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 _{e q}=\frac{\left[ H _2 O \right]}{\left[ H _2\right]\left[ O _2\right]}[/tex]
B. [tex]K _{e q}=\frac{\left[ H _2 O \right]^2}{\left[ H _2\right]^2\left[ O _2\right]}[/tex]
C. [tex]K _{e q}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]}[/tex]
D. [tex]K _{e q}=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2}[/tex]
Answer (2)
Identify the reactants and products in the given reversible reaction.
Determine the stoichiometric coefficients for each reactant and product.
Construct the equilibrium constant expression using the formula K e q = [ R e a c t an t s ] coe ff i c i e n t s [ P ro d u c t s ] coe ff i c i e n t s .
The equilibrium constant expression for the reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) 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 is determined by the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. For the given reaction, 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) , we need to identify the reactants, products, and their corresponding stoichiometric coefficients to construct the correct expression.
Identifying Reactants, Products, and Coefficients In the reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) , H 2 O is the reactant, and H 2 and O 2 are the products. The stoichiometric coefficient for H 2 O is 2, for H 2 is 2, and for O 2 is 1.
Constructing the Equilibrium Constant Expression The equilibrium constant expression, K e q , is given by: K e q = [ R e a c t an t s ] coe ff i c i e n t s [ P ro d u c t s ] coe ff i c i e n t s . Substituting the identified reactants, products, and their stoichiometric coefficients into the expression, we get: K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] 1 .
Selecting the Correct Option Comparing the derived expression with the given options, we find that the correct equilibrium constant expression 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 synthesizing ammonia ( N 2 + 3 H 2 ⇌ 2 N H 3 ), understanding the equilibrium constant helps determine the optimal conditions (temperature, pressure, and reactant ratios) to maximize ammonia production. By manipulating these conditions based on the equilibrium constant, industries can enhance efficiency and reduce waste, leading to cost savings and environmental benefits. Similarly, in environmental science, the equilibrium constant is used to predict the distribution of pollutants in different environmental compartments, such as air, water, and soil.
The equilibrium constant expression for the given reaction 2 H 2 O ( g ) ↔ 2 H 2 ( g ) + O 2 ( g ) is K e q = [ H 2 O ] 2 [ H 2 ] 2 [ O 2 ] , which matches option D. This expression is constructed by taking the products raised to their coefficients divided by the reactants raised to their coefficients. Thus, option D is the correct answer.
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