Hello dear student! Welcome to Unit 4 — Chemical Kinetics. Have you ever wondered why some reactions happen instantly (like an explosion) while others take hours, days, or even years (like rusting)? Chemical kinetics is the branch of chemistry that studies how fast reactions occur and why they occur at different speeds. This unit is very important for your exam, so let’s learn it carefully, step by step.
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4.1 Introduction
In Unit 5, you will study chemical equilibrium — whether a reaction happens at all and how far it proceeds. But here in Unit 4, we focus on speed. Thermodynamics tells us IF a reaction can happen; kinetics tells us HOW FAST it happens.
Think about it this way: just because a reaction is thermodynamically favorable (spontaneous) does NOT mean it will be fast. For example, the conversion of diamond to graphite is spontaneous, but it takes millions of years! So kinetics is a separate and very important topic.
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4.2 The Rate of a Reaction
4.2.1 What Is Rate of Reaction?
The rate of a reaction is the change in concentration of a reactant or product per unit time.
For a general reaction:
$$aA + bB \rightarrow cC + dD$$
The rate can be expressed in terms of any species, but we must account for stoichiometric coefficients:
$$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = \frac{1}{c}\frac{\Delta[C]}{\Delta t} = \frac{1}{d}\frac{\Delta[D]}{\Delta t}$$
Important conventions:
- Negative sign for reactants: their concentration decreases over time.
- Positive sign for products: their concentration increases over time.
- We divide by the coefficient so that the rate has the same value regardless of which species we measure.
Worked Example 4.1: For the reaction $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, if the concentration of N₂ decreases by 0.10 mol/L in 5 seconds, what is the rate of reaction?
Solution:
$$\text{Rate} = -\frac{1}{1}\frac{\Delta[\text{N}_2]}{\Delta t} = -\frac{(-0.10)}{5} = \frac{0.10}{5} = 0.020 \text{ mol/(L·s)}$$
Note: Δ[N₂] = final − initial = −0.10 (negative because N₂ is decreasing), so the negative signs cancel.
Solution:
$$\text{Rate} = -\frac{1}{1}\frac{\Delta[\text{N}_2]}{\Delta t} = -\frac{(-0.10)}{5} = \frac{0.10}{5} = 0.020 \text{ mol/(L·s)}$$
Note: Δ[N₂] = final − initial = −0.10 (negative because N₂ is decreasing), so the negative signs cancel.
Worked Example 4.2: For the same reaction $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with rate = 0.020 mol/(L·s), what is the rate of disappearance of H₂ and the rate of appearance of NH₃?
Solution:
$$-\frac{\Delta[\text{H}_2]}{\Delta t} = 3 \times \text{Rate} = 3 \times 0.020 = 0.060 \text{ mol/(L·s)}$$
$$\frac{\Delta[\text{NH}_3]}{\Delta t} = 2 \times \text{Rate} = 2 \times 0.020 = 0.040 \text{ mol/(L·s)}$$
H₂ disappears 3 times faster than N₂; NH₃ appears 2 times faster than N₂.
Solution:
$$-\frac{\Delta[\text{H}_2]}{\Delta t} = 3 \times \text{Rate} = 3 \times 0.020 = 0.060 \text{ mol/(L·s)}$$
$$\frac{\Delta[\text{NH}_3]}{\Delta t} = 2 \times \text{Rate} = 2 \times 0.020 = 0.040 \text{ mol/(L·s)}$$
H₂ disappears 3 times faster than N₂; NH₃ appears 2 times faster than N₂.
4.2.2 Average Rate vs Instantaneous Rate
Average rate: The rate over a finite time interval.
$$\text{Average rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{[A]_2 – [A]_1}{t_2 – t_1}$$
Instantaneous rate: The rate at a specific instant in time. It is the slope of the tangent to the concentration-vs-time curve at that point.
$$\text{Instantaneous rate} = -\frac{d[A]}{dt}$$
The instantaneous rate at t = 0 is called the initial rate. Initial rates are very important in determining the rate law experimentally.
Key Exam Notes — Rate of Reaction:
• Rate = change in concentration per unit time.
• Always divide by the stoichiometric coefficient when relating rates of different species.
• Negative sign for reactants (disappearing), positive for products (appearing).
• Average rate uses Δ (finite interval); instantaneous rate uses d/dt (at one point).
• The instantaneous rate at t = 0 is the initial rate.
• Units of rate: mol/(L·s) or mol L⁻¹ s⁻¹ or M/s.
• Rate = change in concentration per unit time.
• Always divide by the stoichiometric coefficient when relating rates of different species.
• Negative sign for reactants (disappearing), positive for products (appearing).
• Average rate uses Δ (finite interval); instantaneous rate uses d/dt (at one point).
• The instantaneous rate at t = 0 is the initial rate.
• Units of rate: mol/(L·s) or mol L⁻¹ s⁻¹ or M/s.
Practice Questions:
1. For the reaction $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$, if [SO₂] decreases by 0.04 M in 2 seconds, calculate the rate of reaction and the rate of disappearance of O₂.
2. Why do we divide by the stoichiometric coefficient when expressing rate?
1. For the reaction $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$, if [SO₂] decreases by 0.04 M in 2 seconds, calculate the rate of reaction and the rate of disappearance of O₂.
2. Why do we divide by the stoichiometric coefficient when expressing rate?
Answer 1:
$$\text{Rate} = -\frac{1}{2}\frac{\Delta[\text{SO}_2]}{\Delta t} = -\frac{1}{2}\times\frac{(-0.04)}{2} = \frac{0.02}{2} = 0.010 \text{ mol/(L·s)}$$
Rate of disappearance of O₂: $-\frac{\Delta[\text{O}_2]}{\Delta t} = 1 \times \text{Rate} = 0.010 \text{ mol/(L·s)}$
Answer 2:
Without dividing by the coefficient, the rate would depend on which species you measure. For N₂ + 3H₂ → 2NH₃, H₂ disappears 3 times faster than N₂. Dividing by the coefficient ensures all expressions give the same value for the rate of reaction, regardless of which species is monitored.
$$\text{Rate} = -\frac{1}{2}\frac{\Delta[\text{SO}_2]}{\Delta t} = -\frac{1}{2}\times\frac{(-0.04)}{2} = \frac{0.02}{2} = 0.010 \text{ mol/(L·s)}$$
Rate of disappearance of O₂: $-\frac{\Delta[\text{O}_2]}{\Delta t} = 1 \times \text{Rate} = 0.010 \text{ mol/(L·s)}$
Answer 2:
Without dividing by the coefficient, the rate would depend on which species you measure. For N₂ + 3H₂ → 2NH₃, H₂ disappears 3 times faster than N₂. Dividing by the coefficient ensures all expressions give the same value for the rate of reaction, regardless of which species is monitored.
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4.2.3 Rate Law and Rate Constant
The rate law (rate equation) is an equation that relates the rate of reaction to the concentrations of reactants. For a reaction:
$$aA + bB \rightarrow \text{products}$$
The rate law has the form:
Rate Law: $$\text{Rate} = k[A]^m[B]^n$$
Where:
- k = rate constant (depends on temperature and the nature of the reaction, but NOT on concentrations)
- [A], [B] = concentrations of reactants
- m, n = orders of reaction with respect to A and B (determined EXPERIMENTALLY, not from the balanced equation!)
The overall order of reaction = m + n
Have you noticed something very important? The orders (m and n) are NOT necessarily equal to the coefficients (a and b) from the balanced equation. They must be determined by experiment! This is a very common exam trap.
Worked Example 4.3: For a reaction $A + B \rightarrow C$, experiments give the following data:
| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
| 1 | 0.10 | 0.10 | 0.0020 |
| 2 | 0.20 | 0.10 | 0.0040 |
| 3 | 0.10 | 0.20 | 0.0080 |
Determine the rate law, the order with respect to each reactant, the overall order, and the rate constant k.
