If a reaction vessel at `400^(@)C` is charged with equimolar mixture of `CO` and steam such that `P_(CO)=P_(H_(2))O=4` bar what will be that partial pressure of `H_(2)` at equilibrium
`CO_(if)+H_(2)OhArrCO_(2)+H_(2)K_(P=9`

Options

A. `0.3` bar
B. `0.4` bar
C. `0.2` bar
D. `0.1` bar

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{CO} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + \text{H}_2 \]

### Step 2: Define initial conditions
We are given that the partial pressures of CO and H2O are both 4 bar at the start:
- \( P_{\text{CO}} = 4 \, \text{bar} \)
- \( P_{\text{H}_2\text{O}} = 4 \, \text{bar} \)

### Step 3: Set up the equilibrium expression
The equilibrium constant \( K_p \) is given as 9. The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} \]

### Step 4: Define changes in partial pressures
Let \( x \) be the change in pressure of CO and H2O that reacts to reach equilibrium. The changes in partial pressures will be:
- At equilibrium:
- \( P_{\text{CO}} = 4 - x \)
- \( P_{\text{H}_2\text{O}} = 4 - x \)
- \( P_{\text{CO}_2} = x \)
- \( P_{\text{H}_2} = x \)

### Step 5: Substitute into the equilibrium expression
Substituting the equilibrium pressures into the \( K_p \) expression:
\[ K_p = \frac{x \cdot x}{(4 - x)(4 - x)} = \frac{x^2}{(4 - x)^2} \]

### Step 6: Set the equation equal to \( K_p \)
Set the equation equal to the given \( K_p \):
\[ \frac{x^2}{(4 - x)^2} = 9 \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ x^2 = 9(4 - x)^2 \]
Expanding the right side:
\[ x^2 = 9(16 - 8x + x^2) \]
\[ x^2 = 144 - 72x + 9x^2 \]
Rearranging gives:
\[ 0 = 8x^2 - 72x + 144 \]
Dividing the entire equation by 8:
\[ 0 = x^2 - 9x + 18 \]

### Step 8: Factor or use the quadratic formula
Factoring gives:
\[ (x - 6)(x - 3) = 0 \]
Thus, \( x = 6 \) or \( x = 3 \). Since \( x \) must be less than 4 (the initial pressures), we take \( x = 3 \).

### Step 9: Calculate the partial pressure of \( H_2 \) at equilibrium
At equilibrium, the partial pressure of \( H_2 \) is:
\[ P_{\text{H}_2} = x = 3 \, \text{bar} \]

### Final Answer
The partial pressure of \( H_2 \) at equilibrium is **3 bar**.

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