The reaction, `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)` is at equilibrium at `1300 K` in a `1 L` flask. It also contains `0.30 mol` of `CO, 0.10 mol` of `H_(2)` and `0.02` mol of `H_(2)O` and an unknown amount of `CH_(4)` in the flask. Determine the concentration of `CH_(4)` in the mixture. The equilibrium constant `K_(c )` for the reaction at the given temperature us `3.90`.

Explanation

To determine the concentration of \( CH_4 \) in the mixture at equilibrium, we can follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium constant \( K_c \) for the reaction is given by the expression:
\[ K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3} \]
Where:
- \([CH_4]\) is the concentration of methane,
- \([H_2O]\) is the concentration of water vapor,
- \([CO]\) is the concentration of carbon monoxide,
- \([H_2]\) is the concentration of hydrogen.

### Step 2: Identify known values
From the problem, we know:
- \( K_c = 3.90 \)
- The concentrations in a 1 L flask are:
- \([CO] = 0.30 \, \text{mol/L}\)
- \([H_2] = 0.10 \, \text{mol/L}\)
- \([H_2O] = 0.02 \, \text{mol/L}\)
- Let \([CH_4] = x\) (unknown concentration).

### Step 3: Substitute known values into the equilibrium expression
Substituting the known values into the equilibrium expression gives:
\[ 3.90 = \frac{x \cdot 0.02}{0.30 \cdot (0.10)^3} \]

### Step 4: Calculate \((0.10)^3\)
Calculate \((0.10)^3\):
\[ (0.10)^3 = 0.001 \]

### Step 5: Substitute this value back into the equation
Now substitute this back into the equation:
\[ 3.90 = \frac{x \cdot 0.02}{0.30 \cdot 0.001} \]

### Step 6: Simplify the equation
Now simplify the denominator:
\[ 0.30 \cdot 0.001 = 0.0003 \]
Thus, the equation becomes:
\[ 3.90 = \frac{x \cdot 0.02}{0.0003} \]

### Step 7: Solve for \( x \)
Multiply both sides by \( 0.0003 \):
\[ 3.90 \cdot 0.0003 = x \cdot 0.02 \]
\[ 0.00117 = x \cdot 0.02 \]
Now divide both sides by \( 0.02 \):
\[ x = \frac{0.00117}{0.02} = 0.0585 \]

### Step 8: Final concentration of \( CH_4 \)
Thus, the concentration of \( CH_4 \) is:
\[ [CH_4] = 0.0585 \, \text{mol/L} = 5.85 \times 10^{-2} \, \text{mol/L} \]

### Conclusion
The concentration of \( CH_4 \) in the mixture is \( 5.85 \times 10^{-2} \, \text{mol/L} \).

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