Solution:
General form: Rate = k[A]^m[B]^n
Find m (order w.r.t. A): Compare experiments 1 and 2 (B is constant):
When [A] doubles (0.10→0.20), rate doubles (0.0020→0.0040). So m = 1 (first order in A).
Find n (order w.r.t. B): Compare experiments 1 and 3 (A is constant):
When [B] doubles (0.10→0.20), rate quadruples (0.0020→0.0080). So n = 2 (second order in B).
Rate law: Rate = k[A][B]²
Overall order: 1 + 2 = 3 (third order)
Find k: Use data from experiment 1:
$$0.0020 = k(0.10)(0.10)^2 = k(0.10)(0.01) = k(0.001)$$
$$k = \frac{0.0020}{0.001} = 2.0 \text{ M}^{-2}\text{s}^{-1}$$
| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
| 1 | 0.10 | 0.10 | 0.0020 |
| 2 | 0.20 | 0.10 | 0.0040 |
| 3 | 0.10 | 0.20 | 0.0080 |
Determine the rate law, the order with respect to each reactant, the overall order, and the rate constant k.
Solution:
General form: Rate = k[A]^m[B]^n
Find m (order w.r.t. A): Compare experiments 1 and 2 (B is constant):
When [A] doubles (0.10→0.20), rate doubles (0.0020→0.0040). So m = 1 (first order in A).
Find n (order w.r.t. B): Compare experiments 1 and 3 (A is constant):
When [B] doubles (0.10→0.20), rate quadruples (0.0020→0.0080). So n = 2 (second order in B).
Rate law: Rate = k[A][B]²
Overall order: 1 + 2 = 3 (third order)
Find k: Use data from experiment 1:
$$0.0020 = k(0.10)(0.10)^2 = k(0.10)(0.01) = k(0.001)$$
$$k = \frac{0.0020}{0.001} = 2.0 \text{ M}^{-2}\text{s}^{-1}$$
Worked Example 4.4: For the reaction $2A + B \rightarrow C$, the following data were obtained:
| Exp | [A] (M) | [B] (M) | Rate (M/s) |
| 1 | 0.20 | 0.10 | 0.0060 |
| 2 | 0.40 | 0.10 | 0.0060 |
| 3 | 0.20 | 0.30 | 0.0180 |
Determine the rate law and k.
Solution:
Find m: Experiments 1 and 2: [A] doubles but rate stays the SAME. So m = 0 (zero order in A).
Find n: Experiments 1 and 3: [B] triples (0.10→0.30), rate triples (0.0060→0.0180). So n = 1 (first order in B).
Rate law: Rate = k[B] (A does not appear since m = 0!)
Overall order: 0 + 1 = 1
$$k = \frac{0.0060}{0.10} = 0.060 \text{ s}^{-1}$$
| Exp | [A] (M) | [B] (M) | Rate (M/s) |
| 1 | 0.20 | 0.10 | 0.0060 |
| 2 | 0.40 | 0.10 | 0.0060 |
| 3 | 0.20 | 0.30 | 0.0180 |
Determine the rate law and k.
Solution:
Find m: Experiments 1 and 2: [A] doubles but rate stays the SAME. So m = 0 (zero order in A).
Find n: Experiments 1 and 3: [B] triples (0.10→0.30), rate triples (0.0060→0.0180). So n = 1 (first order in B).
Rate law: Rate = k[B] (A does not appear since m = 0!)
Overall order: 0 + 1 = 1
$$k = \frac{0.0060}{0.10} = 0.060 \text{ s}^{-1}$$
Key Exam Notes — Rate Law:
• Rate = k[A]^m[B]^n — orders m, n are determined EXPERIMENTALLY, not from equation coefficients.
• To find order: change one concentration while keeping others constant, and observe the effect on rate.
• If [A] doubles and rate doubles → m = 1. If rate quadruples → m = 2. If rate doesn’t change → m = 0.
• k is constant for a given reaction at a given temperature — it does NOT change when concentrations change.
• Overall order = sum of individual orders.
• If order = 0 for a reactant, that reactant does NOT appear in the rate law.
• Rate = k[A]^m[B]^n — orders m, n are determined EXPERIMENTALLY, not from equation coefficients.
• To find order: change one concentration while keeping others constant, and observe the effect on rate.
• If [A] doubles and rate doubles → m = 1. If rate quadruples → m = 2. If rate doesn’t change → m = 0.
• k is constant for a given reaction at a given temperature — it does NOT change when concentrations change.
• Overall order = sum of individual orders.
• If order = 0 for a reactant, that reactant does NOT appear in the rate law.
Practice Questions:
1. For a reaction $A + 2B \rightarrow D$, experiments show: when [A] doubles (B constant), rate doubles; when [B] doubles (A constant), rate quadruples. Write the rate law and find the overall order.
2. Why can’t you determine the rate law just by looking at the balanced chemical equation?
1. For a reaction $A + 2B \rightarrow D$, experiments show: when [A] doubles (B constant), rate doubles; when [B] doubles (A constant), rate quadruples. Write the rate law and find the overall order.
2. Why can’t you determine the rate law just by looking at the balanced chemical equation?
Answer 1:
[A] doubles → rate doubles → m = 1 (first order in A)
[B] doubles → rate quadruples → n = 2 (second order in B)
Rate law: Rate = k[A][B]². Overall order = 1 + 2 = 3.
Answer 2:
The balanced equation shows the stoichiometry — how many moles react — but NOT the mechanism (the step-by-step pathway). The rate law depends on the mechanism, specifically the slowest step (rate-determining step). The coefficients in the balanced equation do not necessarily match the orders in the rate law. For example, in Worked Example 4.4, the coefficient of A is 2 but the order is 0!
[A] doubles → rate doubles → m = 1 (first order in A)
[B] doubles → rate quadruples → n = 2 (second order in B)
Rate law: Rate = k[A][B]². Overall order = 1 + 2 = 3.
Answer 2:
The balanced equation shows the stoichiometry — how many moles react — but NOT the mechanism (the step-by-step pathway). The rate law depends on the mechanism, specifically the slowest step (rate-determining step). The coefficients in the balanced equation do not necessarily match the orders in the rate law. For example, in Worked Example 4.4, the coefficient of A is 2 but the order is 0!
4.2.4 Units of the Rate Constant
The units of k depend on the overall order of reaction. Since Rate has units of mol/(L·s):
$$k = \frac{\text{Rate}}{[A]^m[B]^n}$$
| Overall Order | Units of k |
|---|---|
| Zero | mol/(L·s) or M/s |
| First | s⁻¹ |
| Second | L/(mol·s) or M⁻¹s⁻¹ |
| Third | L²/(mol²·s) or M⁻²s⁻¹ |
| n | M^(1−n) · s⁻¹ |
A quick way to remember: $$\text{Units of } k = \text{M}^{1-n} \cdot \text{s}^{-1}$$ where n is the overall order.
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4.2.5 Integrated Rate Laws
The rate law (differential form) tells us rate at any instant. The integrated rate law tells us the concentration at any time. Let’s learn the three most important cases.
First-Order Reactions
For a first-order reaction: Rate = k[A]
Integrated rate law (first order):
$$\ln[A]_t = \ln[A]_0 – kt$$
or equivalently:
$$\log[A]_t = \log[A]_0 – \frac{kt}{2.303}$$
Where [A]₀ = initial concentration, [A]_t = concentration at time t.
Key property of first-order reactions: A plot of ln[A] vs time gives a straight line with slope = −k and intercept = ln[A]₀.
Half-life (t₁/₂) is the time required for the concentration to decrease to half its initial value.
Half-life for first-order reactions:
$$t_{1/2} = \frac{0.693}{k}$$
Notice: for first-order reactions, the half-life is independent of the initial concentration! This is unique to first-order. Whether you start with 1 M or 0.1 M, the time to reach half is the same.
Worked Example 4.5: A first-order reaction has k = 0.0693 s⁻¹. (a) Calculate the half-life. (b) If [A]₀ = 0.80 M, what is [A] after 20 seconds?
