At `87^(@)C` , the following equilibrium is established
`H_(2) (g)+S(s)hArrH_(2)(g), K_(p)=7xx10^(-2)`
If 0.50 mole of hydrogen and 1.0 mole of sulphur are heated to `87^(@)C` in `1 L` vessel. What is the partial pressure of `H_(2)S (g)` at equilibrium?

Options

A. 0.966 atm
B. 1.38n atm
C. 0.0327 atm
D. 1atm

Answer

To solve the problem step by step, we will analyze the equilibrium reaction and calculate the partial pressure of \( H_2S \) at equilibrium.

### Step 1: Write the Equilibrium Reaction
The equilibrium reaction is given as:
\[ H_2(g) + S(s) \rightleftharpoons H_2S(g) \]

### Step 2: Identify Initial Conditions
We are given:
- Initial moles of \( H_2 = 0.50 \) moles
- Initial moles of \( S = 1.0 \) mole (but it is a solid, so it won't affect the equilibrium expression)
- Initial moles of \( H_2S = 0 \) moles
- Volume of the vessel = 1 L

### Step 3: Set Up the Change in Moles at Equilibrium
Let \( x \) be the number of moles of \( H_2 \) that react to form \( H_2S \). At equilibrium, we have:
- Moles of \( H_2 = 0.50 - x \)
- Moles of \( H_2S = x \)

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given as \( 7 \times 10^{-2} \). The expression for \( K_p \) based on the reaction is:
\[ K_p = \frac{P_{H_2S}}{P_{H_2}} \]
Since \( S \) is a solid, it does not appear in the expression.

### Step 5: Calculate Partial Pressures
The partial pressure of a gas can be calculated using the formula:
\[ P = \frac{nRT}{V} \]
Given that the volume \( V = 1 \) L, we can simplify the expressions:
- \( P_{H_2S} = \frac{xRT}{1} = xRT \)
- \( P_{H_2} = \frac{(0.50 - x)RT}{1} = (0.50 - x)RT \)

Substituting these into the \( K_p \) expression:
\[ K_p = \frac{xRT}{(0.50 - x)RT} = \frac{x}{0.50 - x} \]

### Step 6: Substitute the Value of \( K_p \)
Now we substitute \( K_p \):
\[ 7 \times 10^{-2} = \frac{x}{0.50 - x} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 7 \times 10^{-2} (0.50 - x) = x \]
\[ 0.035 - 7 \times 10^{-2} x = x \]
\[ 0.035 = x + 7 \times 10^{-2} x \]
\[ 0.035 = x(1 + 7 \times 10^{-2}) \]
\[ 0.035 = x(1.07) \]
\[ x = \frac{0.035}{1.07} \approx 0.0327 \text{ moles} \]

### Step 8: Calculate the Partial Pressure of \( H_2S \)
Now we can find the partial pressure of \( H_2S \):
\[ P_{H_2S} = xRT \]
Where:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 87^\circ C = 87 + 273 = 360 \, K \)

Substituting the values:
\[ P_{H_2S} = 0.0327 \times 0.0821 \times 360 \]
\[ P_{H_2S} \approx 0.966 \, \text{atm} \]

### Final Answer
The partial pressure of \( H_2S \) at equilibrium is approximately \( 0.966 \, \text{atm} \).

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