Steam is passed over hot carbon to attain the equilibrium at 400k. `C_(s) + H_(2)O_((g)) hArr CO_((g)) + H_(2(g))`. The equilibrium constant k = 1.34 (dimensionless) and the total pressure of the equilibrium mixture is 200 k.pa.
Which one is correct when equilibrium is attained

Options

A. Mole fraction of CO = molefraction of `H_(2) = 0.388`
B. Mole fraction of CO = molefraction of `H_(2)` = 0.0.75
C. Mole fraction of `H_(2)`O = 0.224
D. Mole fraction of `H_(2)`O = 0.85

Answer

To solve the problem, we need to analyze the equilibrium reaction and apply the concepts of equilibrium constant and partial pressures.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction given is:
\[ C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_2_{(g)} \]

2. **Define the Equilibrium Constant (K):**
The equilibrium constant \( K \) for the reaction can be expressed in terms of the partial pressures of the gases involved:
\[ K = \frac{P_{CO} \cdot P_{H_2}}{P_{H_2O}} \]
Since carbon (C) is a solid, its activity is considered to be 1 and does not appear in the expression.

3. **Given Values:**
- \( K = 1.34 \)
- Total pressure \( P_{total} = 200 \, kPa \)

4. **Set Up the Initial Conditions:**
Let:
- Initial moles of \( H_2O \) = 1 (in terms of pressure, assume 1 atm or 100 kPa)
- Initial moles of \( CO \) = 0
- Initial moles of \( H_2 \) = 0

5. **Change in Moles at Equilibrium:**
Let \( x \) be the change in moles of \( H_2O \) that reacts:
- At equilibrium:
- \( P_{H_2O} = 100 - x \)
- \( P_{CO} = x \)
- \( P_{H_2} = x \)

6. **Express the Equilibrium Constant in Terms of x:**
Substitute these values into the expression for \( K \):
\[ K = \frac{x \cdot x}{100 - x} = \frac{x^2}{100 - x} \]
Setting this equal to the given \( K \):
\[ 1.34 = \frac{x^2}{100 - x} \]

7. **Solve for x:**
Rearranging gives:
\[ 1.34(100 - x) = x^2 \]
\[ 134 - 1.34x = x^2 \]
Rearranging further:
\[ x^2 + 1.34x - 134 = 0 \]

8. **Use the Quadratic Formula:**
Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = 1.34, c = -134 \)
\[ x = \frac{-1.34 \pm \sqrt{(1.34)^2 - 4 \cdot 1 \cdot (-134)}}{2 \cdot 1} \]
\[ x = \frac{-1.34 \pm \sqrt{1.7956 + 536}}{2} \]
\[ x = \frac{-1.34 \pm \sqrt{537.7956}}{2} \]
\[ x = \frac{-1.34 \pm 23.2}{2} \]
Taking the positive root:
\[ x \approx 10.43 \, (approximately) \]

9. **Calculate Partial Pressures:**
Now substitute \( x \) back to find the partial pressures:
- \( P_{H_2O} = 100 - 10.43 = 89.57 \, kPa \)
- \( P_{CO} = 10.43 \, kPa \)
- \( P_{H_2} = 10.43 \, kPa \)

10. **Calculate Mole Fractions:**
- Total pressure \( P_{total} = 200 \, kPa \)
- Mole fraction of \( CO \) and \( H_2 \):
\[ \text{Mole fraction of } CO = \frac{P_{CO}}{P_{total}} = \frac{10.43}{200} \approx 0.05215 \]
\[ \text{Mole fraction of } H_2 = \frac{10.43}{200} \approx 0.05215 \]
- Mole fraction of \( H_2O \):
\[ \text{Mole fraction of } H_2O = \frac{P_{H_2O}}{P_{total}} = \frac{89.57}{200} \approx 0.44785 \]

### Conclusion:
From the calculations, we find:
- Mole fraction of \( CO \) and \( H_2 \) is approximately \( 0.052 \).
- Mole fraction of \( H_2O \) is approximately \( 0.447 \).