000161 \]

Now substituting back:
\[ 7.6 = -\frac{\Delta H^\circ}{0.0821} \times (-0.000161) \]

Solving for \( \Delta H^\circ \):
\[ \Delta H^\circ \approx \frac{7.6 \times 0.0821}{0.000161} \]
\[ \Delta H^\circ \approx 3.76 \times 10^3 \text{ J/mol} \]

### Final Answers:
1. Degree of dissociation (α) ≈ 0.0295
2. Density of fluorine gas ≈ 1.85 g/L
3. Enthalpy change (ΔH) ≈ 3760 J/mol