Consider the following reversible reaction.
[tex]2 H_2 O(g) \leftrightarrow 2 H_2(g)+O_2(g)[/tex]
What is the equilibrium constant expression for the given system?
A. [tex]K _{ eq }=\frac{\left[ H _2 O \right]}{\left[ H _2\right]\left[ O _2\right]}[/tex]
B. [tex]K _{ eq }=\frac{\left[ H _2 O \right]^2}{\left[ H _2\right]^2\left[ O _2\right]}[/tex]
C. [tex]K _{ eq }=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]}[/tex]
D. [tex]K _{ eq }=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2}[/tex]
Answer (2)
The equilibrium constant expression relates the concentrations of products to reactants at equilibrium.
For the reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) , identify the reactants ( H 2 O ) and products ( H 2 and O 2 ) and their stoichiometric coefficients.
Construct the equilibrium constant expression using the coefficients as exponents: K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] .
The equilibrium constant expression for the given system is K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] .
Explanation
Understanding Equilibrium Constant Expressions The equilibrium constant expression for a reversible reaction relates the concentrations of reactants and products at equilibrium. For the general reaction a A + b B ⇌ c C + d D the equilibrium constant expression is given by K e q = [ A ] a [ B ] b [ C ] c [ D ] d where [A], [B], [C], and [D] are the equilibrium concentrations of the reactants and products, and a, b, c, and d are their respective stoichiometric coefficients.
Identifying Reactants, Products, and Coefficients For the given reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) the reactants and products are: Reactant: H 2 O Products: H 2 and O 2 The stoichiometric coefficients are: For H 2 O : 2 For H 2 : 2 For O 2 : 1
Constructing the Equilibrium Constant Expression Substituting these into the equilibrium constant expression, we get K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] 1 This matches option K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ]
Final Answer Therefore, the correct equilibrium constant expression for the given system is K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ]
Examples
The equilibrium constant is crucial in various real-world applications, such as optimizing industrial chemical processes. For example, in the Haber-Bosch process for ammonia synthesis, understanding the equilibrium constant helps determine the optimal conditions (temperature and pressure) to maximize ammonia production. Similarly, in environmental science, the equilibrium constant is used to predict the distribution of pollutants in different environmental compartments (air, water, soil). By manipulating reaction conditions to shift the equilibrium, we can control the yield of desired products or minimize the formation of unwanted byproducts, leading to more efficient and sustainable processes.
The equilibrium constant expression for the reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) is given by K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] . This corresponds to option D. It captures the relationship between the concentrations of the products and reactants at equilibrium.
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