# Question #dd381

Mar 16, 2016

The third reaction will have the most stable reactant.

#### Explanation:

The idea here is that the value of the equilibrium constant, ${K}_{c}$, will tell you if the equilibrium favors the reactants or the products.

If the equilibrium constant is smaller than $1$, the equilibrium will favor the reactants, i.e. the reaction vessel will contain more reactants than products at equilibrium.

Simply put, the magnitude of the equilibrium constant tells you how much reactants get converted to products.

This is why the problem asks you about the "most stable reactant" $\to$ you must use the magnitude of ${K}_{c}$ to determine which of the four equilibrium reactions lies furthest to the left, i.e. favors the reactants more.

So, you have four equilibrium reactions

$\text{A"_text((g]) rightleftharpoons "B"_text((g])" " " } {K}_{\textrm{c 1}} = {10}^{- 5}$

$\text{C"_text((g]) rightleftharpoons "D"_text((g])" " " } {K}_{\textrm{c 2}} = {10}^{- 4}$

$\text{M"_text((g]) rightleftharpoons "N"_text((g])" " " } {K}_{\textrm{c 3}} = {10}^{- 9}$

$\text{P"_text((g]) rightleftharpoons "Q"_text((g])" " " } {K}_{\textrm{c 4}} = {10}^{- 3}$

As you can see, the third equilibrium reaction has the smallest ${K}_{c}$, which means that most of the reactant will not be converted to the product.

The reaction vessel will thus contain the least amount of product out of all four equilibrium reactions, i.e. the most amount of reactant.

By definition, the equilibrium constant will be

${K}_{c 3} = \left(\left[\text{N"])/(["M}\right]\right) = {10}^{- 9}$

The ratio that exists between the equilibrium concentration of $\text{N}$ and that of $\text{M}$ tells you that you have significantly more reactant present at equilibrium than product.

By comparison, the least stable reactant will have the largest ${K}_{c}$.

In this case, the fourth equilibrium reaction, which has the largest ${K}_{c}$, will also have the least amount of reactant present at equilibrium.