Chemistry - Chemical Equilibrium

A reversible reaction is one that can proceed in both the forward and reverse directions.

$\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$

Dynamic equilibrium is established when:

1. The reaction is reversible.
2. The system is a closed system (no matter can enter or leave).
3. The forward and reverse rates are equal.
4. The concentrations of all species remain constant (but not necessarily equal).

• At equilibrium, both forward and reverse reactions continue to occur (hence “dynamic”).
• Macroscopic properties (concentration, colour, pressure) are constant.
• The position of equilibrium describes the relative amounts of reactants and products.
• Equilibrium can be approached from either direction.

Caution: Warning (e.g., a gas leaving an open container), equilibrium will never be reached.

If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the System will adjust to oppose the change and restore a new equilibrium.

Worked example 1: For $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)}$What happens if More $\mathrm{N}_2$ is added?

Pressure affects equilibrium only when the number of moles of gas differs between reactants and Products.

Worked example 2: For $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)}$What happens when Pressure is increased?

For the Haber process (exothermic forward reaction):

$\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)} \quad \Delta H = -92 \mathrm{ kJ/mol}$

Increasing temperature shifts equilibrium to the left (endothermic direction), reducing the yield of $\mathrm{NH}_3$. Decreasing temperature increases the yield but slows the rate.

A catalyst increases both the forward and reverse rates equally. It has no effect on the Position of equilibrium or the equilibrium yield. It only helps the system reach equilibrium faster.

For a reversible reaction at equilibrium:

$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$

The equilibrium constant in terms of concentration is:

$K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$

Where all concentrations are equilibrium concentrations in mol/dm$^3$.

1. $K_c$ is constant at a given temperature.
2. $K_c$ is independent of the initial concentrations.
3. $K_c$ changes only with temperature.
4. Pure solids and pure liquids are not included in the $K_c$ expression (their concentrations are effectively constant, incorporated into $K_c$).
5. The units of $K_c$ depend on the stoichiometry of the reaction.

For an exothermic reaction ($\Delta H \lt 0$):

• Increasing temperature: equilibrium shifts left (endothermic), $K_c$ decreases.
• Decreasing temperature: equilibrium shifts right (exothermic), $K_c$ increases.

For an endothermic reaction ($\Delta H \gt 0$):

• Increasing temperature: equilibrium shifts right (endothermic), $K_c$ increases.
• Decreasing temperature: equilibrium shifts left (exothermic), $K_c$ decreases.

For reactions involving solids or pure liquids, only gaseous and aqueous species appear in the $K_c$ Expression.

$\mathrm{CaCO}_{3(s)} \rightleftharpoons \mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)}$

$K_c = [\mathrm{CO}_2]$

ICE (Initial, Change, Equilibrium) tables organise the data for equilibrium calculations.

Worked example 3: 1.00 mol of $\mathrm{H}_2$ and 1.00 mol of $\mathrm{I}_2$ are placed in a 1.00 Dm$^3$ vessel at 450$^\circ$C. At equilibrium, 0.78 mol of $\mathrm{HI}$ has formed. Calculate $K_c$.

$\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2\mathrm{HI}_{(g)}$

Worked example 4: 2.00 mol of $\mathrm{NO}$ and 1.00 mol of $\mathrm{Cl}_2$ are placed in a 2.00 Dm$^3$ flask. At equilibrium, 0.60 mol of $\mathrm{NOCl}$ is present. Calculate $K_c$.

$2\mathrm{NO}_{(g)} + \mathrm{Cl}_{2(g)} \rightleftharpoons 2\mathrm{NOCl}_{(g)}$

Worked example 5: For the reaction $\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)}$ $K_c = 0.0211 \mathrm{ mol/dm}^3$ at a certain temperature. If 1.00 mol of $\mathrm{PCl}_5$ is Placed in a 1.00 dm$^3$ flask, calculate the equilibrium concentrations.

The reaction quotient $Q_c$ has the same form as $K_c$ but uses initial (non-equilibrium) Concentrations.

$Q_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \quad \mathrm{(initial concentrations)}$

Worked example 6: For $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)}$ $K_c = 0.500 \mathrm{ (mol/dm}^3)^{-2}$ at a certain temperature. In a 2.00 dm$^3$ flask, [\mathrm{N}_2] = 1.00$$[\mathrm{H}_2] = 1.00$$[\mathrm{NH}_3] = 0.500 mol/dm$^3$. Will more $\mathrm{NH}_3$ form or will it decompose?

$\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)} \quad \Delta H = -92 \mathrm{ kJ/mol}$

At 450$^\circ$C and 200 atm, the single-pass conversion is only about 15%. However, by continuously Removing $\mathrm{NH}_3$ (by condensation) and recycling unreacted gases, the overall conversion Reaches about 98%.