Solution:
(a) $$t_{1/2} = \frac{0.693}{0.0693} = 10 \text{ s}$$
(b) $$\ln[A]_t = \ln(0.80) – (0.0693)(20)$$
$$\ln[A]_t = -0.2231 – 1.386 = -1.609$$
$$[A]_t = e^{-1.609} = 0.200 \text{ M}$$
Check: after 20 s = 2 half-lives (10 + 10), [A] = 0.80 × (½)² = 0.80 × 0.25 = 0.20 M ✓
Solution:
(a) $$t_{1/2} = \frac{0.693}{0.0693} = 10 \text{ s}$$
(b) $$\ln[A]_t = \ln(0.80) – (0.0693)(20)$$
$$\ln[A]_t = -0.2231 – 1.386 = -1.609$$
$$[A]_t = e^{-1.609} = 0.200 \text{ M}$$
Check: after 20 s = 2 half-lives (10 + 10), [A] = 0.80 × (½)² = 0.80 × 0.25 = 0.20 M ✓
Zero-Order Reactions
For a zero-order reaction: Rate = k[A]⁰ = k (rate is constant!)
Integrated rate law (zero order):
$$[A]_t = [A]_0 – kt$$
Half-life:
$$t_{1/2} = \frac{[A]_0}{2k}$$
Half-life DEPENDS on initial concentration for zero-order (unlike first-order).
Second-Order Reactions
For a second-order reaction with one reactant: Rate = k[A]²
Integrated rate law (second order):
$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$
Half-life:
$$t_{1/2} = \frac{1}{k[A]_0}$$
Half-life depends on initial concentration for second-order. Larger [A]₀ means shorter half-life.
| Property | Zero Order | First Order | Second Order |
|---|---|---|---|
| Rate law | Rate = k | Rate = k[A] | Rate = k[A]² |
| Integrated form | [A] = [A]₀ − kt | ln[A] = ln[A]₀ − kt | 1/[A] = 1/[A]₀ + kt |
| Straight-line plot | [A] vs t | ln[A] vs t | 1/[A] vs t |
| Slope | −k | −k | +k |
| Half-life | [A]₀/(2k) | 0.693/k | 1/(k[A]₀) |
| t₁/₂ depends on [A]₀? | Yes | No | Yes |
Worked Example 4.6: A second-order reaction has k = 0.50 M⁻¹s⁻¹ and [A]₀ = 0.40 M. Find [A] after 3 seconds.
Solution:
$$\frac{1}{[A]_t} = \frac{1}{0.40} + (0.50)(3) = 2.50 + 1.50 = 4.00$$
$$[A]_t = \frac{1}{4.00} = 0.25 \text{ M}$$
Solution:
$$\frac{1}{[A]_t} = \frac{1}{0.40} + (0.50)(3) = 2.50 + 1.50 = 4.00$$
$$[A]_t = \frac{1}{4.00} = 0.25 \text{ M}$$
Key Exam Notes — Integrated Rate Laws:
• First-order: ln[A] vs t is straight line. Half-life = 0.693/k (constant, independent of [A]₀).
• Zero-order: [A] vs t is straight line. Half-life = [A]₀/(2k) (depends on [A]₀).
• Second-order: 1/[A] vs t is straight line. Half-life = 1/(k[A]₀) (depends on [A]₀).
• To determine order graphically: plot ln[A] vs t and 1/[A] vs t. The one that gives a straight line tells you the order.
• Radioactive decay is always first-order.
• After n half-lives: remaining fraction = (½)^n.
• First-order: ln[A] vs t is straight line. Half-life = 0.693/k (constant, independent of [A]₀).
• Zero-order: [A] vs t is straight line. Half-life = [A]₀/(2k) (depends on [A]₀).
• Second-order: 1/[A] vs t is straight line. Half-life = 1/(k[A]₀) (depends on [A]₀).
• To determine order graphically: plot ln[A] vs t and 1/[A] vs t. The one that gives a straight line tells you the order.
• Radioactive decay is always first-order.
• After n half-lives: remaining fraction = (½)^n.
Practice Questions:
1. A first-order reaction has k = 0.0231 min⁻¹. What is its half-life? What fraction remains after 60 minutes?
2. For a second-order reaction with k = 0.10 M⁻¹s⁻¹ and [A]₀ = 0.50 M, calculate the half-life. If [A]₀ were 1.0 M instead, would the half-life be longer or shorter?
1. A first-order reaction has k = 0.0231 min⁻¹. What is its half-life? What fraction remains after 60 minutes?
2. For a second-order reaction with k = 0.10 M⁻¹s⁻¹ and [A]₀ = 0.50 M, calculate the half-life. If [A]₀ were 1.0 M instead, would the half-life be longer or shorter?
Answer 1:
$$t_{1/2} = \frac{0.693}{0.0231} = 30 \text{ min}$$
Number of half-lives in 60 min = 60/30 = 2
Fraction remaining = (½)² = ¼ = 0.25 (25% remains)
Answer 2:
$$t_{1/2} = \frac{1}{(0.10)(0.50)} = \frac{1}{0.05} = 20 \text{ s}$$
If [A]₀ = 1.0 M: t₁/₂ = 1/((0.10)(1.0)) = 10 s. The half-life would be shorter (10 s vs 20 s). For second-order reactions, higher initial concentration means shorter half-life because the reaction proceeds faster at higher concentrations.
$$t_{1/2} = \frac{0.693}{0.0231} = 30 \text{ min}$$
Number of half-lives in 60 min = 60/30 = 2
Fraction remaining = (½)² = ¼ = 0.25 (25% remains)
Answer 2:
$$t_{1/2} = \frac{1}{(0.10)(0.50)} = \frac{1}{0.05} = 20 \text{ s}$$
If [A]₀ = 1.0 M: t₁/₂ = 1/((0.10)(1.0)) = 10 s. The half-life would be shorter (10 s vs 20 s). For second-order reactions, higher initial concentration means shorter half-life because the reaction proceeds faster at higher concentrations.
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4.3 Factors Affecting the Rate of a Chemical Reaction
Several factors influence how fast a reaction proceeds. Understanding these is crucial for both exams and real-world applications.
4.3.1 Nature of Reactants
The type of bonding and the structure of reactant molecules affect the rate. Ionic reactions in solution are usually very fast because ions simply attract each other. Covalent reactions tend to be slower because bonds must be broken and reformed. Reactions involving complex molecules with strong bonds are generally slower than those with simple molecules and weak bonds.
4.3.2 Concentration of Reactants
In general, increasing the concentration of reactants increases the rate of reaction. Why? Because more reactant particles are present per unit volume, so collisions between them occur more frequently. This is reflected in the rate law — higher [A] means higher rate (for reactions where the order with respect to A is positive).
Note: For zero-order reactions, rate is independent of concentration. But this is a special case.
4.3.3 Surface Area
For reactions involving solids, increasing the surface area increases the rate. A powdered solid reacts much faster than a single lump because more particles are exposed to the other reactant.
Example: A lump of coal burns slowly, but coal dust can cause an explosion because the enormous surface area allows rapid reaction with oxygen.
4.3.4 Temperature
This is one of the most important factors. Increasing temperature generally increases the rate of reaction. But why exactly?
Collision Theory
For a reaction to occur, reactant particles must:
- Collide with each other.
- Collide with sufficient energy (equal to or greater than the activation energy, Eₐ).
- Collide with the correct orientation (proper geometry).
Only collisions that satisfy ALL three conditions are called effective collisions.
Activation Energy (Eₐ)
The activation energy is the minimum energy that colliding molecules must have for a reaction to occur. Think of it as an energy barrier that must be overcome.
Energy
^
| *
| * * (activated complex /
| * * transition state)
|——-*—–*——- Ea (activation energy)
| * *
| * *
| * * (products)
| * reactants *
| * *
+————————-> Reaction progress
The activated complex (or transition state) is the high-energy, unstable arrangement of atoms at the top of the energy barrier. It is not a real molecule — it has partial bonds being broken and formed.
