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 step by step, we will follow the outlined procedure based on the information provided in the question and the video transcript.

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

### Step 2: Identify initial moles and conditions
- Initial moles of \( \text{H}_2 \) = 0.2 mol
- Initial moles of \( \text{S} \) = 2.0 mol (solid, does not affect equilibrium)
- Volume of the vessel = 1.0 L
- Temperature = 90°C = 363 K

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the moles of \( \text{H}_2 \) that react at equilibrium. Therefore:
- Moles of \( \text{H}_2 \) at equilibrium = \( 0.2 - x \)
- Moles of \( \text{H}_2\text{S} \) formed = \( x \)

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

### Step 5: Relate partial pressures to moles
Using the ideal gas law, we can express the partial pressures in terms of moles and volume:
- \( P_{\text{H}_2} = \frac{n_{\text{H}_2}}{V}RT = \frac{0.2 - x}{1}RT \)
- \( P_{\text{H}_2\text{S}} = \frac{n_{\text{H}_2\text{S}}}{V}RT = \frac{x}{1}RT \)

### Step 6: Substitute into the \( K_p \) expression
Substituting the expressions for partial pressures into the \( K_p \) equation:
\[ K_p = \frac{x}{0.2 - x} \]
Given that \( K_p = 6.78 \times 10^{-2} \):
\[ 6.78 \times 10^{-2} = \frac{x}{0.2 - x} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 6.78 \times 10^{-2} (0.2 - x) = x \]
Expanding this:
\[ 0.01356 - 6.78 \times 10^{-2} x = x \]
Rearranging gives:
\[ 0.01356 = x + 6.78 \times 10^{-2} x \]
\[ 0.01356 = x(1 + 6.78 \times 10^{-2}) \]
\[ x = \frac{0.01356}{1.0678} \approx 0.0127 \, \text{mol} \]

### Step 8: Calculate the partial pressure of \( \text{H}_2\text{S} \)
Since \( x \) is the number of moles of \( \text{H}_2\text{S} \) formed, the concentration of \( \text{H}_2\text{S} \) is:
\[ [\text{H}_2\text{S}] = x = 0.0127 \, \text{mol/L} \]
Using the ideal gas law to find the partial pressure:
\[ P_{\text{H}_2\text{S}} = [\text{H}_2\text{S}]RT \]
Where \( R = 0.0821 \, \text{L atm/(K mol)} \) and \( T = 363 \, \text{K} \):
\[ P_{\text{H}_2\text{S}} = 0.0127 \times 0.0821 \times 363 \]
Calculating gives:
\[ P_{\text{H}_2\text{S}} \approx 0.38 \, \text{atm} \]

### Final Answer
The partial pressure of \( \text{H}_2\text{S} \) formed is approximately **0.38 atm**.

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