$2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{SO}_{3(g)} \quad \Delta H = -198 \mathrm{ kJ/mol}$

1. Sulfur burn: Sulfur or metal sulfide ores are burned in air to produce $\mathrm{SO}_2$.

$\mathrm{S}_{(s)} + \mathrm{O}_{2(g)} \to \mathrm{SO}_{2(g)}$

1. Purification: $\mathrm{SO}_2$ is purified to remove impurities that could poison the catalyst.

2. Catalytic oxidation: $\mathrm{SO}_2$ is oxidised to $\mathrm{SO}_3$ over a $\mathrm{V}_2\mathrm{O}_5$ catalyst.

3. Absorption: $\mathrm{SO}_3$ is dissolved in concentrated $\mathrm{H}_2\mathrm{SO}_4$ to form oleum ($\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7$), which is then diluted to give $\mathrm{H}_2\mathrm{SO}_4$.

$\mathrm{SO}_{3(g)} + \mathrm{H}_2\mathrm{SO}_{4(l)} \to \mathrm{H}_2\mathrm{S}_2\mathrm{O}_{7(l)}$

$\mathrm{H}_2\mathrm{S}_2\mathrm{O}_{7(l)} + \mathrm{H}_2\mathrm{O}_{(l)} \to 2\mathrm{H}_2\mathrm{SO}_{4(l)}$

• Higher pressure would favour $\mathrm{SO}_3$ (3 mol gas to 2 mol gas), but the improvement is small and not worth the cost of high-pressure equipment.
• Lower temperature would favour $\mathrm{SO}_3$ (exothermic), but 400—450$^\circ$C is needed for an acceptable rate with the $\mathrm{V}_2\mathrm{O}_5$ catalyst.
• Excess $\mathrm{O}_2$ is used to shift equilibrium to the right and improve $\mathrm{SO}_2$ conversion.

$4\mathrm{NH}_{3(g)} + 5\mathrm{O}_{2(g)} \rightleftharpoons 4\mathrm{NO}_{(g)} + 6\mathrm{H}_2\mathrm{O}_{(g)} \quad \Delta H = -905 \mathrm{ kJ/mol}$

• Catalyst: Platinum-rhodium alloy at 850$^\circ$C, 8 atm.
• The $\mathrm{NO}$ produced is further oxidised to $\mathrm{NO}_2$ Then absorbed in water to form $\mathrm{HNO}_3$.

$\mathrm{C}_2\mathrm{H}_{4(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightleftharpoons \mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(g)} \quad \Delta H = -46 \mathrm{ kJ/mol}$

• Catalyst: Phosphoric acid on silica support.
• Conditions: 300$^\circ$C, 60—70 atm.
• Excess steam shifts equilibrium to the right.

Worked example 7: For $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)}$ $K_c = 6.00 \times 10^{-2} \mathrm{ (mol/dm}^3)^{-2}$ at 400$^\circ$C. If 1.00 mol of $\mathrm{N}_2$ And 3.00 mol of $\mathrm{H}_2$ are placed in a 2.00 dm$^3$ flask, calculate the equilibrium Concentrations and the percentage conversion of $\mathrm{N}_2$.

For gaseous equilibria, $K_p$ uses partial pressures instead of concentrations.

$K_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}$

The relationship between $K_c$ and $K_p$ is:

$K_p = K_c(RT)^{\Delta n}$

Where $\Delta n = (\mathrm{moles of gaseous products}) - (\mathrm{moles of gaseous reactants})$.

For $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)}$: $\Delta n = 2 - 4 = -2$.

$K_p = K_c(RT)^{-2} = \frac{K_c}{(RT)^2}$

If $\Delta n = 0$ Then $K_p = K_c$.

Worked example 8: For $\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2\mathrm{HI}_{(g)}$, $K_c = 50.0$ at 700 K. Calculate $K_p$.

A tug-of-war that never ends: Chemical equilibrium is like two teams pulling on a rope — both forward and reverse reactions happen at the same rate, so concentrations don’t change, but reactions haven’t stopped.

Why it matters: From industrial ammonia production to blood oxygen transport, equilibrium determines how much product you get. Le Chatelier’s principle predicts how systems respond to disturbances.

The key insight: Changing concentration or pressure shifts equilibrium, but changing temperature changes the equilibrium constant itself — this distinction is crucial.

1. Including solids and liquids in $K_c$: Only aqueous and gaseous species appear in the expression. Solids and pure liquids have constant effective concentrations.

2. Confusing $K_c$ and $Q_c$: $K_c$ uses equilibrium concentrations; $Q_c$ uses current (non-equilibrium) concentrations. Compare $Q_c$ to $K_c$ to determine direction.

3. Changing concentration does not change $K_c$: Adding or removing reactants/products shifts the equilibrium position but does not change $K_c$ (at constant temperature).

4. Catalysts and equilibrium: A catalyst does NOT change the position of equilibrium or $K_c$. It only speeds up attainment of equilibrium.

5. Units of $K_c$: Always include units. For $\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3$The units are $(\mathrm{mol/dm}^3)^{-2}$.

6. Pressure changes only affect gaseous equilibria when $\Delta n \neq 0$: If the moles of gas are the same on both sides, changing pressure has no effect on equilibrium position.

7. Reversing the reaction inverts $K_c$: For the reverse reaction, $K_c' = 1/K_c$.

8. Multiplying the equation by $n$ raises $K_c$ to the $n$Th power: If the equation is multiplied by 2, $K_c' = (K_c)^2$.

At a certain temperature, $K_c = 4.00$ for the reaction:

$\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2\mathrm{HI}_{(g)}$

If 2.00 mol of $\mathrm{H}_2$ and 2.00 mol of $\mathrm{I}_2$ are placed in a 1.00 dm$^3$ flask, Calculate the equilibrium concentrations of all species.

For the reaction $\mathrm{CO}_{(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightleftharpoons \mathrm{CO}_{2(g)} + \mathrm{H}_{2(g)}$ $K_c = 1.60$ at 900 K. If 1.00 mol of $\mathrm{CO}$ and 1.00 mol of $\mathrm{H}_2\mathrm{O}$ are Placed in a 2.00 dm$^3$ container, calculate the equilibrium concentrations and the percentage of $\mathrm{CO}$ that has reacted.

For the exothermic reaction $\mathrm{A}_{(g)} + \mathrm{B}_{(g)} \rightleftharpoons 2\mathrm{C}_{(g)}$Explain the effect of Each of the following changes on (i) the equilibrium position, (ii) the value of $K_c$ And (iii) The rate of attainment of equilibrium:

(a) Increasing temperature (b) Adding more $\mathrm{A}$ (c) Adding a catalyst (d) Decreasing the Volume of the container

At 500 K, $K_c = 0.0400 \mathrm{ (mol/dm}^3)$ for the reaction:

$\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)}$

0.800 mol of $\mathrm{PCl}_5$ is placed in a 5.00 dm$^3$ flask. Calculate the equilibrium Concentrations and the percentage dissociation of $\mathrm{PCl}_5$.

Explain why in the Haber process, a temperature of 450$^\circ$C is used instead of room temperature, Even though a lower temperature would give a higher equilibrium yield of ammonia.

For the reaction $\mathrm{N}_2\mathrm{O}_{4(g)} \rightleftharpoons 2\mathrm{NO}_{2(g)}$ $K_c = 0.361 \mathrm{ mol/dm}^3$ at 373 K. If 0.500 mol of $\mathrm{N}_2\mathrm{O}_4$ is placed in a 2.00 dm$^3$ flask at 373 K, calculate the equilibrium concentrations and the total pressure of the Gas mixture at equilibrium.

For the reaction $2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{SO}_{3(g)}$ at 500$^\circ$C, $K_c = 2.50 \times 10^{10} \mathrm{ (mol/dm}^3)^{-1}$. Explain why this very large Value of $K_c$ does NOT mean that $\mathrm{SO}_2$ and $\mathrm{O}_2$ cannot be present at Equilibrium.

When a disturbance is applied to an equilibrium system, the system shifts to partially oppose the Change. The new equilibrium can be estimated using $K_c$.

Worked example 9: For $\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2\mathrm{HI}_{(g)}$, $K_c = 49.0$ at a Certain temperature. At equilibrium, $[\mathrm{H}_2] = [\mathrm{I}_2] = 0.100$ and $[\mathrm{HI}] = 0.700$ mol/dm$^3$. If 0.200 mol/dm$^3$ of $\mathrm{HI}$ is suddenly added, what are The new equilibrium concentrations?

The equilibrium constant is related to the standard Gibbs free energy change:

$\Delta G^\circ = -RT \ln K$

Where $R = 8.314 \mathrm{ J/(mol K)}$ and $K$ is the equilibrium constant (dimensionless, or use $K_c$ with appropriate standard state of 1 mol/dm$^3$).

At equilibrium, $\Delta G = 0$ (not $\Delta G^\circ = 0$).

$\Delta G = \Delta G^\circ + RT \ln Q$

When $\Delta G = 0$: $0 = \Delta G^\circ + RT \ln K$Giving $\Delta G^\circ = -RT \ln K$.

Worked example 10: $\Delta G^\circ = -5.40 \mathrm{ kJ/mol}$ for a reaction at 298 K. Calculate $K$.

flowchart TD
    A[1_Chemical Equilibrium] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]