How Temperature Affects Rate
When temperature increases:
- Particles move faster → more collisions per unit time.
- More importantly: a larger FRACTION of molecules have energy ≥ Eₐ. This is the main reason rate increases.
As a general rule, increasing temperature by 10 °C roughly doubles the reaction rate.
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The Arrhenius Equation
The quantitative relationship between the rate constant k and temperature was developed by Svante Arrhenius:
Arrhenius Equation:
$$k = A e^{-E_a/(RT)}$$
Where:
- k = rate constant at temperature T
- A = frequency factor (related to collision frequency and orientation; sometimes called the pre-exponential factor)
- Eₐ = activation energy (J/mol)
- R = gas constant = 8.314 J/(mol·K)
- T = absolute temperature (K)
The Arrhenius equation shows that k increases exponentially with temperature and decreases with higher activation energy.
Taking the natural log of both sides gives the logarithmic form:
$$\ln k = \ln A – \frac{E_a}{RT}$$
This is in the form y = mx + c, so a plot of ln k vs 1/T gives a straight line with:
- Slope = −Eₐ/R
- Intercept = ln A
Two-Temperature Form of Arrhenius Equation
If we know k at two different temperatures:
Two-temperature Arrhenius Equation:
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} – \frac{1}{T_2}\right)$$
or using log base 10:
$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} – \frac{1}{T_2}\right)$$
Worked Example 4.7: A reaction has k = 0.050 s⁻¹ at 300 K and k = 0.200 s⁻¹ at 320 K. Calculate the activation energy.
Solution:
$$\ln\frac{0.200}{0.050} = \frac{E_a}{8.314}\left(\frac{1}{300} – \frac{1}{320}\right)$$
$$\ln(4) = \frac{E_a}{8.314}\left(\frac{320 – 300}{300 \times 320}\right)$$
$$1.386 = \frac{E_a}{8.314}\left(\frac{20}{96000}\right)$$
$$1.386 = \frac{E_a}{8.314}(0.0002083)$$
$$E_a = \frac{1.386 \times 8.314}{0.0002083} = \frac{11.523}{0.0002083} = 55,320 \text{ J/mol} = 55.3 \text{ kJ/mol}$$
Solution:
$$\ln\frac{0.200}{0.050} = \frac{E_a}{8.314}\left(\frac{1}{300} – \frac{1}{320}\right)$$
$$\ln(4) = \frac{E_a}{8.314}\left(\frac{320 – 300}{300 \times 320}\right)$$
$$1.386 = \frac{E_a}{8.314}\left(\frac{20}{96000}\right)$$
$$1.386 = \frac{E_a}{8.314}(0.0002083)$$
$$E_a = \frac{1.386 \times 8.314}{0.0002083} = \frac{11.523}{0.0002083} = 55,320 \text{ J/mol} = 55.3 \text{ kJ/mol}$$
Worked Example 4.8: A reaction has Eₐ = 75.0 kJ/mol and k = 0.030 s⁻¹ at 298 K. What is k at 350 K?
Solution:
$$\ln\frac{k_2}{0.030} = \frac{75000}{8.314}\left(\frac{1}{298} – \frac{1}{350}\right)$$
$$= 9021.4 \times \left(\frac{350 – 298}{298 \times 350}\right)$$
$$= 9021.4 \times \frac{52}{104300}$$
$$= 9021.4 \times 0.000499 = 4.500$$
$$\frac{k_2}{0.030} = e^{4.500} = 90.0$$
$$k_2 = 90.0 \times 0.030 = 2.70 \text{ s}^{-1}$$
Solution:
$$\ln\frac{k_2}{0.030} = \frac{75000}{8.314}\left(\frac{1}{298} – \frac{1}{350}\right)$$
$$= 9021.4 \times \left(\frac{350 – 298}{298 \times 350}\right)$$
$$= 9021.4 \times \frac{52}{104300}$$
$$= 9021.4 \times 0.000499 = 4.500$$
$$\frac{k_2}{0.030} = e^{4.500} = 90.0$$
$$k_2 = 90.0 \times 0.030 = 2.70 \text{ s}^{-1}$$
Key Exam Notes — Arrhenius Equation:
• k = Ae^(−Eₐ/RT) — higher T → larger k → faster reaction.
• Higher Eₐ → smaller k → slower reaction.
• Plot of ln k vs 1/T: slope = −Eₐ/R, intercept = ln A.
• Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂).
• R must be 8.314 J/(mol·K) (not 0.0821!). Eₐ must be in J/mol in the calculation.
• Remember to convert Eₐ from kJ/mol to J/mol (multiply by 1000) before using in the equation!
• The 10 °C rule is approximate — the Arrhenius equation is exact.
• k = Ae^(−Eₐ/RT) — higher T → larger k → faster reaction.
• Higher Eₐ → smaller k → slower reaction.
• Plot of ln k vs 1/T: slope = −Eₐ/R, intercept = ln A.
• Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂).
• R must be 8.314 J/(mol·K) (not 0.0821!). Eₐ must be in J/mol in the calculation.
• Remember to convert Eₐ from kJ/mol to J/mol (multiply by 1000) before using in the equation!
• The 10 °C rule is approximate — the Arrhenius equation is exact.
Practice Questions:
1. A reaction has Eₐ = 50.0 kJ/mol. If the temperature increases from 300 K to 310 K, by what factor does the rate constant increase?
2. For a reaction, a plot of ln k vs 1/T gives a straight line with slope = −6000 K. Calculate Eₐ.
1. A reaction has Eₐ = 50.0 kJ/mol. If the temperature increases from 300 K to 310 K, by what factor does the rate constant increase?
2. For a reaction, a plot of ln k vs 1/T gives a straight line with slope = −6000 K. Calculate Eₐ.
Answer 1:
$$\ln\frac{k_2}{k_1} = \frac{50000}{8.314}\left(\frac{1}{300} – \frac{1}{310}\right)$$
$$= 6015.5 \times \frac{10}{93000} = 6015.5 \times 0.0001075 = 0.647$$
$$\frac{k_2}{k_1} = e^{0.647} = 1.91$$
The rate constant increases by a factor of about 1.91 (approximately doubles).
Answer 2:
Slope = −Eₐ/R
$$-6000 = \frac{-E_a}{8.314}$$
$$E_a = 6000 \times 8.314 = 49,884 \text{ J/mol} = 49.9 \text{ kJ/mol}$$
$$\ln\frac{k_2}{k_1} = \frac{50000}{8.314}\left(\frac{1}{300} – \frac{1}{310}\right)$$
$$= 6015.5 \times \frac{10}{93000} = 6015.5 \times 0.0001075 = 0.647$$
$$\frac{k_2}{k_1} = e^{0.647} = 1.91$$
The rate constant increases by a factor of about 1.91 (approximately doubles).
Answer 2:
Slope = −Eₐ/R
$$-6000 = \frac{-E_a}{8.314}$$
$$E_a = 6000 \times 8.314 = 49,884 \text{ J/mol} = 49.9 \text{ kJ/mol}$$
4.3.5 Catalyst
A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction. It provides an alternative reaction pathway with a lower activation energy.
Energy
^
| * (uncatalyzed path)
| * *
| * *
| * *
| * *
| * * * * * (catalyzed path)
| * * * *
|——-*———–*—– Ea (uncatalyzed)
| * Ea(cat) *
| * *
| * *
| * reactants * products
+———————————> Reaction progress
Key points about catalysts:
- They lower the activation energy — both forward and reverse reactions.
- They do NOT change the thermodynamics (ΔH, equilibrium position, or equilibrium constant remain the same).
- They increase the rate of both forward and reverse reactions equally.
- They are not consumed — they appear in the mechanism but not in the overall equation.
- They provide an alternative pathway, not more energy.
- Examples: enzymes in biological systems, platinum in catalytic converters, MnO₂ in decomposition of H₂O₂.
