Consider the following reversible reaction.
[tex]CO(g)+2 H_2(g) \longleftrightarrow CH_3 OH(g)[/tex]
What is the equilibrium constant expression for the given system?
A. [tex]Keq =\frac{[ CO ][ H 2]^2}{\left[ CH _3 OH \right]}[/tex]
B. [tex]Keq =\frac{\left[ CH _3 OH \right]}{[ CO ][ H 2]^2}[/tex]
C. [tex]Keq -\frac{[ CO ][ H 2]}{\left[ CH _3 OH \right]}[/tex]
D. [tex]Keq =\frac{\left[ CH _3 OH \right]}{[ CO ][ H 2]}[/tex]
Answer (1)
The equilibrium constant expression, K e q , is products over reactants.
For the reaction a A + b B ⟷ c C + d D , K e q = [ A ] a [ B ] b [ C ] c [ D ] d .
For CO ( g ) + 2 H 2 ( g ) ⟷ C H 3 O H ( g ) , the equilibrium constant expression is K e q = [ CO ] [ H 2 ] 2 [ C H 3 O H ] .
The equilibrium constant expression for the given system is K e q = [ CO ] [ H 2 ] 2 [ C H 3 O H ] .
Explanation
Understanding the Problem The problem asks us to identify the correct equilibrium constant expression for the given reversible reaction. The general form of the equilibrium constant expression for a reversible reaction is products over reactants, with each concentration raised to the power of its stoichiometric coefficient.
General Equilibrium Constant Expression 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] represent the equilibrium concentrations of reactants A, B, and products C, D, respectively, and a, b, c, and d are their stoichiometric coefficients in the balanced chemical equation.
Applying to the Given Reaction For the given reaction, CO ( g ) + 2 H 2 ( g ) ⟷ C H 3 O H ( g ) , the reactants are CO and H 2 , and the product is C H 3 O H . The stoichiometric coefficient for CO is 1, for H 2 is 2, and for C H 3 O H is 1. Therefore, the equilibrium constant expression is: K e q = [ CO ] [ H 2 ] 2 [ C H 3 O H ]
Final Answer Therefore, the correct equilibrium constant expression for the given system is: K e q = [ CO ] [ H 2 ] 2 [ C H 3 O H ] .
Examples
Understanding equilibrium constants is crucial in various real-world applications, such as optimizing industrial chemical processes. For example, in the production of ammonia via the Haber-Bosch process, controlling the equilibrium constant allows manufacturers to maximize ammonia yield while minimizing energy consumption and waste. By manipulating factors like temperature and pressure, engineers can shift the equilibrium to favor product formation, making the process more efficient and cost-effective. This principle extends to many other chemical reactions, from drug synthesis to polymer production, highlighting the practical importance of equilibrium constants in chemical engineering.
