Equilibrium Constants and Their Applications
The Equilibrium Constant ($K$)
- As discussed in the previous section, the equilibrium constant $K$ is a quantitative measure of the position of equilibrium for a chemical reaction.
- It is expressed as:
$$
K = \frac{[\text{Products}]^{\text{stoichiometric coefficients}}}{[\text{Reactants}]^{\text{stoichiometric coefficients}}}
$$
Note
The magnitude of $K$ provides valuable information about the extent of a reaction at equilibrium.
- When $K << 1$: Reactants are Favored
- A very small $K$ value (e.g., $K = 10^{-5}$) indicates that the equilibrium position is heavily skewed toward the reactants.
- Only a small proportion of reactants has been converted into products.
- When $K \approx 1$: Comparable Amounts of Reactants and Products
- When $K$ is close to 1 (e.g., $K = 0.8$), there are significant amounts of both reactants and products at equilibrium.
- Neither direction of the reaction is strongly favored.
- When $K >> 1$: Products are Favored
- A large $K$ value (e.g., $K = 10^5$) means the equilibrium position is strongly shifted toward the products.
- Most reactants are converted into products.
Example
Dissociation of Water:
- In the reaction $\text{H}_2\text{O} \leftrightharpoons \text{H}^+ + \text{OH}^- $, $K \approx 10^{-14}$ at 25°C.
- This small value shows that water largely remains undissociated.
Combustion of Methane:
- For the reaction $ \text{CH}_4 + 2\text{O}_2 \leftrightharpoons \text{CO}_2 + 2\text{H}_2\text{O} $, $K$ is very large.
- This indicates that nearly all methane and oxygen are converted into carbon dioxide and water.
Tip
If $K$ is close to 1, small changes in external conditions (e.g., temperature or pressure) can significantly shift the equilibrium position.
Relationship Between Forward and Reverse ($K$)
- Reversible reactions can proceed in both forward and reverse directions.
- The equilibrium constant for the reverse reaction is the reciprocal of the forward reaction: $$
K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}
$$
This relationship arises because reversing the reaction inverts the roles of products and reactants, thereby inverting the ratio.
Example
Calculating $ K_{\text{reverse}} $
For the reaction $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}2(g) $, if $ K{\text{forward}} = 0.59 $, then:
$$
K_{\text{reverse}} = \frac{1}{0.59} \approx 1.7
$$
Common Mistake
- Students often forget to adjust $K$ when the stoichiometric coefficients of a reaction are changed.
- For example, doubling all coefficients requires squaring $K$, while halving them requires taking the square root of $K$.
Manipulating $K$ for different reactions
| Effect on $K_{\mathrm{c}}$ | |
|---|---|
| Reversing the reaction | $\frac{1}{K_c}$ |
| Doubling the reaction coefficients | ${K_{\mathrm{c}}}^2$ |
| Halving the reaction coefficients | $\sqrt{K_c}$ |
| Adding together two reactions | $K_{\mathrm{c}} (1) \times K_{\mathrm{c}} (2)$ |
Self review
- What does a $ K $ value of 0.01 indicate about the position of equilibrium?
- How does increasing temperature affect $ K $ for an endothermic reaction?
- If $ K_{\text{forward}} = 4.0 $, what is $ K_{\text{reverse}} $?
