`H_(2)S_((g))` intially at a pressure of 10 atm anda temperature of 800K, dissociates as `2H_(2)S_((g)) hArr 2H_(2) +S_(2(g))` At equilibrium, the partial pressure of S2 vapour is 0.02 atm. Thus, `K_(p)` is

Options

A. `3.23 xx 10^(-7)`
B. `6.45 xx 10^(-7)`
C. `1.55 xx 10^(6)`
D. `6.2 xx 10^(-7)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of hydrogen sulfide (H₂S) is given by the equation:
\[ 2 \text{H}_2\text{S} (g) \rightleftharpoons 2 \text{H}_2 (g) + \text{S}_2 (g) \]

### Step 2: Define initial conditions
Initially, we have:
- Pressure of H₂S = 10 atm
- Pressure of H₂ = 0 atm
- Pressure of S₂ = 0 atm

### Step 3: Define changes in pressure at equilibrium
Let \( x \) be the change in pressure of H₂S that dissociates. According to the stoichiometry of the reaction:
- For every 2 moles of H₂S that dissociate, 2 moles of H₂ and 1 mole of S₂ are produced.
- Thus, at equilibrium, the changes in pressure will be:
- Pressure of H₂S = \( 10 - x \)
- Pressure of H₂ = \( 0 + x \)
- Pressure of S₂ = \( 0 + \frac{x}{2} \)

### Step 4: Use the given information
We are given that the partial pressure of S₂ at equilibrium is 0.02 atm. Therefore:
\[ \frac{x}{2} = 0.02 \]
From this, we can find \( x \):
\[ x = 0.04 \text{ atm} \]

### Step 5: Calculate the equilibrium pressures
Now we can find the equilibrium pressures:
- Pressure of H₂S = \( 10 - x = 10 - 0.04 = 9.96 \text{ atm} \)
- Pressure of H₂ = \( x = 0.04 \text{ atm} \)
- Pressure of S₂ = \( 0.02 \text{ atm} \) (as given)

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{H}_2})^2 \cdot (P_{\text{S}_2)} }{(P_{\text{H}_2\text{S}})^2} \]

### Step 7: Substitute the equilibrium pressures into the \( K_p \) expression
Substituting the equilibrium pressures:
\[ K_p = \frac{(0.04)^2 \cdot (0.02)}{(9.96)^2} \]

### Step 8: Calculate \( K_p \)
Calculating the values:
- \( (0.04)^2 = 0.0016 \)
- \( (9.96)^2 \approx 99.2 \)

Now substituting these values into the equation:
\[ K_p = \frac{0.0016 \cdot 0.02}{99.2} \]
\[ K_p = \frac{0.000032}{99.2} \approx 3.23 \times 10^{-7} \]

### Final Answer
Thus, the value of \( K_p \) is approximately \( 3.23 \times 10^{-7} \).

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