consider the following reversible reaction.\n\n$2h_{2}o(g) leftrightarrow 2h_{2}(g) + o_{2}(g)$\n\nwhat is…

consider the following reversible reaction.\n\n$2h_{2}o(g) leftrightarrow 2h_{2}(g) + o_{2}(g)$\n\nwhat is the equilibrium constant expression for the given system?\n\n$k_{eq} = \\frac{h_{2}o}{h_{2}o_{2}}$\n\n$k_{eq} = \\frac{h_{2}o^{2}}{h_{2}^{2}o_{2}}$\n\n$k_{eq} = \\frac{h_{2}^{2}o_{2}}{h_{2}o}$\n\n$k_{eq} = \\frac{h_{2}^{2}o_{2}}{h_{2}o^{2}}$

consider the following reversible reaction.\n\n$2h_{2}o(g) leftrightarrow 2h_{2}(g) + o_{2}(g)$\n\nwhat is the equilibrium constant expression for the given system?\n\n$k_{eq} = \\frac{h_{2}o}{h_{2}o_{2}}$\n\n$k_{eq} = \\frac{h_{2}o^{2}}{h_{2}^{2}o_{2}}$\n\n$k_{eq} = \\frac{h_{2}^{2}o_{2}}{h_{2}o}$\n\n$k_{eq} = \\frac{h_{2}^{2}o_{2}}{h_{2}o^{2}}$

Answer

Explanation:

Step1: Identify the chemical equation

The given reversible reaction is: $$2H_{2}O(g) \rightleftharpoons 2H_{2}(g) + O_{2}(g)$$

Step2: Define the equilibrium constant expression

The equilibrium constant $K_{eq}$ is the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. $$K_{eq} = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}$$

Step3: Apply the formula to the reaction

The products are $H_{2}$ and $O_{2}$ with coefficients $2$ and $1$. The reactant is $H_{2}O$ with coefficient $2$. $$K_{eq} = \frac{[H_{2}]^{2}[O_{2}]^{1}}{[H_{2}O]^{2}}$$

Answer:

$K_{eq} = \frac{[H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$