To determine the equilibrium concentration of methane [tex]\(CH_4\)[/tex], we can start with the equilibrium expression based on the reaction:
[tex]\[
\text{CO} (g) + 3 \text{H}_2 (g) \longleftrightarrow \text{CH}_4 (g) + \text{H}_2\text{O} (g)
\][/tex]
The equilibrium constant expression for this reaction is given by:
[tex]\[
K_{eq} = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}
\][/tex]
We are provided with:
- [tex]\(K_{eq} = 3.90\)[/tex]
- [tex]\([\text{CO}] = 0.30 \, M\)[/tex]
- [tex]\([\text{H}_2] = 0.10 \, M\)[/tex]
- [tex]\([\text{H}_2\text{O}] = 0.020 \, M\)[/tex]
We need to find the equilibrium concentration of [tex]\(\text{CH}_4\)[/tex], denoted as [tex]\([CH_4] = x\)[/tex].
Rearranging the equilibrium expression to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{K_{eq} \times [\text{CO}] \times [\text{H}_2]^3}{[\text{H}_2\text{O}]}
\][/tex]
Substituting the known values into the equation:
[tex]\[
x = \frac{3.90 \times 0.30 \times (0.10)^3}{0.020}
\][/tex]
Breaking it down step-by-step:
1. Calculate [tex]\((0.10)^3\)[/tex]:
[tex]\[
(0.10)^3 = 0.001
\][/tex]
2. Multiply [tex]\(3.90 \times 0.30 \times 0.001\)[/tex]:
[tex]\[
3.90 \times 0.30 \times 0.001 = 0.00117
\][/tex]
3. Divide by [tex]\(0.020\)[/tex]:
[tex]\[
\frac{0.00117}{0.020} = 0.0585
\][/tex]
Thus, the equilibrium concentration of [tex]\(\text{CH}_4\)[/tex] is:
[tex]\[
\boxed{0.0585}
\][/tex]
Expressing this in scientific notation:
[tex]\[
\boxed{5.9 \times 10^{-2}}
\][/tex]
