What is the Keq expression for the following reaction?
[tex]A(s)+B(s) \leftrightharpoons C(s)+2 D(g)[/tex]
A. [tex]Keq =\frac{1}{[D]^2}[/tex]
B. [tex]Keq =[ D ]^2[/tex]
C. [tex]Keq =\frac{[C][D]^2}{[A][B]}[/tex]
D. [tex]Keq =\frac{[C][D]}{[A][B]}[/tex]
Answer (1)
Solids do not appear in the K e q expression.
Write the general form of the equilibrium constant expression, K e q .
Consider only the gaseous species.
The correct K e q expression is K e q = [ D ] 2 .
Explanation
Understanding the Problem We are asked to find the equilibrium constant expression, K e q , for the reaction: A ( s ) + B ( s ) ⇋ C ( s ) + 2 D ( g ) Remember that the equilibrium constant expression relates the concentrations of products to reactants at equilibrium. However, there's a key rule: solids and pure liquids do not appear in the K e q expression because their concentrations are essentially constant.
Excluding Solids Since A, B, and C are solids, they do not contribute to the K e q expression. Only the gaseous product D matters. The general form of K e q is: K e q = [ R e a c t an t s ] [ P ro d u c t s ] In this specific case, since A, B, and C are solids, they are excluded from the expression.
Writing the Keq Expression The product D has a stoichiometric coefficient of 2, which means it will be raised to the power of 2 in the K e q expression. Therefore, the correct K e q expression is: K e q = [ D ] 2
Final Answer Thus, the K e q expression for the given reaction is: K e q = [ D ] 2
Examples
Consider a scenario where you're producing a gas (D) from solid reactants (A and B) in a closed container. The equilibrium constant, K e q , helps you understand the relationship between the amount of gas produced and the equilibrium state of the reaction. For instance, if K e q is large, it means that at equilibrium, there's a high concentration of gas D compared to the 'effective concentration' of the solid reactants. This is crucial in industrial processes to optimize gas production. By understanding K e q , you can manipulate reaction conditions (like temperature) to favor the formation of the desired gaseous product, making the process more efficient and cost-effective.
