To determine the equilibrium constant expression for the given reversible reaction:
[tex]\[
2 H_2O(g) \leftrightarrow 2 H_2(g) + O_2(g)
\][/tex]
we need to follow these steps:
1. Write the balanced chemical equation: This is already given as:
[tex]\[
2 H_2O(g) \leftrightarrow 2 H_2(g) + O_2(g)
\][/tex]
2. Identify the reactants and products: In this reaction, the reactant is [tex]\( H_2O(g) \)[/tex] and the products are [tex]\( H_2(g) \)[/tex] and [tex]\( O_2(g) \)[/tex].
3. Write the general form of the equilibrium constant expression ( [tex]\( K_{\text{eq}} \)[/tex] ): For a general reaction of the form
[tex]\[
aA + bB \leftrightarrow cC + dD
\][/tex]
the equilibrium constant expression is:
[tex]\[
K_{\text{eq}} = \frac{[C]^c[D]^d}{[A]^a[B]^b}
\][/tex]
4. Apply this to the specific reaction:
For the reaction
[tex]\[
2 H_2O(g) \leftrightarrow 2 H_2(g) + O_2(g)
\][/tex]
the equilibrium constant expression would be:
[tex]\[
K_{\text{eq}} = \frac{[H_2]^2[O_2]}{[H_2O]^2}
\][/tex]
This is because the coefficients in the balanced equation determine the exponents for the concentrations of the substances.
5. Check the given options to find the correct expression:
- [tex]\( K_{\text{eq}} = \frac{[H_2O]}{[H_2][O_2]} \)[/tex]
- [tex]\( K_{\text{eq}} = \frac{[H_2O]^2}{[H_2]^2[O_2]} \)[/tex]
- [tex]\( K_{\text{eq}} = \frac{[H_2]^2[O_2]}{[H_2O]} \)[/tex]
- [tex]\( K_{\text{eq}} = \frac{[H_2]^2[O_2]}{[H_2O]^2} \)[/tex]
6. Select the correct option: The correct equilibrium constant expression is
[tex]\[
K_{\text{eq}} = \frac{[H_2]^2[O_2]}{[H_2O]^2}
\][/tex]
Therefore, the correct option is:
[tex]\( 4 \)[/tex]
