Question

carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o).
co(g) + 3h₂(g) \longleftrightarrow ch₄(g) + h₂o(g)
the reaction is at equilibrium at 1,000 k. the equilibrium constant of the reaction is 3.90. at equilibrium, the concentrations are as follows.
co = 0.30 m
h₂ = 0.10 m
h₂o = 0.020 m
what is the equilibrium concentration of ch₄ expressed in scientific notation?
\bigcirc 0059
\bigcirc 5.9 \times 10^{-2}
\bigcirc 0.059
\bigcirc 5.9 \times 10^{2}

Explanation:

Step1: Recall the equilibrium constant expression

For the reaction \( \text{CO}(g) + 3\text{H}_2(g)
ightleftharpoons \text{CH}_4(g) + \text{H}_2\text{O}(g) \), the equilibrium constant \( K \) is given by:
\[ K = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \]
We know \( K = 3.90 \), \( [\text{CO}] = 0.30 \, \text{M} \), \( [\text{H}_2] = 0.10 \, \text{M} \), and \( [\text{H}_2\text{O}] = 0.020 \, \text{M} \). We need to solve for \( [\text{CH}_4] \).

Step2: Rearrange the formula to solve for \( [\text{CH}_4] \)

\[ [\text{CH}_4] = \frac{K [\text{CO}][\text{H}_2]^3}{[\text{H}_2\text{O}]} \]

Step3: Substitute the known values into the formula

First, calculate \( [\text{H}_2]^3 \):
\[ [\text{H}_2]^3 = (0.10)^3 = 0.001 \, \text{M}^3 \]
Then substitute \( K = 3.90 \), \( [\text{CO}] = 0.30 \, \text{M} \), \( [\text{H}_2]^3 = 0.001 \, \text{M}^3 \), and \( [\text{H}_2\text{O}] = 0.020 \, \text{M} \) into the formula:
\[ [\text{CH}_4] = \frac{3.90 \times 0.30 \times 0.001}{0.020} \]
Calculate the numerator:
\[ 3.90 \times 0.30 \times 0.001 = 3.90 \times 0.0003 = 0.00117 \]
Then divide by the denominator:
\[ [\text{CH}_4] = \frac{0.00117}{0.020} = 0.0585 \approx 0.059 \]
In scientific notation, \( 0.059 = 5.9 \times 10^{-2} \)

Answer:

\( 5.9 \times 10^{-2} \) (corresponding to the option "5.9 x 10⁻²")