Question

For the above reaction:
(A) $K_{eq}$ > 1
(B) $K_{eq}$ <1
(C) $K_{eq}$ = 1
(D) $K_{eq}$ = 0

Answer 1

Hint: $K_{eq}$ is known as equilibrium constant. If it is greater than 1 then the reaction will move forward, less than 1 then the reaction will move backward and when it is equal to one then the current state of reaction is the equilibrium state.

Complete step by step answer:
First of all, let's get a brief knowledge about the equilibrium constant.
The equilibrium constant of a chemical reaction is a value of its reaction quotient when the reaction is at chemical equilibrium, a state approached by a dynamic chemical system after a sufficient amount of time has elapsed at which its composition has no measurable tendency towards further change. For a given reaction conditions, the equilibrium constant (K) is independent of the initial concentrations of the reactant and product species in the mixture. Thus, when we have the initial composition of a system, known equilibrium constant values can be used to determine the composition of the system when it will be at equilibrium. However, reaction parameters like temperature, solvent, and ionic strength may all influence the value of the equilibrium constant from its original value.
You might be thinking that what does equilibrium constant tell us about the reaction. Let's check it out.
- When $K_{eq}$ is very large, the concentration of the products will be much greater than the concentration of the reactants. In this condition, the reaction essentially "goes to completion"; all (or most of) of the reactants are used up to form the products.
 - When $K_{eq}$is very small, the concentration of the reactant will be much greater than the concentration of the products. The reaction does not occur to any great extent. Most of the reactants remain unchanged, and there are only a few products produced.
- When $K_{eq}$ is neither very large nor very small (close to 1) then in this condition roughly equal amounts of reactants and products are present at equilibrium state.
When we write the equilibrium constant of the above reaction, we will get.
$K={\displaystyle \frac{[OA{c}^{-}]}{[HOAc][O{H}^{-}]}}$
Now, multiplying and dividing by [$H^{+}$] in the above equation, we get
$K={\displaystyle \frac{[OA{c}^{-}]H^{+}}{[HOAc][O{H}^{-}]H^{+}}}$
Here, we can see that${\displaystyle \frac{[OA{c}^{-}]H^{+}}{[HOAc]}}$ is equal to equilibrium constant of an acid and$[O{H}^{-}]H^{+}$ is the equilibrium constant of water.
Therefore, the above equation may be written as follows:
$K_{eq}={\displaystyle \frac{K_a}{K_w}}$
Since, it is a weak acid. So, $K_w<K_a$.
Therefore $K_{eq}$ > 1.

So, the correct answer is (A)- $K_{eq}$ > 1.

Note:
Remember that the equilibrium constant value is the ratio of the concentrations of the products over the reactants. This means that we can use the value of K to predict whether there are more products or reactants at equilibrium for a given reaction only.