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 if `K_(P)=9`
`CO+H_(2)OhArrCO_(2)+H_(2)`

Options

A. `3` bar
B. `4` bar
C. `2` bar
D. `1` bar

Explanation

To solve the problem step by step, we will follow the equilibrium expression and the given conditions.

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

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

At the start, there are no products:
- \( P_{\text{CO}_2} = 0 \)
- \( P_{\text{H}_2} = 0 \)

### Step 3: Set up the change in pressures
Let \( x \) be the change in pressure of CO and H2O that reacts to form products. Therefore, 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 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{x \cdot x}{(4 - x)(4 - x)} = \frac{x^2}{(4 - x)^2} \]

### Step 5: Substitute the given \( K_p \) value
We are given that \( K_p = 9 \):
\[ 9 = \frac{x^2}{(4 - x)^2} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 9(4 - x)^2 = x^2 \]
Expanding the left side:
\[ 9(16 - 8x + x^2) = x^2 \]
This simplifies to:
\[ 144 - 72x + 9x^2 = x^2 \]
Rearranging gives:
\[ 8x^2 - 72x + 144 = 0 \]

### Step 7: Solve the quadratic equation
Dividing the entire equation by 8:
\[ x^2 - 9x + 18 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ x = \frac{9 \pm \sqrt{81 - 72}}{2} \]
\[ x = \frac{9 \pm \sqrt{9}}{2} \]
\[ x = \frac{9 \pm 3}{2} \]
Calculating the two possible values:
1. \( x = \frac{12}{2} = 6 \)
2. \( x = \frac{6}{2} = 3 \)

### Step 8: Determine the valid solution
Since \( x \) represents the pressure change and must be less than the initial pressures (4 bar), we take \( x = 3 \) bar.

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

### Final Answer
The partial pressure of hydrogen at equilibrium is **3 bar**.

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