Question #dd381
1 Answer
The third reaction will have the most stable reactant.
Explanation:
The idea here is that the value of the equilibrium constant,
If the equilibrium constant is smaller than
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"
So, you have four equilibrium reactions
#"A"_text((g]) rightleftharpoons "B"_text((g])" " " "K_text(c1) = 10^(-5)#
#"C"_text((g]) rightleftharpoons "D"_text((g])" " " "K_text(c2) = 10^(-4)#
#"M"_text((g]) rightleftharpoons "N"_text((g])" " " "K_text(c3) = 10^(-9)#
#"P"_text((g]) rightleftharpoons "Q"_text((g])" " " "K_text(c4) = 10^(-3)#
As you can see, the third equilibrium reaction has the smallest
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_(c3) = (["N"])/(["M"]) = 10^(-9)#
The ratio that exists between the equilibrium concentration of
By comparison, the least stable reactant will have the largest
In this case, the fourth equilibrium reaction, which has the largest