平衡

A reversible reaction 可逆反应 can go both ways. The forward reaction 正反应 makes products; the reverse reaction 逆反应 turns products back into reactants. We show this with the sign $\rightleftharpoons$.

In a closed system 封闭系统 (nothing enters or leaves), the reaction reaches a dynamic equilibrium 动态平衡 when:

• the rate of the forward reaction equals the rate of the reverse reaction.
• the concentrations of reactants and products stay constant.

It is called dynamic because both reactions are still happening — they just cancel out. A closed system is needed, or products would escape and equilibrium could never be reached.

Le Chatelier's principle 勒夏特列原理 says: if you change a system at equilibrium, the position of equilibrium 平衡 moves to oppose (reduce) that change.

A catalyst lets equilibrium be reached faster, but it does not change the position of equilibrium.

Explaining a shift with $K_c$. At A level, "it opposes the change" is not enough; explain it with the constant. Take $\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})$, where the forward reaction is endothermic and

• Concentration change. At equilibrium $[\text{N}_2\text{O}_4] = 0.40$ and $[\text{NO}_2] = 0.20\ \text{mol dm}^{-3}$, so $K_c = 0.20^2/0.40 = 0.10$. Extra $\text{N}_2\text{O}_4$ raises its concentration to $0.70$. Now $\dfrac{0.20^2}{0.70} = 0.057$, which is smaller than $K_c$. The temperature has not changed, so $K_c$ is still $0.10$: some $\text{N}_2\text{O}_4$ must turn into $\text{NO}_2$ until the fraction is $0.10$ again. The equilibrium moves to the right.
• Temperature change. Raising the temperature favours the endothermic direction, here the forward one. This time $K_c$ itself changes (from $0.10$ to about $0.20$), so the mixture moves to the right until it fits the new value.

Reading and sketching these graphs.

• A substance that is added or removed jumps vertically at that moment; the other lines do not jump.
• After the jump, the lines bend towards new steady values and become flat again: a new equilibrium.
• The changes follow the equation: here $[\text{NO}_2]$ rises by twice as much as $[\text{N}_2\text{O}_4]$ falls (1 : 2).
• A temperature change gives no jump, only gradual changes.
• An increase in pressure made by compressing the gas makes every gas line jump up, because all the concentrations rise together. Then the shift towards fewer gas molecules follows.

Common mistakes.

• Saying that adding a reactant "increases $K_c$". Only temperature changes $K_c$; a concentration or pressure change moves the position.
• Saying the system goes back to the original concentrations. The shift only partly undoes the change: in the graph $[\text{N}_2\text{O}_4]$ ends at $0.67$, not $0.40$.
• Saying a catalyst "gives more product". It gives the same equilibrium mixture, only sooner.
• Predicting the temperature effect without checking the sign of $\Delta H$. Find which direction is endothermic first.

For a reaction at equilibrium, the equilibrium constant 平衡常数 links the amounts of products and reactants. For the reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:

where the square brackets mean concentration in $\text{mol dm}^{-3}$.

Worked example. At equilibrium the mixture $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ has $[\text{H}_2] = 0.20$, $[\text{I}_2] = 0.20$ and $[\text{HI}] = 1.6\ \text{mol dm}^{-3}$. Find $K_c$.

(no units here, because the concentration powers cancel top and bottom).

Worked example (ICE table). $1.00\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ are sealed in a $1.00\ \text{dm}^3$ flask; at equilibrium $0.20\ \text{mol}$ of $\text{H}_2$ remains. Set out an ICE table (Initial, Change, Equilibrium):

$0.80\ \text{mol}$ of $\text{H}_2$ reacted, so $2\times0.80 = 1.60\ \text{mol}$ HI formed. In the $1.00\ \text{dm}^3$ flask the concentrations equal the moles, so $K_c = \dfrac{1.60^2}{0.20\times0.20} = 64$.

