If 0.2 mol of `H_(2)(g)` and 2.0 mol of S(s) are mixed in a 1.0 L vessel at `90^(@)C`, the partial pressure of `H_(2)S(g)` formed according to the reaction `H_(2)(g)+S(s)hArr H_(2)S(g),K_(p)` at `363K=6.78xx10^(-2)` would be

Options

A. 0.19 atm
B. 0.38 atm
C. 0.6 atm
D. `6.8xx10^(2)"atm/"(0.2xx2)`

Answer

To solve the problem of determining the partial pressure of H₂S formed from the reaction of H₂ and S in a 1.0 L vessel at 90 °C (363 K), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is:
\[ \text{H}_2(g) + \text{S}(s) \rightleftharpoons \text{H}_2\text{S}(g) \]

### Step 2: Identify the initial moles of reactants
From the problem, we have:
- Moles of H₂ = 0.2 mol
- Moles of S = 2.0 mol (but S is a solid and does not affect the equilibrium expression)

### Step 3: Set up an ICE table (Initial, Change, Equilibrium)
We can set up an ICE table to track the changes in concentration:

| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---------------|---------------|--------------|--------------------|
| H₂ | 0.2 | -X | 0.2 - X |
| S (s) | 2.0 | 0 | 2.0 |
| H₂S | 0 | +X | X |

### Step 4: Write the expression for Kp
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{\text{H}_2\text{S}}}{P_{\text{H}_2}} \]
Since S is a solid, it does not appear in the expression.

### Step 5: Relate Kp to the concentrations
Given \( K_p = 6.78 \times 10^{-2} \), we can express it in terms of the moles at equilibrium:
\[ K_p = \frac{(X)}{(0.2 - X)} \]

### Step 6: Substitute Kp into the equation
Substituting the value of \( K_p \):
\[ 6.78 \times 10^{-2} = \frac{X}{0.2 - X} \]

### Step 7: Solve for X
Cross-multiplying gives:
\[ 6.78 \times 10^{-2} (0.2 - X) = X \]
Expanding this:
\[ 1.356 \times 10^{-2} - 6.78 \times 10^{-2}X = X \]
Rearranging gives:
\[ 1.356 \times 10^{-2} = X + 6.78 \times 10^{-2}X \]
\[ 1.356 \times 10^{-2} = X(1 + 6.78 \times 10^{-2}) \]
\[ X = \frac{1.356 \times 10^{-2}}{1 + 6.78 \times 10^{-2}} \]
Calculating the denominator:
\[ 1 + 0.0678 = 1.0678 \]
Thus:
\[ X = \frac{1.356 \times 10^{-2}}{1.0678} \approx 1.27 \times 10^{-2} \text{ mol} \]

### Step 8: Calculate the partial pressure of H₂S
Using the ideal gas law, we can find the partial pressure:
\[ P = \frac{nRT}{V} \]
Where:
- \( n = X = 1.27 \times 10^{-2} \text{ mol} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 363 \text{ K} \)
- \( V = 1.0 \text{ L} \)

Substituting the values:
\[ P_{\text{H}_2\text{S}} = \frac{(1.27 \times 10^{-2})(0.0821)(363)}{1.0} \]
Calculating:
\[ P_{\text{H}_2\text{S}} \approx 0.38 \text{ atm} \]

### Final Answer
The partial pressure of H₂S formed is approximately **0.38 atm**.

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