# What is the value of Kc of the overall reaction after combining the two individual reactions?

##### 1 Answer

#### Answer:

#### Explanation:

The key here is to examine the expression of the equilibrium constant for all three reactions.

You know that for this reaction

#"A"_ ((g)) + "B"_ ((g)) rightleftharpoons "AB"_ ((g))#

you have.

#K_ ("c 1") = (["AB"])/(["A"] * ["B"])#

Similarly, for this reaction

#"AB"_ ((g)) + "A" _ ((g)) rightleftharpoons "A"_ 2"B"_ ((g))#

you have

#K_("c 2") = (["A"_2"B"])/(["AB"] * ["A"])#

Now, for the overall reaction

#\ \ \ \ "A"_ ((g)) + "B"_ ((g)) rightleftharpoons color(red)(cancel(color(black)("AB"_ ((g)))))#

#color(red)(cancel(color(black)("AB"_ ((g))))) + "A"_ ((g)) rightleftharpoons "A"_ 2"B"_ ((g))#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#\ \ \ 2"A"_ ((g)) + "B"_ ((g)) rightleftharpoons "A"_ 2"B"_ ((g))#

You have

#K_ "c overall" = (["A"_ 2"B"])/(["A"]^2 * ["B"])#

Now, notice that you can get the expression of the equilibrium constant for the overall reaction by **multiplying** the equilibrium constants of the two individual reactions.

#K_("c 1") * K_("c 2") = (color(red)(cancel(color(black)(["AB"]))))/(["A"] * ["B"]) * (["A"_2"B"])/(color(red)(cancel(color(black)(["AB"]))) * ["A"])#

#color(white)(aaaaaaaa) = (["A"_ 2"B"])/(["A"]^2 * ["B"])#

#color(white)(aaaaaaaa) = K_ "c overall"#

Plug in your values to find

#K_ "c overall" = 0.24 * 3.8 = color(darkgreen)(ul(color(black)(0.91)))#

The answer is rounded to two **sig figs**.

So remember, when you're **adding** two individual equilibrium reactions, you need to **multiply** their respective equilibrium constants to get the equilibrium constant of the *overall reaction*.