Equilibrium

A reversible reaction 可逆反应 runs both ways. Chemical equilibrium 化学平衡 is reached when the forward and reverse rates become equal, so concentrations stop changing. Equilibrium is dynamic – both reactions continue, but at matching rates, so nothing appears to change.

Dynamic equilibrium: the forward and reverse rates become equal

The rates graph above explains why concentrations settle; this next graph shows what the concentrations do – reactants fall and products rise until, once the rates match, both hold constant (they need not be equal).

Reactant and product concentrations change, then hold constant once equilibrium is reached

At equilibrium the amounts of reactants and products are fixed but usually not equal. Whether the mixture favors products or reactants depends on the reaction; you compare the current state to the equilibrium condition to predict which way it shifts.

The reaction quotient 反应商 $Q$ has the same form as the equilibrium expression but uses current concentrations:

At equilibrium $Q$ equals the equilibrium constant 平衡常数 $K$. Comparing them predicts direction: $Q shifts forward (toward products); $Q>K$ shifts reverse; $Q=K$ means already at equilibrium.

Write $K$ from the balanced equation (pure solids and liquids are left out). From equilibrium concentrations (or partial pressures for $K_p$), plug in and compute. $K_c$ uses molarities; $K_p$ uses pressures.

Worked example. For $\text{N}_2\text{O}_4\rightleftharpoons2\text{NO}_2$, the equilibrium concentrations are $[\text{N}_2\text{O}_4]=0.20\ \text{M}$ and $[\text{NO}_2]=0.10\ \text{M}$. Then

The NO$_2$ coefficient of $2$ becomes the power; N$_2$O$_4$ (coefficient $1$) is just to the first power.

• $K\gg 1$: products strongly favored (reaction nearly complete).
• $K\ll 1$: reactants favored (little reaction).
• $K\approx 1$: significant amounts of both.

The size of $K$ tells you where the equilibrium "sits."

$K$ changes in predictable ways: reversing a reaction inverts $K$ ($1/K$); multiplying coefficients by $n$ raises $K$ to the $n$th power; adding reactions multiplies their $K$ values. Only temperature changes the value of $K$ itself.

Use an ICE table (Initial, Change, Equilibrium): fill in starting amounts, express the change with $x$ using the mole ratios, then substitute into the $K$ expression and solve for $x$. When $K$ is small, the approximation "$x$ is negligible" often simplifies the algebra.

Worked example. For $\text{H}_2+\text{I}_2\rightleftharpoons2\text{HI}$ with $K_c=50$, start with $0.100\ \text{M}$ each of H$_2$ and I$_2$. The ICE table gives equilibrium $[\text{H}_2]=[\text{I}_2]=0.100-x$ and $[\text{HI}]=2x$. Substitute:

so $[\text{HI}]=2x=0.156\ \text{M}$. Taking the square root of both sides works here because the expression is a perfect square.

A particulate picture at equilibrium shows a fixed mix of reactant and product particles. Graphs of concentration vs time level off (never reaching zero) once equilibrium is reached – both curves flatten at the same moment.

Le Chatelier's principle 勒沙特列原理: if you disturb a system at equilibrium, it shifts to partly counteract the change. Adding a reactant shifts forward; removing product shifts forward; increasing pressure (by reducing volume) shifts toward the side with fewer gas moles; raising temperature shifts in the endothermic direction.

Le Chatelier's principle: the equilibrium shifts to oppose the change made

You can justify every Le Chatelier shift with $Q$ versus $K$: a disturbance changes $Q$, and the system reacts to bring $Q$ back to $K$. This quantitative view backs up the qualitative rule and is the fuller answer on the exam.

For a slightly soluble salt, the solubility product 溶度积 $K_{sp}$ is the equilibrium constant for its dissolving:

A smaller $K_{sp}$ means less soluble. The common-ion effect – adding an ion already in the salt – pushes the equilibrium back and lowers solubility.

Worked example. Silver chloride has $K_{sp}=1.8\times10^{-10}$. If its molar solubility is $s$, then $[\text{Ag}^+]=[\text{Cl}^-]=s$, so $K_{sp}=s^2$ and

Adding NaCl (a common ion) would raise $[\text{Cl}^-]$, forcing $s$ down – far less AgCl would dissolve.

• At equilibrium the forward and reverse rates are equal, but the amounts are usually not.
• Leave pure solids and liquids out of the $K$ expression; a large $K$ favours products, a small $K$ favours reactants.
• Apply Le Chatelier: add reactant → shift forward; raise pressure → shift toward fewer gas moles; raise temperature → shift in the endothermic direction.
• A catalyst does not shift the position of equilibrium — it only speeds the approach.
• Use an ICE table for equilibrium concentrations, and the small-$x$ approximation when $K$ is small.