To determine the correct equilibrium constant expression for the given reaction, we start by analyzing the balanced chemical equation:
[tex]\[
2 H_2O (g) \leftrightarrow 2 H_2(g) + O_2(g)
\][/tex]
The general form of the equilibrium constant expression, [tex]\( K_{eq} \)[/tex], for a reaction
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
is given by:
[tex]\[
K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\][/tex]
where:
- [tex]\( [A] \)[/tex], [tex]\( [B] \)[/tex], [tex]\( [C] \)[/tex], and [tex]\( [D] \)[/tex] are the molar concentrations of the reactants and products.
- [tex]\( a \)[/tex], [tex]\( b \)[/tex], [tex]\( c \)[/tex], and [tex]\( d \)[/tex] are their respective stoichiometric coefficients from the balanced equation.
Applying this principle to our reaction, we substitute the reactants and products along with their coefficients:
[tex]\[
K_{eq} = \frac{[H_2]^2 [O_2]^1}{[H_2O]^2}
\][/tex]
Simplifying the expression, we get:
[tex]\[
K_{eq} = \frac{[H_2]^2 [O_2]}{[H_2O]^2}
\][/tex]
Now we compare this result with the given options:
1. [tex]\( K _{ eq }=\frac{\left[ H _2 O \right]}{\left[ H _2\right]\left[ O _2\right]} \)[/tex]
2. [tex]\( K _{ eq }=\frac{\left[ H _2 O \right]^2}{\left[ H _2\right]^2\left[ O _2\right]} \)[/tex]
3. [tex]\( K _{ eq }=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]} \)[/tex]
4. [tex]\( K _{ eq }=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2} \)[/tex]
The correct equilibrium constant expression that matches our derivation is:
[tex]\[
K_{eq} = \frac{[H_2]^2 [O_2]}{[H_2O]^2}
\][/tex]
This corresponds to option 4.
Therefore, the correct answer is:
[tex]\[
K _{\text {eq }}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2}
\][/tex]
