At a certain temperature , the equilibrium constant `(K_(c))` is `4//9` for the reaction `:`
`CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)`
If we take 10 mole of each of the four gases in a one `-` litre container, what would be the equilibrium mole percent of `H_(2)(g)` ?

Options

A. 20
B. 40
C. 60
D. 80

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Determine initial concentrations
Given that we have 10 moles of each gas in a 1-liter container, the initial concentrations are:
- \([CO] = 10 \, \text{mol/L}\)
- \([H_2O] = 10 \, \text{mol/L}\)
- \([CO_2] = 10 \, \text{mol/L}\)
- \([H_2] = 10 \, \text{mol/L}\)

### Step 3: Calculate the reaction quotient \(Q_c\)
The expression for \(Q_c\) for the reaction is:
\[ Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} \]
Substituting the initial concentrations:
\[ Q_c = \frac{(10)(10)}{(10)(10)} = 1 \]

### Step 4: Compare \(Q_c\) with \(K_c\)
Given \(K_c = \frac{4}{9}\):
Since \(Q_c (1) > K_c \left(\frac{4}{9}\right)\), the reaction will shift to the left to reach equilibrium.

### Step 5: Set up the change in concentrations
Let \(x\) be the amount of CO and H2O that reacts. At equilibrium:
- \([CO] = 10 - x\)
- \([H_2O] = 10 - x\)
- \([CO_2] = 10 + x\)
- \([H_2] = 10 + x\)

### Step 6: Write the equilibrium expression
The equilibrium expression can be written as:
\[ K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(10 + x)(10 + x)}{(10 - x)(10 - x)} = \frac{4}{9} \]

### Step 7: Solve for \(x\)
Cross-multiplying gives:
\[ 9(10 + x)^2 = 4(10 - x)^2 \]
Expanding both sides:
\[ 9(100 + 20x + x^2) = 4(100 - 20x + x^2) \]
\[ 900 + 180x + 9x^2 = 400 - 80x + 4x^2 \]
Rearranging gives:
\[ 5x^2 + 260x + 500 = 0 \]
Dividing by 5:
\[ x^2 + 52x + 100 = 0 \]

### Step 8: Use the quadratic formula to find \(x\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-52 \pm \sqrt{52^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{-52 \pm \sqrt{2704 - 400}}{2} = \frac{-52 \pm \sqrt{2304}}{2} = \frac{-52 \pm 48}{2} \]
Calculating the two possible values for \(x\):
1. \(x = \frac{-4}{2} = -2\) (not valid)
2. \(x = \frac{-100}{2} = -50\) (not valid)

Thus, we take \(x = 2\).

### Step 9: Calculate equilibrium concentrations
At equilibrium:
- \([CO] = 10 - 2 = 8 \, \text{mol/L}\)
- \([H_2O] = 10 - 2 = 8 \, \text{mol/L}\)
- \([CO_2] = 10 + 2 = 12 \, \text{mol/L}\)
- \([H_2] = 10 + 2 = 12 \, \text{mol/L}\)

### Step 10: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 8 + 8 + 12 + 12 = 40 \, \text{mol} \]

### Step 11: Calculate mole percent of \(H_2\)
Mole percent of \(H_2\):
\[ \text{Mole percent of } H_2 = \left(\frac{12}{40}\right) \times 100 = 30\% \]

### Final Answer
The equilibrium mole percent of \(H_2(g)\) is **30%**.

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