Worked Example 4.9: A reaction has Eₐ = 80 kJ/mol without a catalyst. With a catalyst, Eₐ = 50 kJ/mol. By what factor does the rate constant increase at 300 K? (Assume A is the same.)
Solution:
$$\frac{k_{\text{cat}}}{k_{\text{uncat}}} = \frac{Ae^{-E_{a,\text{cat}}/RT}}{Ae^{-E_{a,\text{uncat}}/RT}} = e^{(E_{a,\text{uncat}} – E_{a,\text{cat}})/RT}$$
$$= e^{(80000 – 50000)/(8.314 \times 300)} = e^{30000/2494.2} = e^{12.03}$$
$$= 167,700$$
The rate constant increases by a factor of about 168,000! Even a small decrease in Eₐ causes an enormous increase in rate. This shows the incredible power of catalysts.
Solution:
$$\frac{k_{\text{cat}}}{k_{\text{uncat}}} = \frac{Ae^{-E_{a,\text{cat}}/RT}}{Ae^{-E_{a,\text{uncat}}/RT}} = e^{(E_{a,\text{uncat}} – E_{a,\text{cat}})/RT}$$
$$= e^{(80000 – 50000)/(8.314 \times 300)} = e^{30000/2494.2} = e^{12.03}$$
$$= 167,700$$
The rate constant increases by a factor of about 168,000! Even a small decrease in Eₐ causes an enormous increase in rate. This shows the incredible power of catalysts.
Key Exam Notes — Catalysts:
• Catalyst lowers Eₐ → increases k → increases rate.
• Does NOT change ΔH, equilibrium constant K, or equilibrium position.
• Speeds up BOTH forward and reverse reactions equally.
• Not consumed in the reaction (regenerated at the end of the mechanism).
• Provides an alternative pathway with lower energy barrier.
• Enzymes are biological catalysts — highly specific and very efficient.
• Catalyst lowers Eₐ → increases k → increases rate.
• Does NOT change ΔH, equilibrium constant K, or equilibrium position.
• Speeds up BOTH forward and reverse reactions equally.
• Not consumed in the reaction (regenerated at the end of the mechanism).
• Provides an alternative pathway with lower energy barrier.
• Enzymes are biological catalysts — highly specific and very efficient.
Practice Questions:
1. Does a catalyst change the enthalpy change (ΔH) of a reaction? Explain.
2. Why does lowering the activation energy have such a large effect on the rate constant?
1. Does a catalyst change the enthalpy change (ΔH) of a reaction? Explain.
2. Why does lowering the activation energy have such a large effect on the rate constant?
Answer 1:
No. The enthalpy change (ΔH) is the difference in energy between reactants and products. A catalyst only changes the PATHWAY (lowers the energy barrier), but the starting point (reactants) and ending point (products) remain the same. Therefore ΔH is unchanged. You can see this in the energy diagram — the catalyzed and uncatalyzed paths start and end at the same energy levels.
Answer 2:
Because Eₐ appears in the EXPONENT of the Arrhenius equation (k = Ae^(−Eₐ/RT)). Even a small change in Eₐ causes a large change in the exponent, which then causes an exponential change in k. For example, at 300 K, decreasing Eₐ by just 10 kJ/mol increases k by a factor of about e^(10000/2494) ≈ 55. This exponential sensitivity makes activation energy the most important factor controlling reaction rate.
No. The enthalpy change (ΔH) is the difference in energy between reactants and products. A catalyst only changes the PATHWAY (lowers the energy barrier), but the starting point (reactants) and ending point (products) remain the same. Therefore ΔH is unchanged. You can see this in the energy diagram — the catalyzed and uncatalyzed paths start and end at the same energy levels.
Answer 2:
Because Eₐ appears in the EXPONENT of the Arrhenius equation (k = Ae^(−Eₐ/RT)). Even a small change in Eₐ causes a large change in the exponent, which then causes an exponential change in k. For example, at 300 K, decreasing Eₐ by just 10 kJ/mol increases k by a factor of about e^(10000/2494) ≈ 55. This exponential sensitivity makes activation energy the most important factor controlling reaction rate.
4.3.6 Summary of All Factors
| Factor | Effect on Rate | Reason |
|---|---|---|
| Concentration ↑ | Rate ↑ | More particles → more frequent collisions |
| Temperature ↑ | Rate ↑ (significantly) | More molecules have energy ≥ Eₐ |
| Surface area ↑ | Rate ↑ | More exposed particles for collision |
| Catalyst added | Rate ↑ | Lowers activation energy Eₐ |
| Nature of reactants | Varies | Bond strength, molecular complexity |
Quick Revision Notes — Exam Focus
1. Important Definitions
• Chemical kinetics: Study of the speed of chemical reactions and the factors affecting it.
• Rate of reaction: Change in concentration of reactant or product per unit time.
• Rate law: Equation relating rate to concentrations: Rate = k[A]^m[B]^n.
• Order of reaction: The exponent to which concentration is raised in the rate law (determined experimentally).
• Rate constant (k): Proportionality constant in the rate law; depends on T and the reaction, not on concentrations.
• Half-life (t₁/₂): Time for concentration to decrease to half its initial value.
• Activation energy (Eₐ): Minimum energy required for a reaction to occur.
• Activated complex: Unstable, high-energy arrangement of atoms at the peak of the energy barrier.
• Catalyst: Substance that increases rate by lowering Eₐ without being consumed.
• Effective collision: A collision with sufficient energy (≥ Eₐ) and correct orientation.
• Rate of reaction: Change in concentration of reactant or product per unit time.
• Rate law: Equation relating rate to concentrations: Rate = k[A]^m[B]^n.
• Order of reaction: The exponent to which concentration is raised in the rate law (determined experimentally).
• Rate constant (k): Proportionality constant in the rate law; depends on T and the reaction, not on concentrations.
• Half-life (t₁/₂): Time for concentration to decrease to half its initial value.
• Activation energy (Eₐ): Minimum energy required for a reaction to occur.
• Activated complex: Unstable, high-energy arrangement of atoms at the peak of the energy barrier.
• Catalyst: Substance that increases rate by lowering Eₐ without being consumed.
• Effective collision: A collision with sufficient energy (≥ Eₐ) and correct orientation.
2. All Key Formulas
$$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = \frac{1}{c}\frac{\Delta[C]}{\Delta t}$$
$$\text{Rate} = k[A]^m[B]^n$$
$$\text{Units of } k: \text{M}^{1-n}\text{s}^{-1} \quad (n = \text{overall order})$$
$$\ln[A]_t = \ln[A]_0 – kt \quad \text{(first order)}$$ $$t_{1/2} = \frac{0.693}{k} \quad \text{(first order only!)}$$
$$[A]_t = [A]_0 – kt \quad \text{(zero order)}$$ $$t_{1/2} = \frac{[A]_0}{2k} \quad \text{(zero order)}$$
$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \quad \text{(second order)}$$ $$t_{1/2} = \frac{1}{k[A]_0} \quad \text{(second order)}$$
$$k = Ae^{-E_a/(RT)}$$
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} – \frac{1}{T_2}\right)$$
$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} – \frac{1}{T_2}\right)$$
3. Graphical Methods Summary
| Plot | Straight Line If | Slope | Intercept |
|---|---|---|---|
| [A] vs t | Zero order | −k | [A]₀ |
| ln[A] vs t | First order | −k | ln[A]₀ |
| 1/[A] vs t | Second order | +k | 1/[A]₀ |
| ln k vs 1/T | Always (Arrhenius) | −Eₐ/R | ln A |
4. Half-Life Comparison
Order | Formula | Depends on [A]₀?
———-|——————-|——————
Zero | [A]₀/(2k) | YES
First | 0.693/k | NO (constant!)
Second | 1/(k[A]₀) | YES
5. Collision Theory Essentials
• Three requirements for effective collision: (1) collision, (2) energy ≥ Eₐ, (3) correct orientation.