For reactions of gases, we use partial pressure 分压 instead of concentration. The partial pressure of a gas is the share of the total pressure that it provides. It is found from the mole fraction 摩尔分数 (the fraction of all the moles that are that gas):

The constant written with partial pressures is $K_p$.

Worked example. A gas mixture holds $2.0\ \text{mol}$ of $\text{N}_2$ and $6.0\ \text{mol}$ of $\text{H}_2$ at a total pressure of $200\ \text{kPa}$. Find the partial pressure of each gas.

There are $8.0\ \text{mol}$ in total, so

Worked example (a numeric $K_p$). For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the equilibrium partial pressures are $p(\text{N}_2)=20$, $p(\text{H}_2)=40$ and $p(\text{NH}_3)=10\ \text{kPa}$. Then

Only temperature changes the value of $K_c$ or $K_p$. Changing concentration or pressure shifts the position of equilibrium but leaves the constant unchanged; adding a catalyst changes neither the constant nor the position — it only makes equilibrium arrive faster.

These two industrial processes are chosen by balancing yield, rate and cost using Le Chatelier's principle.

• the Haber process 哈伯法 makes ammonia: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic). It uses about $450\,°\text{C}$, $200\ \text{atm}$ and an iron catalyst. A low temperature would give more ammonia but too slowly, so a moderate temperature is a compromise.
• the Contact process 接触法 makes sulfur trioxide: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ (exothermic). It uses about $450\,°\text{C}$, near $1$–$2\ \text{atm}$ and a vanadium(V) oxide catalyst.

A proton 质子 is simply an $\text{H}^+$ ion. The Brønsted–Lowry theory defines acids and bases by what they do with protons:

• an acid 酸 is a proton donor 质子供体 (it gives away $\text{H}^+$).
• a base 碱 is a proton acceptor 质子受体 (it takes $\text{H}^+$). A base that dissolves in water is called an alkali.

Common acids you must know: hydrochloric acid ($\text{HCl}$), sulfuric acid ($\text{H}_2\text{SO}_4$), nitric acid ($\text{HNO}_3$) and ethanoic acid ($\text{CH}_3\text{COOH}$). Common alkalis: sodium hydroxide ($\text{NaOH}$), potassium hydroxide ($\text{KOH}$) and ammonia ($\text{NH}_3$).

This is about how fully an acid or base splits up in water — not how concentrated it is.

• a strong acid 强酸 or strong base 强碱 is fully dissociated 解离 in water (almost every molecule splits into ions).
• a weak acid 弱酸 or weak base 弱碱 is only partly dissociated (most molecules stay whole).

The pH scale measures how acidic a solution is: pure water is pH 7, acids are below 7, and alkalis are above 7. A strong acid has a lower pH than a weak acid of the same concentration.

You can tell strong and weak acids apart by:

• reaction with a reactive metal: a strong acid fizzes faster.
• pH: measured with a pH meter or universal indicator 通用指示剂.
• electrical conductivity: a strong acid conducts better, because it has more ions.

Neutralisation 中和 happens when the hydrogen ions from an acid react with the hydroxide ions from an alkali:

A salt 盐 is also formed, from the rest of the acid and base.

In a titration 滴定 you add one solution to another and follow the pH. The titration curve 滴定曲线 has a steep, almost vertical jump near the end point. The exact shape depends on whether each reactant is strong or weak.

To pick an indicator 指示剂, choose one whose colour change falls inside that steep jump. For a strong acid with a strong base most indicators work; for a weak acid with a strong base you need one that changes in the higher pH range.

• Define dynamic equilibrium with all three points: forward and reverse rates equal, concentrations constant, closed system.
• Answer Le Chatelier questions by stating the change, the direction of shift, and the reason; a catalyst does not shift the position (it speeds both rates).
• Write $K_c$ as products over reactants, each raised to its balancing number; include state symbols to decide what appears.
• Justify Haber/Contact conditions as a compromise between yield, rate and cost — do not just quote them.
• Brønsted: acid = proton donor, base = proton acceptor; conjugate pairs differ by one $\text{H}^+$.