To identify the correct equilibrium constant expression for the given reaction:
[tex]\[ 2 H_2O(g) \leftrightharpoons 2 H_2(g) + O_2(g) \][/tex]
we need to use the general form of the equilibrium constant expression for a reversible reaction:
For a generic reaction of the form:
[tex]\[ aA + bB \leftrightharpoons cC + dD \][/tex]
the equilibrium constant expression (\( K_\text{eq} \)) is given by:
[tex]\[ K_\text{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]
Here, \( [A] \), \( [B] \), \( [C] \), and \( [D] \) represent the molar concentrations of the reactants and products, and \( a \), \( b \), \( c \), and \( d \) are their respective stoichiometric coefficients.
Applying this to our given reaction:
[tex]\[ 2 H_2O(g) \leftrightharpoons 2 H_2(g) + O_2(g) \][/tex]
we can identify the following:
- \( [H_2O] \) with a coefficient of 2 on the reactant side
- \( [H_2] \) with a coefficient of 2 on the product side
- \( [O_2] \) with a coefficient of 1 on the product side
Thus, the equilibrium constant expression (\( K_\text{eq} \)) for this reaction is:
[tex]\[ K_\text{eq} = \frac{[H_2]^2 [O_2]}{[H_2O]^2} \][/tex]
Comparing this with the given multiple choices:
1. \( K_{\text {eop }}=\frac{\left[ H _2 O \right]}{\left[ H _2\right]\left[ O _2\right]} \)
2. \( K_{S O}=\frac{\left[ H _2 O \right]^2}{\left[ H _2\right]^2\left[ O _2\right]} \)
3. \( K_{\infty 0}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]} \)
4. \( K_{e Q}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2} \)
The correct equilibrium constant expression is:
[tex]\[ K_{e Q}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2} \][/tex]
Therefore, the correct choice is [tex]\( K_{e Q} \)[/tex].
