# 6.3: Manipulating Equilibrium Constants

We will take advantage of two useful relationships when working with equilibrium constants. First, if we reverse a reaction’s direction, the equilibrium constant for the new reaction is simply the inverse of that for the original reaction. For example, the equilibrium constant for the reaction

$\textrm A+2\textrm B\rightleftharpoons A\textrm B_2\hspace{5mm}K_1=\mathrm{\dfrac{[AB_2]}{[A][B]^2}}$

is the inverse of that for the reaction

$A\textrm B_2\rightleftharpoons\textrm A + 2\textrm B\hspace{5mm}K_2=(K_1)^{-1}=\dfrac{[A][\textrm B]^2}{[\textrm{AB}_2]}$

Second, if we add together two reactions to obtain a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions.

$\mathrm{A+C\rightleftharpoons AC}\hspace{4mm}K_3=\mathrm{\dfrac{[AC]}{[A][C]}}$

$\mathrm{AC+C\rightleftharpoons AC_2}\hspace{5mm}K_4=\mathrm{\dfrac{[AC_2]}{[AC][C]}}$

$\mathrm{A+2C\rightleftharpoons AC_2}\hspace{5mm}K_5=K_3\times K_4=\mathrm{\dfrac{[AC]}{[A][C]}\times\dfrac{[AC_2]}{[AC][C]}=\dfrac{[AC_2]}{[A][C]^2}}$

Example 6.1

Calculate the equilibrium constant for the reaction

$\mathrm{2A+B\rightleftharpoons C+3D}$

given the following information

\begin{align} &\textrm{Rxn 1: A + B} \rightleftharpoons \textrm D &&K_1 = 0.40\\ &\textrm{Rxn 2: A + E} \rightleftharpoons \textrm{C + D + F} &&K_2 = 0.10\\ &\textrm{Rxn 3: C + E} \rightleftharpoons \textrm B &&K_3 = 2.0\\ &\textrm{Rxn 4: F + C} \rightleftharpoons \textrm{D + B} &&K_4 = 5.0 \end{align}

Solution

The overall reaction is equivalent to

$\textrm{Rxn 1 + Rxn 2} - \textrm{Rxn 3 + Rxn 4}$

Subtracting a reaction is equivalent to adding the reverse reaction; thus, the overall equilibrium constant is

$K=\dfrac{K_1\times K_2\times K_4}{K_3}=\dfrac{0.40\times0.10\times5.0}{2.0}=0.10$

Practice Exercise 6.1

Calculate the equilibrium constant for the reaction

$\mathrm{C+D+F\rightleftharpoons 2A+3B}$

using the equilibrium constants from Example 6.1.