Question

consider the following equilibrium:
co(g) + 2h₂(g) ⇌ ch₃oh(g) k_eq = 26

which statement about this system is true?

if the equilibrium concentration of co is 3.4 × 10⁻¹ m, and the equilibrium concentration of h₂ is 6.8 × 10⁻¹ m, the equilibrium concentration of ch₃oh is
done

Explanation:

Step1: Recall Equilibrium Constant Expression

For the reaction \( \text{CO}(g) + 2\text{H}_2(g)
ightleftharpoons \text{CH}_3\text{OH}(g) \), the equilibrium constant expression is \( K_{\text{eq}}=\frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} \).

Step2: Rearrange Formula to Solve for \([\text{CH}_3\text{OH}]\)

Multiply both sides by \([\text{CO}][\text{H}_2]^2\) to get \([\text{CH}_3\text{OH}]=K_{\text{eq}} \times [\text{CO}] \times [\text{H}_2]^2\).

Step3: Substitute Known Values

Given \( K_{\text{eq}} = 26 \), \([\text{CO}]=3.4\times 10^{-1}\ \text{M}\), and \([\text{H}_2]=6.8\times 10^{-1}\ \text{M}\). First, calculate \([\text{H}_2]^2=(6.8\times 10^{-1})^2 = 4.624\times 10^{-1}\ \text{M}^2\). Then, multiply by \([\text{CO}]\): \(3.4\times 10^{-1}\times4.624\times 10^{-1}=1.57216\times 10^{-1}\ \text{M}^3\). Finally, multiply by \(K_{\text{eq}}\): \(26\times1.57216\times 10^{-1}=4.087616\approx 4.1\ \text{M}\).

Answer:

The equilibrium concentration of \( \text{CH}_3\text{OH} \) is approximately \( 4.1\ \text{M} \) (or more precisely \( 4.09\ \text{M} \) depending on rounding steps).