A mixture of one mole of `CO_(2)` and "mole" of `H_(2)` attains equilibrium at a temperature of `250^(@)C` and a total pressure of `0.1` atm for the change `CO_(2)(g)+H_(2)(g) hArr CO(g)+H_(2)O(g)`. Calculate `K_(p)` if the analysis of final reaction mixture shows `0.16` volume percent of CO.

Explanation

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction:

\[ CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g) \]

Given:
- Initial moles of \( CO_2 \) = 1 mole
- Initial moles of \( H_2 \) = 1 mole
- Total pressure = 0.1 atm
- Volume percent of \( CO \) = 0.16%

### Step 1: Determine the change in moles at equilibrium

Let \( x \) be the moles of \( CO \) and \( H_2O \) formed at equilibrium. The moles of \( CO_2 \) and \( H_2 \) will decrease by \( x \).

At equilibrium:
- Moles of \( CO_2 \) = \( 1 - x \)
- Moles of \( H_2 \) = \( 1 - x \)
- Moles of \( CO \) = \( x \)
- Moles of \( H_2O \) = \( x \)

### Step 2: Calculate the total moles at equilibrium

The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - x) + (1 - x) + x + x = 2 \]

### Step 3: Relate volume percent of \( CO \) to moles

The volume percent of \( CO \) is given as 0.16%. This means:
\[ \frac{x}{\text{Total moles}} \times 100 = 0.16 \]
Substituting the total moles:
\[ \frac{x}{2} \times 100 = 0.16 \]
Solving for \( x \):
\[ x = 0.16 \times \frac{2}{100} = 0.0032 \text{ moles} \]

### Step 4: Calculate the equilibrium concentrations

At equilibrium:
- Moles of \( CO_2 \) = \( 1 - 0.0032 = 0.9968 \)
- Moles of \( H_2 \) = \( 1 - 0.0032 = 0.9968 \)
- Moles of \( CO \) = \( 0.0032 \)
- Moles of \( H_2O \) = \( 0.0032 \)

### Step 5: Write the expression for \( K_c \)

The expression for \( K_c \) is:
\[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.0032)(0.0032)}{(0.9968)(0.9968)} \]

### Step 6: Calculate \( K_c \)

Calculating \( K_c \):
\[ K_c = \frac{0.00001024}{0.9936} \approx 1.03 \times 10^{-5} \]

### Step 7: Convert \( K_c \) to \( K_p \)

Using the relation:
\[ K_p = K_c (RT)^{\Delta n} \]
Where \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) and \( T = 250 + 273 = 523 \, K \).

Calculate \( \Delta n \):
- Moles of products = 2 (1 mole of \( CO \) + 1 mole of \( H_2O \))
- Moles of reactants = 2 (1 mole of \( CO_2 \) + 1 mole of \( H_2 \))
- \( \Delta n = 2 - 2 = 0 \)

Thus,
\[ K_p = K_c (RT)^0 = K_c \]

### Final Answer

\[ K_p \approx 1.03 \times 10^{-5} \]