• Temperature affects rate mainly by changing the FRACTION of molecules with energy ≥ Eₐ, not just by increasing collision frequency.
• Not all collisions lead to reaction — only effective ones do.
• The Maxwell-Boltzmann distribution curve shifts right and flattens at higher T, showing more molecules with high energy.
• Temperature affects rate mainly by changing the FRACTION of molecules with energy ≥ Eₐ, not just by increasing collision frequency.
• Not all collisions lead to reaction — only effective ones do.
• The Maxwell-Boltzmann distribution curve shifts right and flattens at higher T, showing more molecules with high energy.
6. Common Mistakes to Avoid
❌ Using coefficients from the balanced equation as orders in the rate law.
❌ Forgetting to convert Eₐ from kJ/mol to J/mol in the Arrhenius equation (multiply by 1000!).
❌ Using R = 0.0821 in the Arrhenius equation — use R = 8.314 J/(mol·K).
❌ Forgetting that Celsius must be converted to Kelvin in all calculations.
❌ Confusing average rate (over an interval) with instantaneous rate (at one point).
❌ Using the wrong half-life formula — first-order is the ONLY one where t₁/₂ = 0.693/k.
❌ Forgetting negative sign for reactants in rate expressions.
❌ Thinking a catalyst changes ΔH or the equilibrium position — it does NOT.
❌ Confusing activation energy with enthalpy change (ΔH). Eₐ is the energy barrier; ΔH is the difference between products and reactants.
❌ Forgetting to convert Eₐ from kJ/mol to J/mol in the Arrhenius equation (multiply by 1000!).
❌ Using R = 0.0821 in the Arrhenius equation — use R = 8.314 J/(mol·K).
❌ Forgetting that Celsius must be converted to Kelvin in all calculations.
❌ Confusing average rate (over an interval) with instantaneous rate (at one point).
❌ Using the wrong half-life formula — first-order is the ONLY one where t₁/₂ = 0.693/k.
❌ Forgetting negative sign for reactants in rate expressions.
❌ Thinking a catalyst changes ΔH or the equilibrium position — it does NOT.
❌ Confusing activation energy with enthalpy change (ΔH). Eₐ is the energy barrier; ΔH is the difference between products and reactants.
Challenge Exam Questions
Test yourself thoroughly! Try each question before checking the answer.
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Section A: Multiple Choice Questions
Q1. For the reaction $2A + B \rightarrow 3C$, if the rate of disappearance of B is 0.30 M/s, the rate of appearance of C is:
A) 0.10 M/s B) 0.30 M/s C) 0.60 M/s D) 0.90 M/s
A) 0.10 M/s B) 0.30 M/s C) 0.60 M/s D) 0.90 M/s
Answer: D
$$\text{Rate} = -\frac{\Delta[B]}{\Delta t} = \frac{1}{3}\frac{\Delta[C]}{\Delta t}$$
$$0.30 = \frac{1}{3}\frac{\Delta[C]}{\Delta t}$$
$$\frac{\Delta[C]}{\Delta t} = 0.30 \times 3 = 0.90 \text{ M/s}$$
$$\text{Rate} = -\frac{\Delta[B]}{\Delta t} = \frac{1}{3}\frac{\Delta[C]}{\Delta t}$$
$$0.30 = \frac{1}{3}\frac{\Delta[C]}{\Delta t}$$
$$\frac{\Delta[C]}{\Delta t} = 0.30 \times 3 = 0.90 \text{ M/s}$$
Q2. The units of the rate constant for a third-order reaction are:
A) s⁻¹ B) M⁻¹s⁻¹ C) M⁻²s⁻¹ D) M/s
A) s⁻¹ B) M⁻¹s⁻¹ C) M⁻²s⁻¹ D) M/s
Answer: C
Units of k = M^(1−n) · s⁻¹ where n = 3.
= M^(1−3) · s⁻¹ = M⁻²s⁻¹
Units of k = M^(1−n) · s⁻¹ where n = 3.
= M^(1−3) · s⁻¹ = M⁻²s⁻¹
Q3. For a first-order reaction, the half-life is 20 minutes. How long will it take for 87.5% of the reactant to decompose?
A) 40 min B) 60 min C) 80 min D) 100 min
A) 40 min B) 60 min C) 80 min D) 100 min
Answer: B
87.5% decomposed means 12.5% remains = 1/8 = (1/2)³.
So 3 half-lives are needed: 3 × 20 = 60 minutes.
87.5% decomposed means 12.5% remains = 1/8 = (1/2)³.
So 3 half-lives are needed: 3 × 20 = 60 minutes.
Q4. Which of the following does NOT affect the rate constant k?
A) Temperature B) Activation energy C) Concentration of reactants D) Presence of a catalyst
A) Temperature B) Activation energy C) Concentration of reactants D) Presence of a catalyst
Answer: C
The rate constant k depends on temperature (Arrhenius equation), activation energy, and the presence of a catalyst (which changes Eₐ). It does NOT depend on the concentration of reactants. Changing concentration changes the RATE, not k.
The rate constant k depends on temperature (Arrhenius equation), activation energy, and the presence of a catalyst (which changes Eₐ). It does NOT depend on the concentration of reactants. Changing concentration changes the RATE, not k.
Q5. A catalyst increases the rate of a reaction by:
A) Increasing the temperature B) Increasing the activation energy C) Decreasing the activation energy D) Increasing the concentration
A) Increasing the temperature B) Increasing the activation energy C) Decreasing the activation energy D) Increasing the concentration
Answer: C
A catalyst provides an alternative reaction pathway with a lower activation energy. It does not change temperature, does not increase Eₐ (it decreases it), and does not change concentrations.
A catalyst provides an alternative reaction pathway with a lower activation energy. It does not change temperature, does not increase Eₐ (it decreases it), and does not change concentrations.
Section B: Fill in the Blanks
Q6. The minimum energy that colliding molecules must possess for a reaction to occur is called ________.
Answer: Activation energy (Eₐ). It is the energy barrier that must be overcome for reactants to be converted to products.
Q7. For a ________-order reaction, the half-life is independent of the initial concentration.
Answer: First. Only for first-order reactions is t₁/₂ = 0.693/k, which does not contain [A]₀. For zero-order and second-order reactions, t₁/₂ depends on [A]₀.
Q8. According to the Arrhenius equation, a plot of ________ versus ________ gives a straight line with slope equal to −Eₐ/R.
Answer: ln k versus 1/T. From ln k = ln A − Eₐ/(RT), this is in the form y = mx + c where y = ln k, x = 1/T, slope m = −Eₐ/R, and intercept c = ln A.
Q9. The unstable arrangement of atoms at the maximum of the potential energy barrier is called the ________ complex or ________ state.
Answer: Activated complex or transition state. It is not a true molecule — it has partial bonds being broken and formed, and it exists only momentarily at the peak of the energy barrier.
Q10. In the rate law Rate = k[A]²[B], the overall order of the reaction is ________.
Answer: 3 (second order in A + first order in B = 2 + 1 = 3). The overall order is the sum of all individual orders.
Section C: Short Answer Questions
Q11. Explain the three conditions that must be met for a collision between reactant molecules to result in a reaction.
Answer:
For a collision to be effective (result in reaction), three conditions must be satisfied:
(1) Collision: The reactant molecules must physically collide with each other. No collision means no reaction.
(2) Sufficient energy: The colliding molecules must have kinetic energy equal to or greater than the activation energy (Eₐ). If the energy is too low, the molecules simply bounce apart without reacting.
(3) Correct orientation: The molecules must collide with the proper geometry. Even if they have enough energy, colliding from the wrong angle may not allow the necessary bonds to break and form. For example, in the reaction H₂ + I₂ → 2HI, the H–H and I–I bonds must be aligned properly for the exchange to occur.
For a collision to be effective (result in reaction), three conditions must be satisfied:
(1) Collision: The reactant molecules must physically collide with each other. No collision means no reaction.
