At a certain temperature, the equilibrium constant for the chemical reaction shown is 2.84×10−3 . At equilibrium, the concentration of AB is 2.425 M, the concentration of BC is 2.525 M, and the concentration of AC is 0.290 M. Calculate the concentration of B at equilibrium. AB(aq)+BC(aq)↽−−⇀AC(aq)+2B(aq) AB ( aq ) + BC ( aq ) ↽ − − ⇀ AC ( aq ) + 2 B ( aq )

Chemical equilibrium is called the state of a system where no changes are observed in the concentration of reagents or products, over time, that is, they remain constant. This occurs in reversible reactions, where the reaction rate of reagents to products is the same as that of products to reagents.

So, Chemical Equilibrium is the state in which the direct and indirect reaction have the same reaction rate. Each equilibrium reaction has its constant, which relates the speed of a direct reaction and its inverse.

Being:

aA + bB ⇔ cC + dD

constant Kc is:

[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

The constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.

In this case, being:

AB(aq)+BC(aq) ⇔ AC(aq)+2 B(aq)

constant Kc is:

[tex]Kc=\frac{[AC] *[B]^{2} }{[AB]*[BC] }[/tex]

So, you know:

Kc=2.84*10⁻³
[AB]= 2.425 M
[BC]= 2.525 M
[AC]=0.290 M
[B]= ?

Replacing:

[tex]2.84*10^{-3} =\frac{0.290*[B]^{2} }{2.425*2.525}[/tex]

Solving:

[tex]\sqrt{\frac{2.84*10^{-3}*2.425*2.525}{0.290} } =[B][/tex]

[B]=0.245 M

The concentration of B at equilibrium is 0.245 M