Consider the equation below.
[tex]$CaCO _3(s) \longleftrightarrow CaO ( s )+ CO _2(g)$[/tex]
What is the equilibrium constant expression for the given reaction?
A. [tex]$K _{\infty}=\frac{[ CaO ]}{\left[ CaCO _3\right]}$[/tex]
B. [tex]$K _{\theta q}=\frac{\left[ CO _2\right][ CaO ]}{\left[ CaCO _3\right]}$[/tex]
C. [tex]$K _{ eq }=\left[ CO _2\right]$[/tex]
D. [tex]$K_{e q}=\frac{1}{\left[ CO _2\right]}$[/tex]
Answer (2)
The reaction is C a C O 3 ( s ) ⟷ C a O ( s ) + C O 2 ( g ) .
Solids ( C a C O 3 and C a O ) do not appear in the equilibrium constant expression.
The equilibrium constant expression only includes the concentration of the gaseous product, C O 2 .
The equilibrium constant expression is K e q = [ C O 2 ] .
K e q = [ C O 2 ]
Explanation
Understanding the Reaction The reaction is given by C a C O 3 ( s ) ⟷ C a O ( s ) + C O 2 ( g ) We are asked to find the equilibrium constant expression K e q for this reaction.
Equilibrium Constant Expression The general form of the equilibrium constant expression is given by the ratio of the product of the activities of the products to the product of the activities of the reactants, each raised to the power of their stoichiometric coefficients. However, the activities of solids and pure liquids are defined as 1. Therefore, they do not appear in the equilibrium constant expression.
Identifying Solids and Gases In the given reaction, C a C O 3 ( s ) and C a O ( s ) are solids, and C O 2 ( g ) is a gas. Therefore, the equilibrium constant expression will only include the concentration of C O 2 ( g ) .
Writing the Equilibrium Constant Expression The equilibrium constant expression is given by K e q = [ C O 2 ]
Final Answer Therefore, the correct answer is K e q = [ C O 2 ] .
Examples
The equilibrium constant is crucial in various real-world applications, such as predicting the extent of chemical reactions in industrial processes. For instance, in the production of lime (CaO) from limestone ( C a C O 3 ), understanding the equilibrium constant helps optimize reaction conditions to maximize the yield of lime and carbon dioxide. By controlling factors like temperature and pressure, industries can manipulate the equilibrium to favor the desired products, making the process more efficient and cost-effective.
For example, if we know the value of K e q at a certain temperature, we can predict the concentration of C O 2 at equilibrium. If K e q = 0.5 , then at equilibrium, [ C O 2 ] = 0.5 . This information is vital for designing and operating chemical reactors.
The equilibrium constant expression for the reaction C a C O 3 ( s ) ⟷ C a O ( s ) + C O 2 ( g ) is simply K e q = [ C O 2 ] because only the concentration of the gas is included, as solids do not contribute to the equilibrium expression. Therefore, the correct answer is option C: K e q = [ C O 2 ] .
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