(2) Sufficient energy: The colliding molecules must have kinetic energy equal to or greater than the activation energy (Eₐ). If the energy is too low, the molecules simply bounce apart without reacting.
(3) Correct orientation: The molecules must collide with the proper geometry. Even if they have enough energy, colliding from the wrong angle may not allow the necessary bonds to break and form. For example, in the reaction H₂ + I₂ → 2HI, the H–H and I–I bonds must be aligned properly for the exchange to occur.
Q12. Distinguish between the rate of reaction and the rate constant. How does each change when the concentration of a reactant is doubled (for a first-order reaction)?
Answer:
Rate of reaction: The speed at which reactants are consumed or products are formed. It depends on concentrations: Rate = k[A] for first-order.
Rate constant (k): The proportionality factor in the rate law. It is constant for a given reaction at a given temperature and does NOT depend on concentrations.
When [A] is doubled for a first-order reaction:
• Rate doubles (Rate = k[A] → Rate_new = k[2A] = 2 × k[A] = 2 × Rate).
• k remains UNCHANGED. The rate constant is independent of concentration.
Rate of reaction: The speed at which reactants are consumed or products are formed. It depends on concentrations: Rate = k[A] for first-order.
Rate constant (k): The proportionality factor in the rate law. It is constant for a given reaction at a given temperature and does NOT depend on concentrations.
When [A] is doubled for a first-order reaction:
• Rate doubles (Rate = k[A] → Rate_new = k[2A] = 2 × k[A] = 2 × Rate).
• k remains UNCHANGED. The rate constant is independent of concentration.
Q13. Why does increasing temperature have a much greater effect on reaction rate than increasing concentration?
Answer:
Increasing concentration increases the NUMBER of collisions, which increases rate proportionally (linearly). For example, doubling [A] roughly doubles the rate (for first-order).
Increasing temperature has TWO effects: (1) slightly more collisions (minor effect), and (2) a much larger FRACTION of molecules have energy ≥ Eₐ (major effect). Because Eₐ appears in the EXPONENT of the Arrhenius equation, even a small temperature increase causes an exponential increase in the rate constant. For example, a 10 °C increase typically doubles the rate — a much more dramatic effect than doubling concentration.
Increasing concentration increases the NUMBER of collisions, which increases rate proportionally (linearly). For example, doubling [A] roughly doubles the rate (for first-order).
Increasing temperature has TWO effects: (1) slightly more collisions (minor effect), and (2) a much larger FRACTION of molecules have energy ≥ Eₐ (major effect). Because Eₐ appears in the EXPONENT of the Arrhenius equation, even a small temperature increase causes an exponential increase in the rate constant. For example, a 10 °C increase typically doubles the rate — a much more dramatic effect than doubling concentration.
Section D: Calculation Questions
Q14. For the reaction $2NO + O_2 \rightarrow 2NO_2$, the following data were obtained:
| Exp | [NO] (M) | [O₂] (M) | Initial Rate (M/s) |
| 1 | 0.010 | 0.010 | 2.5 × 10⁻³ |
| 2 | 0.020 | 0.010 | 1.0 × 10⁻² |
| 3 | 0.010 | 0.020 | 5.0 × 10⁻³ |
(a) Determine the rate law.
(b) Calculate the rate constant k with units.
(c) What is the rate when [NO] = 0.030 M and [O₂] = 0.015 M?
| Exp | [NO] (M) | [O₂] (M) | Initial Rate (M/s) |
| 1 | 0.010 | 0.010 | 2.5 × 10⁻³ |
| 2 | 0.020 | 0.010 | 1.0 × 10⁻² |
| 3 | 0.010 | 0.020 | 5.0 × 10⁻³ |
(a) Determine the rate law.
(b) Calculate the rate constant k with units.
(c) What is the rate when [NO] = 0.030 M and [O₂] = 0.015 M?
Answer:
(a) Find order w.r.t. NO: Exps 1 and 2 (O₂ constant): [NO] doubles (0.010→0.020), rate quadruples (2.5×10⁻³→1.0×10⁻²). So m = 2 (second order in NO).
Find order w.r.t. O₂: Exps 1 and 3 (NO constant): [O₂] doubles (0.010→0.020), rate doubles. So n = 1 (first order in O₂).
Rate law: Rate = k[NO]²[O₂]
Overall order = 2 + 1 = 3
(b) Using experiment 1:
$$2.5 \times 10^{-3} = k(0.010)^2(0.010) = k(1.0 \times 10^{-6})$$
$$k = \frac{2.5 \times 10^{-3}}{1.0 \times 10^{-6}} = 2500 \text{ M}^{-2}\text{s}^{-1}$$
(Units: M^(1−3)s⁻¹ = M⁻²s⁻¹ ✓)
(c) Rate = (2500)(0.030)²(0.015) = (2500)(9.0×10⁻⁴)(0.015) = (2500)(1.35×10⁻⁵) = 0.0338 M/s
(a) Find order w.r.t. NO: Exps 1 and 2 (O₂ constant): [NO] doubles (0.010→0.020), rate quadruples (2.5×10⁻³→1.0×10⁻²). So m = 2 (second order in NO).
Find order w.r.t. O₂: Exps 1 and 3 (NO constant): [O₂] doubles (0.010→0.020), rate doubles. So n = 1 (first order in O₂).
Rate law: Rate = k[NO]²[O₂]
Overall order = 2 + 1 = 3
(b) Using experiment 1:
$$2.5 \times 10^{-3} = k(0.010)^2(0.010) = k(1.0 \times 10^{-6})$$
$$k = \frac{2.5 \times 10^{-3}}{1.0 \times 10^{-6}} = 2500 \text{ M}^{-2}\text{s}^{-1}$$
(Units: M^(1−3)s⁻¹ = M⁻²s⁻¹ ✓)
(c) Rate = (2500)(0.030)²(0.015) = (2500)(9.0×10⁻⁴)(0.015) = (2500)(1.35×10⁻⁵) = 0.0338 M/s
Q15. The decomposition of hydrogen peroxide is first-order with k = 3.5 × 10⁻³ min⁻¹. (a) What is the half-life? (b) How long will it take for 75% of the H₂O₂ to decompose? (c) What percentage remains after 200 minutes?
Answer:
(a) $$t_{1/2} = \frac{0.693}{3.5 \times 10^{-3}} = 198 \text{ min}$$
(b) 75% decomposed means 25% remains = [A]_t/[A]₀ = 0.25 = (1/2)².
So 2 half-lives: 2 × 198 = 396 min.
Or using the equation: $$\ln(0.25) = -kt \Rightarrow -1.386 = -(3.5 \times 10^{-3})t \Rightarrow t = 396 \text{ min}$$
(c) $$\ln\frac{[A]_t}{[A]_0} = -kt = -(3.5 \times 10^{-3})(200) = -0.70$$
$$\frac{[A]_t}{[A]_0} = e^{-0.70} = 0.497$$
Percentage remaining = 49.7% ≈ 50%
(a) $$t_{1/2} = \frac{0.693}{3.5 \times 10^{-3}} = 198 \text{ min}$$
(b) 75% decomposed means 25% remains = [A]_t/[A]₀ = 0.25 = (1/2)².
So 2 half-lives: 2 × 198 = 396 min.
Or using the equation: $$\ln(0.25) = -kt \Rightarrow -1.386 = -(3.5 \times 10^{-3})t \Rightarrow t = 396 \text{ min}$$
(c) $$\ln\frac{[A]_t}{[A]_0} = -kt = -(3.5 \times 10^{-3})(200) = -0.70$$
$$\frac{[A]_t}{[A]_0} = e^{-0.70} = 0.497$$
Percentage remaining = 49.7% ≈ 50%
Q16. A reaction has rate constant k = 0.025 M⁻¹s⁻¹ at 300 K and k = 0.150 M⁻¹s⁻¹ at 350 K. (a) Calculate the activation energy. (b) Calculate the frequency factor A. (c) What is k at 400 K?
