Show that `K_p` for the reaction,
`2H_2S(g) hArr 2H_2(g) +S_2(g)` is given by the expression
`K_p=(alpha^3P)/((2+alpha)(1-alpha)^2)`
where `alpha` is the degree of dissociation and P is the total equilibrium pressure. Calculate `K_c` of the reaction if `alpha` at 298 k and 1 atm . pressure is 0.055.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction and Identify Variables
The reaction given is:
\[ 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \]

Here, we need to find the equilibrium constant \( K_p \) expressed in terms of the degree of dissociation \( \alpha \) and total pressure \( P \).

### Step 2: Determine the Change in Moles
In the reaction:
- Moles of reactants = 2 (from \( 2H_2S \))
- Moles of products = 2 (from \( 2H_2 \)) + 1 (from \( S_2 \)) = 3

Thus, the change in moles (\( \Delta n \)) is:
\[ \Delta n = \text{Moles of products} - \text{Moles of reactants} = 3 - 2 = 1 \]

### Step 3: Write the Expression for \( K_p \)
The expression for \( K_p \) in terms of \( \alpha \) and \( P \) is given by:
\[ K_p = \frac{\alpha^3 P}{(2 + \alpha)(1 - \alpha)^2} \]

### Step 4: Substitute Known Values
Given:
- \( \alpha = 0.055 \)
- \( P = 1 \, \text{atm} \)

Substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(0.055)^3 \cdot 1}{(2 + 0.055)(1 - 0.055)^2} \]

### Step 5: Calculate \( K_p \)
Calculating the numerator:
\[ (0.055)^3 = 0.000166375 \]

Calculating the denominator:
- \( 2 + 0.055 = 2.055 \)
- \( 1 - 0.055 = 0.945 \)
- \( (0.945)^2 = 0.893025 \)

Now, substituting these into the denominator:
\[ (2.055)(0.893025) = 1.836 \]

Now, substituting back to find \( K_p \):
\[ K_p = \frac{0.000166375}{1.836} \approx 9.06 \times 10^{-5} \]

### Step 6: Calculate \( K_c \)
Using the relation between \( K_p \) and \( K_c \):
\[ K_p = K_c \cdot RT^{\Delta n} \]

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 298 \, \text{K} \)
- \( \Delta n = 1 \)

Rearranging for \( K_c \):
\[ K_c = \frac{K_p}{RT} \]

Substituting the known values:
\[ K_c = \frac{9.06 \times 10^{-5}}{(0.0821)(298)} \]

Calculating \( RT \):
\[ RT = 0.0821 \times 298 \approx 24.4758 \]

Now substituting back to find \( K_c \):
\[ K_c = \frac{9.06 \times 10^{-5}}{24.4758} \approx 3.70 \times 10^{-6} \]

### Final Answer
Thus, the value of \( K_c \) is approximately:
\[ K_c \approx 3.70 \times 10^{-6} \]

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