To determine the equilibrium constant expression for the given reaction:
[tex]\[ 4 HCl (aq) + O_2 (g) \leftrightarrow 2 H_2O (l) + 2 Cl_2 (g) \][/tex]
we start by recalling that the equilibrium constant expression is determined in terms of the concentrations of the reactants and the products. The products appear in the numerator, while the reactants appear in the denominator.
For a general chemical equilibrium:
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
the equilibrium constant expression [tex]\( K_{eq} \)[/tex] is:
[tex]\[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]
Now, applying this to our specific reaction:
[tex]\[ 4 HCl (aq) + O_2 (g) \leftrightarrow 2 H_2O (l) + 2 Cl_2 (g) \][/tex]
1. Identify the reactants: [tex]\( HCl \)[/tex] and [tex]\( O_2 \)[/tex].
2. Identify the products: [tex]\( H_2O \)[/tex] and [tex]\( Cl_2 \)[/tex].
3. The coefficients of the reactants are: [tex]\( HCl \)[/tex] has a coefficient of 4, [tex]\( O_2 \)[/tex] has a coefficient of 1.
4. The coefficients of the products are: [tex]\( H_2O \)[/tex] has a coefficient of 2, [tex]\( Cl_2 \)[/tex] has a coefficient of 2.
Note that water ([tex]\( H_2O \)[/tex]) is in the liquid phase and it is typically not included in the equilibrium constant expression for reactions occurring in aqueous solution.
Hence, the equilibrium constant expression becomes:
[tex]\[ K_{eq} = \frac{[Cl_2]^2}{[HCl]^4 [O_2]} \][/tex]
So, the correct answer is:
[tex]\[ K_{eq} = \frac{\left[Cl_2\right]^2}{\left[HCl\right]^4\left[O_2\right]} \][/tex]
