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

Mar 7, 2018

$0.91$

#### Explanation:

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

You know that for this reaction

${\text{A"_ ((g)) + "B"_ ((g)) rightleftharpoons "AB}}_{\left(g\right)}$

you have.

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

Similarly, for this reaction

${\text{AB"_ ((g)) + "A" _ ((g)) rightleftharpoons "A"_ 2"B}}_{\left(g\right)}$

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))
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
$\setminus \setminus \setminus 2 {\text{A"_ ((g)) + "B"_ ((g)) rightleftharpoons "A"_ 2"B}}_{\left(g\right)}$

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"])

$\textcolor{w h i t e}{a a a a a a a a} = \left(\left[\text{A"_ 2"B"])/(["A"]^2 * ["B}\right]\right)$

$\textcolor{w h i t e}{a a a a a a a a} = {K}_{\text{c overall}}$

Plug in your values to find

${K}_{\text{c overall}} = 0.24 \cdot 3.8 = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.91}}}$

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.