Константа рівноваги йонних реакцій

We've been studying equilibrium! We've learned how to calculate equilibrium constants for any chemical reaction. All you need is either concentration or pressure. But, there are also a few mathematical tricks you can use. Today, we'll look at how you can manipulate equilibrium constants.

Tools for Manipulating $K_{eq}$

The reverse reaction has the inverse of $K_{eq}$.

$$\begin{array}{r|l} Reaction & K_{eq} \\ \hline A+B \rightleftharpoons C & K_{eq} \\ C \rightleftharpoons A+B & \frac{1}{K_{eq}} \end{array}$$

Multiplying the mole ratio by a coefficient raises the equilibrium constant to that power.

$$\begin{array}{r|l} Reaction & K_{eq} \\ \hline A+B \rightleftharpoons C & K_{eq} \\ 2A+2B \rightleftharpoons 2C & K_{eq}^{2} \end{array}$$

When adding reaction mechanisms, multiply their equilibrium constants.

$$\begin{array}{r|l} Reaction & K_{eq} \\ \hline Step\, 1: A+B \rightleftharpoons C & k_{1} \\ Step\, 2: C \rightleftharpoons E & k_{2} \\ Overall: A+B+\cancel{C}\rightleftharpoons \cancel{C}+E & k_{1} \times k_{2} \end{array}$$

Example Problem

Carbon dioxide gas reacts with water to form carbonic acid. It's the reason for ocean acidification. Ocean acidification is slowly killing many crustaceans that have calcium carbonate shells. Given the reactions below, find $K_{eq}$ for the reaction of calcium carbonate with water.

$Equation\, 1:\, CO_{2} + H_{2}O \rightleftharpoons H^{+} + HCO_{3}^{-}$ $K_{eq}=4.5\times 10^-7$

$Equation\, 2:\, HCO_{3}^{-} \rightleftharpoons H^{+} + CO_{3}^{2-}$ $K_{eq}=4.7\times 10^{-11}$

$Equation\, 3:\,CaCO_{3} \rightleftharpoons Ca^{+} + CO_{3}^{2-}$ $K_{eq}=3.36\times 10^-9$

First, we need to write the balanced equation for the reaction between calcium carbonate and an acid.

$CaCO_{3} + H^{+} \rightleftharpoons Ca^{2+} + HCO_{3}^{-}$ $K_{eq}=?$

Now we can construct the reaction mechanism. We'll need to reverse equation 2. Then, we'll add it to equation 3.

$Step\, 1:\, H^{+} + CO_{3}^{2-} \rightleftharpoons HCO_{3}^{-}$ $K_{eq}=\frac{1}{4.7\times 10^{-11}}$.

$Step\,2:\,CaCO_{3} \rightleftharpoons Ca^{+} + CO_{3}^{2-}$ $K_{eq}=3.36\times 10^{-9}$

$Overall: CaCO_{3} + H^{+} \rightleftharpoons Ca^{2+} + HCO_{3}^{-}$

$K_{eq}=\frac{1}{4.7\times 10^{-11}} \times (3.36\times10^{-9})=75.16$

Based on the magnitude of the equilibrium constant, the reaction favors the products.

(Video) Adding and Manipulating K (Equilibrium Constants)

By Denovo Tutor

In this video, Denovo Tutor explains how to manipulate equilibrium constants. He shows you an example problem where you have to contrive three different reactions! In the end, you've solved for the net reaction's equilibrium constant. This is an excellent video for learning how to manipulate equilibrium reactions.