Notes
Dynamic equilibrium
At equilibrium, the forward and reverse rates are equal, so concentrations stay constant (not necessarily equal). Reactions still happen in both directions.
The equilibrium constant
For $aA + bB \rightleftharpoons cC + dD$:
- Leave out pure solids and liquids. $K_p$ uses partial pressures.
- $K \gg 1$: products favored. $K \ll 1$: reactants favored.
- Reverse the reaction: $\displaystyle K' = \frac{1}{K}$. Multiply by $n$: $K' = K^n$. Add reactions: multiply the $K$ values.
The reaction quotient $Q$
$Q$ has the same form as $K$ but uses current concentrations.
- $Q < K$: the reaction shifts right (toward products).
- $Q > K$: it shifts left.
- $Q = K$: at equilibrium.
ICE tables
Set up Initial, Change and Equilibrium rows. Use the changes from stoichiometry and solve for $x$. If $K$ is very small, you can often assume $x$ is negligible compared with the initial value.
Le Châtelier's principle
A system at equilibrium shifts to partly counteract a stress.
- Add reactant or remove product: shift right.
- Increase pressure by decreasing volume: shift toward fewer moles of gas.
- Raise temperature: shift in the endothermic direction. Only temperature changes the value of $K$.
- Adding a catalyst or an inert gas at constant volume: no shift.
Solubility
$K_{sp}$ for $AgCl(s) \rightleftharpoons Ag^+ + Cl^-$ is $[Ag^+][Cl^-] = s^2$. A common ion lowers solubility.
Topics in this unit
Each topic has its own short notes and practice.
- 7.1Introduction to Equilibrium
- 7.2Direction of Reversible Reactions
- 7.3Reaction Quotient and Equilibrium Constant
- 7.4Calculating the Equilibrium Constant
- 7.5Magnitude of the Equilibrium Constant
- 7.6Properties of the Equilibrium Constant
- 7.7Calculating Equilibrium Concentrations
- 7.8Representations of Equilibrium
- 7.9Introduction to Le Châtelier's Principle
- 7.10Reaction Quotient and Le Châtelier's Principle
- 7.11Introduction to Solubility Equilibria
- 7.12Common-Ion Effect
Practice
Equilibrium is dynamic: rates balance, concentrations stay constant.
Pure solids are left out.
$Q < K$, so it forms more product.
Higher pressure favors the side with fewer moles of gas (2 vs 4).
Heat acts like a product; adding it shifts left and lowers $K$.
Reverse: $\displaystyle \frac{1}{4}$. Double: $\displaystyle \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
$s^2 = 1.8 \times 10^{-10}$, so $s = 1.3 \times 10^{-5}$.
The common ion $\mathrm{Cl^-}$ shifts the equilibrium toward solid AgCl.
Free response
Write your answer first, then open the scoring guide.
Free-response question
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ has $K_c = 49$ at a certain temperature. 1.00 mol of $\mathrm{H_2}$ and 1.00 mol of $\mathrm{I_2}$ are placed in a 1.00 L flask.
(a) Write the $K_c$ expression.
Show the answer and scoring guide
$\displaystyle K_c = \frac{[HI]^2}{[H_2][I_2]}$.
Check off the points you earned:
(b) Calculate the equilibrium concentration of HI.
Show the answer and scoring guide
ICE: $[H_2] = [I_2] = 1.00 - x$, $[HI] = 2x$. $\displaystyle \frac{(2x)^2}{(1.00 - x)^2} = 49 \Rightarrow \frac{2x}{1.00 - x} = 7 \Rightarrow x = \frac{7}{9} = 0.778$. $[HI] = 1.56$ M.
Check off the points you earned:
(c) More $\mathrm{H_2}$ is added after equilibrium. Explain the effect on [HI] using $Q$.
Show the answer and scoring guide
Adding $\mathrm{H_2}$ increases the denominator, so $Q < K$. The reaction shifts right, and [HI] increases.
Check off the points you earned: