(I once asked a class, "What is the most important question in chemistry?" Some students said, "What reacts with what?" Others said, "What is the product?" Then I said, "No. The two most important questions are: how fast, and how far? How fast is kinetics. How far is equilibrium. Together, they determine everything about a reaction — whether it will happen in a second or a century, whether it will go to completion or stop halfway, and how we can control it. In this chapter, we go deeper into both questions than ever before. We will learn about rate equations, order of reaction, and the Arrhenius equation. We will explore dynamic equilibrium, equilibrium constants, and the factors that shift equilibrium. By the end, you will understand the two most fundamental questions in chemistry — how fast and how far. Let us begin.) In your earlier studies, you learned that reaction rates depend on concentration, temperature, surface area, pressure, and catalysts. You learned about collision theory and activation energy. You also learned about dynamic equilibrium and Le Chatelier's principle. That is a solid foundation — but at the advanced level, we go deeper. We ask: How exactly does concentration affect rate? What is the order of a reaction, and how do we determine it? How does temperature affect the rate constant? How do we write an equilibrium constant expression? What does the size of the equilibrium constant tell us? These are the questions that this chapter answers. We will explore rate equations, rate constants, order of reaction, and the Arrhenius equation. We will learn about the equilibrium constant (Kc and Kp), the reaction quotient (Q), and how to use them to predict the direction of a reaction. By the end, you will have a deep understanding of reaction kinetics and equilibria — the two pillars of chemical dynamics. Let us begin.
The rate of a reaction is the change in concentration of a reactant or product per unit time. For a general reaction:
aA + bB → cC + dD
The rate can be expressed as:
Rate = −(1/a)(Δ[A]/Δt) = −(1/b)(Δ[B]/Δt) = +(1/c)(Δ[C]/Δt) = +(1/d)(Δ[D]/Δt)
Example: For the reaction N₂ + 3H₂ → 2NH₃, the rate can be expressed as:
Rate = −(Δ[N₂]/Δt) = −(1/3)(Δ[H₂]/Δt) = +(1/2)(Δ[NH₃]/Δt)
Important: The rate of a reaction changes over time. It is usually fastest at the beginning (when reactant concentrations are highest) and slows down as the reactants are used up.
For many reactions, the rate is proportional to the concentration of the reactants raised to some power. This gives the rate equation (or rate law):
Rate = k[A]^m[B]^n
Where:
The overall order of reaction is m + n.
| Order | Effect of Doubling Concentration | Example Rate Equation |
|---|---|---|
| Zero order (0) | No effect on rate | Rate = k |
| First order (1) | Rate doubles | Rate = k[A] |
| Second order (2) | Rate quadruples | Rate = k[A]² |
| Third order (3) | Rate increases by factor of 8 | Rate = k[A]³ |
Important: The order of a reaction must be determined experimentally. It cannot be deduced from the balanced equation.
There are several methods for determining the order of a reaction:
Measure the initial rate of the reaction at different initial concentrations of the reactants. Compare the rates to determine the order.
Example: If doubling [A] doubles the rate, the reaction is first order with respect to A. If doubling [A] quadruples the rate, it is second order.
Plot concentration against time and analyse the shape of the curve:
| Order | Graph of Concentration vs Time | Graph of Rate vs Concentration |
|---|---|---|
| Zero | Straight line (constant slope) | Horizontal line |
| First | Exponential decay curve | Straight line through origin |
| Second | Curve that approaches zero more slowly | Parabolic curve |
For a first-order reaction, the half-life is constant and independent of the initial concentration. For other orders, the half-life changes with concentration.
The Arrhenius equation relates the rate constant (k) to temperature:
k = A × e^(−Ea/RT)
Where:
Taking the natural logarithm of both sides gives:
ln(k) = ln(A) − Ea/(RT)
This is the equation of a straight line: plotting ln(k) against 1/T gives a straight line with slope −Ea/R and intercept ln(A).
Importance: The Arrhenius equation shows that a small increase in temperature causes a large increase in the rate constant, because the exponential term is very sensitive to temperature.
(The Arrhenius equation is one of the most useful equations in chemistry. It tells you exactly how much faster a reaction will go if you raise the temperature. It is the reason a 10°C rise can double or triple a reaction rate — and it is the reason food cooks faster at higher temperatures.)
A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. The catalyst is not used up in the reaction.
| Type of Catalysis | Description | Example |
|---|---|---|
| Homogeneous | Catalyst and reactants are in the same phase. | Acid catalysis in esterification (all liquids). |
| Heterogeneous | Catalyst and reactants are in different phases. | Iron in the Haber process (solid catalyst, gaseous reactants). |
| Enzyme catalysis | Biological catalysts (proteins) that are highly specific. | Amylase, catalase, proteases. |
How a catalyst works: A catalyst provides a surface or a chemical environment where the reactants can come together with a lower activation energy. For heterogeneous catalysts, the reactants adsorb onto the catalyst surface, where they can react more easily. For homogeneous catalysts, the catalyst reacts with a reactant to form an intermediate, which then reacts further to give the product and regenerate the catalyst.
Catalyst poisoning: Some substances (called poisons) bind strongly to the catalyst and prevent it from working. For example, lead poisons the catalyst in a catalytic converter.
In a reversible reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the backward reaction. The concentrations of reactants and products remain constant — but they are not necessarily equal.
Key features of dynamic equilibrium:
For a general reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant (Kc) is defined as:
Kc = [C]^c [D]^d / [A]^a [B]^b
Where the concentrations are the equilibrium concentrations.
For reactions involving gases, we can use partial pressures instead of concentrations:
Kp = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b
Relationship between Kc and Kp: Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants).
| Value of K | Meaning |
|---|---|
| K >> 1 | Equilibrium lies to the right — mostly products. |
| K | Equilibrium lies to the left — mostly reactants. |
| K ≈ 1 | Significant amounts of both reactants and products at equilibrium. |
Important: The equilibrium constant is only affected by temperature. Changing concentration, pressure, or adding a catalyst does not change K — it only changes the position of equilibrium.
The reaction quotient (Q) has the same form as the equilibrium constant, but uses the current concentrations (not necessarily at equilibrium).
Q = [C]^c [D]^d / [A]^a [B]^b
By comparing Q with K, we can predict the direction in which the reaction will proceed:
| Comparison | Direction of Reaction |
|---|---|
| Q | Reaction proceeds to the right (towards products). |
| Q > K | Reaction proceeds to the left (towards reactants). |
| Q = K | System is at equilibrium. |
Le Chatelier's principle states that if a system at equilibrium is subjected to a change, the system will shift in the direction that tends to counteract the change and restore equilibrium.
| Change | Effect on Equilibrium | Effect on K |
|---|---|---|
| Increase concentration of reactant | Shifts to the right | No change |
| Increase concentration of product | Shifts to the left | No change |
| Increase temperature (exothermic reaction) | Shifts to the left | Decreases |
| Increase temperature (endothermic reaction) | Shifts to the right | Increases |
| Increase pressure (fewer gas moles on product side) | Shifts to the right | No change |
| Increase pressure (more gas moles on product side) | Shifts to the left | No change |
| Add catalyst | No shift | No change |
| Add inert gas at constant volume | No shift | No change |
Important: Only temperature changes affect the value of the equilibrium constant.
The principles of kinetics and equilibria are used extensively in industry to optimise chemical processes.
| Process | Key Conditions | Reasoning |
|---|---|---|
| Haber process (ammonia) | 450°C, 200 atm, iron catalyst | Compromise between yield (favoured by low temperature, high pressure) and rate (favoured by high temperature, catalyst). |
| Contact process (sulphuric acid) | 450°C, 1–2 atm, vanadium(V) oxide catalyst | Compromise between yield and rate. High pressure not used because yield is already high at low pressure. |
| Ostwald process (nitric acid) | 900°C, platinum-rhodium catalyst | High temperature and catalyst used to increase rate; equilibrium not a major concern because the reaction goes essentially to completion. |
| Esterification | Reflux with acid catalyst | Catalyst speeds up the reaction; excess alcohol or removal of water shifts equilibrium to the right. |
The effect of temperature on equilibrium depends on whether the forward reaction is exothermic or endothermic:
This is consistent with Le Chatelier's principle: the system tries to absorb the added heat by favouring the endothermic direction.
Kinetics and equilibrium are closely related:
(This relationship between kinetics and equilibrium is one of the deepest ideas in chemistry. It tells us that equilibrium is not a static state — it is a dynamic balance between two opposing reactions. And the ratio of their rates determines where the balance lies. Once you understand this, you understand the heart of chemical dynamics.)
Question: For the reaction A + B → C, the following initial rates were measured:
| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 0.02 |
| 2 | 0.2 | 0.1 | 0.04 |
| 3 | 0.1 | 0.2 | 0.08 |
Determine the order with respect to A and B, and write the rate equation.
Solution: Compare experiments 1 and 2: [A] doubles, [B] stays the same, rate doubles → first order with respect to A. Compare experiments 1 and 3: [B] doubles, [A] stays the same, rate quadruples → second order with respect to B. Rate equation: Rate = k[A][B]². Overall order = 1 + 2 = 3.
Question: For the reaction H₂ + I₂ ⇌ 2HI, the equilibrium concentrations are [H₂] = 0.2 M, [I₂] = 0.2 M, and [HI] = 1.6 M. Calculate Kc.
Solution: Kc = [HI]² / ([H₂][I₂]) = (1.6)² / (0.2 × 0.2) = 2.56 / 0.04 = 64.
Question: For the reaction N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.5 at a certain temperature. If the current concentrations are [N₂] = 0.1 M, [H₂] = 0.1 M, and [NH₃] = 0.05 M, in which direction will the reaction proceed?
Solution: Q = [NH₃]² / ([N₂][H₂]³) = (0.05)² / (0.1 × 0.1³) = 0.0025 / (0.1 × 0.001) = 0.0025 / 0.0001 = 25. Since Q (25) > K (0.5), the reaction will proceed to the left (towards reactants) to reach equilibrium.
Question: The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Calculate the activation energy. (R = 8.314 J/K·mol)
Solution: Using the Arrhenius equation in the form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). ln(2) = (Ea/8.314)(1/300 − 1/310). 0.693 = (Ea/8.314)(0.00333 − 0.00323). 0.693 = (Ea/8.314)(0.0001). Ea = 0.693 × 8.314 / 0.0001 = 57,600 J/mol = 57.6 kJ/mol.
1. Define rate of reaction.
The rate of a reaction is the change in concentration of a reactant or product per unit time.
2. What is the order of a reaction?
The order of a reaction is the power to which the concentration of a reactant is raised in the rate equation. It must be determined experimentally.
3. State the Arrhenius equation.
k = A × e^(−Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
4. Write the equilibrium constant expression for the reaction N₂ + 3H₂ ⇌ 2NH₃.
Kc = [NH₃]² / ([N₂][H₂]³).
5. What does a large equilibrium constant (K >> 1) indicate?
A large K indicates that the equilibrium lies to the right — mostly products at equilibrium.
6. How can you predict the direction of a reaction using Q and K?
If Q K, the reaction proceeds to the left (towards reactants). If Q = K, the system is at equilibrium.
7. Explain why a catalyst does not change the position of equilibrium.
A catalyst speeds up the forward and backward reactions equally, so it does not change the relative rates or the position of equilibrium. It only helps the system reach equilibrium faster.
Question 1 [10 marks]
(a) Define rate of reaction and describe how it can be measured. [3]
(b) Describe the collision theory and explain the role of activation energy. [4]
(c) Explain, using collision theory, why increasing temperature increases the rate of reaction. [3]
(a) Rate of reaction is the change in concentration of a reactant or product per unit time. It can be measured by measuring the volume of gas produced, the loss in mass, the change in colour, or the time for a precipitate to form. (3 marks: 1 for definition, 2 for methods.)
(b) Collision theory states that for a reaction to occur, particles must collide with enough energy and the correct orientation. Activation energy is the minimum energy required for a collision to result in a reaction. (4 marks: 2 for collision theory, 2 for activation energy.)
(c) Increasing temperature gives particles more kinetic energy. They move faster and collide more frequently. More importantly, a greater proportion of collisions have enough energy to overcome the activation energy. This means more successful collisions per unit time, so the rate increases. (3 marks: 1 for more collisions, 1 for more energy, 1 for "more successful collisions".)
Question 2 [10 marks]
(a) Define equilibrium constant and write the expression for Kc for the reaction: 2SO₂ + O₂ ⇌ 2SO₃. [3]
(b) At a certain temperature, the equilibrium concentrations for the reaction are [SO₂] = 0.2 M, [O₂] = 0.1 M, and [SO₃] = 0.4 M. Calculate Kc. [3]
(c) Predict the effect on the equilibrium position of: (i) increasing the temperature (the forward reaction is exothermic), (ii) increasing the pressure, (iii) adding a catalyst. [4]
(a) The equilibrium constant (Kc) is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. For 2SO₂ + O₂ ⇌ 2SO₃: Kc = [SO₃]² / ([SO₂]²[O₂]). (3 marks: 2 for definition, 1 for expression.)
(b) Kc = (0.4)² / ((0.2)² × 0.1) = 0.16 / (0.04 × 0.1) = 0.16 / 0.004 = 40. (3 marks: 1 for substitution, 1 for calculation, 1 for answer.)
(c) (i) Increasing temperature shifts equilibrium to the left (endothermic direction), reducing SO₃ yield. (ii) Increasing pressure shifts equilibrium to the right (fewer moles of gas), increasing SO₃ yield. (iii) Adding a catalyst has no effect on the equilibrium position; it only speeds up the approach to equilibrium. (4 marks: 1 for each correct prediction, 1 for explanation.)
Question 3 [10 marks]
(a) Describe how the order of a reaction can be determined experimentally. [4]
(b) State the Arrhenius equation and explain the significance of each term. [4]
(c) Explain why a small increase in temperature can cause a large increase in reaction rate. [2]
(a) The order of a reaction can be determined by the initial rates method (measuring initial rates at different concentrations and comparing them), by graphical methods (plotting concentration vs time and analysing the shape of the curve), or by the half-life method (for first-order reactions, the half-life is constant). (4 marks: 2 for initial rates, 1 for graphical, 1 for half-life.)
(b) k = A × e^(−Ea/RT). k is the rate constant; A is the pre-exponential factor (frequency of collisions with correct orientation); Ea is the activation energy; R is the gas constant; T is the temperature in Kelvin. (4 marks: 1 for equation, 1 for k, 1 for A, 1 for Ea/R/T.)
(c) A small increase in temperature causes a large increase in reaction rate because the exponential term in the Arrhenius equation is very sensitive to temperature. The proportion of molecules with energy greater than the activation energy increases exponentially with temperature. (2 marks: 1 for exponential sensitivity, 1 for "more molecules exceed Ea".)
Challenge 1: Explain why a catalyst increases the rate of a reaction but does not change the position of equilibrium.
A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. This lower activation energy is available to both the forward and the backward reactions. Because both reactions are sped up equally, the relative rates remain the same, and the equilibrium position is unchanged. The catalyst only helps the system reach equilibrium faster. The equilibrium position depends on the thermodynamics of the reaction — the relative energies of reactants and products — not on the kinetics. So a catalyst affects the rate (kinetics) but not the position (thermodynamics).
Challenge 2: Explain how the equilibrium constant (K) changes with temperature for an exothermic reaction and an endothermic reaction.
For an exothermic reaction (ΔH negative), increasing the temperature shifts the equilibrium to the left (towards reactants) according to Le Chatelier's principle. This means the concentration of products decreases and the concentration of reactants increases, so the equilibrium constant K decreases. For an endothermic reaction (ΔH positive), increasing the temperature shifts the equilibrium to the right (towards products). This means the concentration of products increases and the concentration of reactants decreases, so the equilibrium constant K increases. In both cases, the change in K with temperature is a direct consequence of Le Chatelier's principle. Temperature is the only factor that changes the value of K.
Challenge 3: Explain why the reaction between hydrogen and oxygen to form water is spontaneous but does not occur at room temperature without a spark.
The reaction between hydrogen and oxygen to form water is highly spontaneous — the Gibbs free energy change (ΔG) is very negative. The products (water) are much more stable than the reactants (hydrogen and oxygen gas). However, the reaction does not occur at room temperature without a spark because the activation energy is very high. The H–H and O=O bonds are strong and require a lot of energy to break. At room temperature, very few collisions have enough energy to overcome this activation energy, so the reaction rate is negligible. A spark provides the necessary activation energy to start the reaction. Once started, the reaction is highly exothermic and self-sustaining. This is a classic example of the difference between thermodynamic spontaneity (ΔG negative) and kinetic rate (slow due to high activation energy).
"How fast, how far — the two questions that define every reaction. Now you understand the kinetics and equilibria that answer them."
