Question

consider the following reversible reaction.
2h₂o(g) ⇌ 2h₂(g) + o₂(g)
what is the equilibrium constant expression for the given system?
(\bigcirc) ( k_{eq} = \frac{\text{h}_2\text{o}}{\text{h}_2\text{o}_2} )
(\bigcirc) ( k_{eq} = \frac{\text{h}_2\text{o}^2}{\text{h}_2^2\text{o}_2} )
(\bigcirc) ( k_{eq} = \frac{\text{h}_2^2\text{o}_2}{\text{h}_2\text{o}} )
(\bigcirc) ( k_{eq} = \frac{\text{h}_2^2\text{o}_2}{\text{h}_2\text{o}^2} )

Explanation:

Step1: Recall Equilibrium Constant Formula

For a reversible reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K_{eq} \) is given by \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the brackets represent the molar concentrations of the substances, and the exponents are the stoichiometric coefficients from the balanced reaction.

Step2: Apply Formula to Given Reaction

The given reaction is \( 2H_2O(g)
ightleftharpoons 2H_2(g) + O_2(g) \). Here, the reactant is \( H_2O \) with a stoichiometric coefficient of 2, and the products are \( H_2 \) (stoichiometric coefficient 2) and \( O_2 \) (stoichiometric coefficient 1).

Using the equilibrium constant formula, we place the products in the numerator and the reactant in the denominator, each raised to their respective stoichiometric coefficients. So, the numerator will be \( [H_2]^2[O_2]^1 \) (or just \( [H_2]^2[O_2] \)) and the denominator will be \( [H_2O]^2 \).

Thus, the equilibrium constant expression is \( K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2} \).

Answer:

\( K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2} \) (the last option among the given choices)