Answer:
(a) $$\ln\frac{0.150}{0.025} = \frac{E_a}{8.314}\left(\frac{1}{300} – \frac{1}{350}\right)$$
$$\ln(6) = \frac{E_a}{8.314}\left(\frac{50}{105000}\right)$$
$$1.7918 = \frac{E_a}{8.314}(0.0004762)$$
$$E_a = \frac{1.7918 \times 8.314}{0.0004762} = \frac{14.90}{0.0004762} = 31,300 \text{ J/mol} = 31.3 \text{ kJ/mol}$$
(b) Using ln k = ln A − Eₐ/(RT) with data from 300 K:
$$\ln(0.025) = \ln A – \frac{31300}{8.314 \times 300}$$
$$-3.689 = \ln A – 12.55$$
$$\ln A = -3.689 + 12.55 = 8.861$$
$$A = e^{8.861} = 7,070 \text{ M}^{-1}\text{s}^{-1}$$
(c) $$\ln k = 8.861 – \frac{31300}{8.314 \times 400} = 8.861 – 9.41 = -0.549$$
$$k = e^{-0.549} = 0.578 \text{ M}^{-1}\text{s}^{-1}$$
(a) $$\ln\frac{0.150}{0.025} = \frac{E_a}{8.314}\left(\frac{1}{300} – \frac{1}{350}\right)$$
$$\ln(6) = \frac{E_a}{8.314}\left(\frac{50}{105000}\right)$$
$$1.7918 = \frac{E_a}{8.314}(0.0004762)$$
$$E_a = \frac{1.7918 \times 8.314}{0.0004762} = \frac{14.90}{0.0004762} = 31,300 \text{ J/mol} = 31.3 \text{ kJ/mol}$$
(b) Using ln k = ln A − Eₐ/(RT) with data from 300 K:
$$\ln(0.025) = \ln A – \frac{31300}{8.314 \times 300}$$
$$-3.689 = \ln A – 12.55$$
$$\ln A = -3.689 + 12.55 = 8.861$$
$$A = e^{8.861} = 7,070 \text{ M}^{-1}\text{s}^{-1}$$
(c) $$\ln k = 8.861 – \frac{31300}{8.314 \times 400} = 8.861 – 9.41 = -0.549$$
$$k = e^{-0.549} = 0.578 \text{ M}^{-1}\text{s}^{-1}$$
Q17. For a second-order reaction $A \rightarrow$ products with k = 0.20 M⁻¹s⁻¹ and [A]₀ = 0.50 M:
(a) Calculate [A] after 5 seconds.
(b) Calculate the half-life.
(c) How long does it take for [A] to reach 0.10 M?
(a) Calculate [A] after 5 seconds.
(b) Calculate the half-life.
(c) How long does it take for [A] to reach 0.10 M?
Answer:
(a) $$\frac{1}{[A]_t} = \frac{1}{0.50} + (0.20)(5) = 2.0 + 1.0 = 3.0$$
$$[A]_t = \frac{1}{3.0} = 0.333 \text{ M}$$
(b) $$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{(0.20)(0.50)} = \frac{1}{0.10} = 10 \text{ s}$$
(c) $$\frac{1}{0.10} = \frac{1}{0.50} + (0.20)t$$
$$10.0 = 2.0 + 0.20t$$
$$0.20t = 8.0$$
$$t = 40 \text{ s}$$
(a) $$\frac{1}{[A]_t} = \frac{1}{0.50} + (0.20)(5) = 2.0 + 1.0 = 3.0$$
$$[A]_t = \frac{1}{3.0} = 0.333 \text{ M}$$
(b) $$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{(0.20)(0.50)} = \frac{1}{0.10} = 10 \text{ s}$$
(c) $$\frac{1}{0.10} = \frac{1}{0.50} + (0.20)t$$
$$10.0 = 2.0 + 0.20t$$
$$0.20t = 8.0$$
$$t = 40 \text{ s}$$
Q18. In an experiment, the concentration of a reactant was measured at different times:
| Time (s) | [A] (M) |
| 0 | 0.400 |
| 100 | 0.200 |
| 200 | 0.100 |
| 300 | 0.050 |
(a) Show that this reaction is first-order.
(b) Calculate k.
(c) What is the half-life?
| Time (s) | [A] (M) |
| 0 | 0.400 |
| 100 | 0.200 |
| 200 | 0.100 |
| 300 | 0.050 |
(a) Show that this reaction is first-order.
(b) Calculate k.
(c) What is the half-life?
Answer:
(a) Method 1 — Check half-lives:
0.400 → 0.200: takes 100 s
0.200 → 0.100: takes 100 s (200 − 100)
0.100 → 0.050: takes 100 s (300 − 200)
The half-life is constant (100 s) regardless of concentration → first-order.
Method 2 — Check ln[A] vs t is linear:
ln(0.400) = −0.916; ln(0.200) = −1.609; ln(0.100) = −2.303; ln(0.050) = −2.996
Differences: −0.693, −0.694, −0.693 (constant difference per 100 s) → straight line → first-order ✓
(b) $$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{100} = 0.00693 \text{ s}^{-1}$$
Or from the slope: k = 0.693/100 = 0.00693 s⁻¹
(c) $$t_{1/2} = 100 \text{ s}$$ (already determined above)
(a) Method 1 — Check half-lives:
0.400 → 0.200: takes 100 s
0.200 → 0.100: takes 100 s (200 − 100)
0.100 → 0.050: takes 100 s (300 − 200)
The half-life is constant (100 s) regardless of concentration → first-order.
Method 2 — Check ln[A] vs t is linear:
ln(0.400) = −0.916; ln(0.200) = −1.609; ln(0.100) = −2.303; ln(0.050) = −2.996
Differences: −0.693, −0.694, −0.693 (constant difference per 100 s) → straight line → first-order ✓
(b) $$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{100} = 0.00693 \text{ s}^{-1}$$
Or from the slope: k = 0.693/100 = 0.00693 s⁻¹
(c) $$t_{1/2} = 100 \text{ s}$$ (already determined above)
Q19. A reaction has an activation energy of 60.0 kJ/mol. By what factor does the rate constant increase when the temperature rises from 298 K to 308 K? Repeat the calculation for a reaction with Eₐ = 120 kJ/mol. What conclusion can you draw?
Answer:
For Eₐ = 60 kJ/mol:
$$\ln\frac{k_2}{k_1} = \frac{60000}{8.314}\left(\frac{1}{298} – \frac{1}{308}\right) = 7217 \times \frac{10}{91784} = 0.786$$
$$\frac{k_2}{k_1} = e^{0.786} = 2.19$$
For Eₐ = 120 kJ/mol:
$$\ln\frac{k_2}{k_1} = \frac{120000}{8.314}\left(\frac{1}{298} – \frac{1}{308}\right) = 14434 \times 0.000109 = 1.572$$
$$\frac{k_2}{k_1} = e^{1.572} = 4.82$$
Conclusion: A 10 °C temperature increase causes a larger rate increase for reactions with higher activation energy. The reaction with Eₐ = 120 kJ/mol is about 4.8 times faster, while the one with Eₐ = 60 kJ/mol is only about 2.2 times faster. This is because Eₐ appears in the numerator of the Arrhenius exponent — higher Eₐ means greater sensitivity to temperature changes.
For Eₐ = 60 kJ/mol:
$$\ln\frac{k_2}{k_1} = \frac{60000}{8.314}\left(\frac{1}{298} – \frac{1}{308}\right) = 7217 \times \frac{10}{91784} = 0.786$$
$$\frac{k_2}{k_1} = e^{0.786} = 2.19$$
For Eₐ = 120 kJ/mol:
$$\ln\frac{k_2}{k_1} = \frac{120000}{8.314}\left(\frac{1}{298} – \frac{1}{308}\right) = 14434 \times 0.000109 = 1.572$$
$$\frac{k_2}{k_1} = e^{1.572} = 4.82$$
Conclusion: A 10 °C temperature increase causes a larger rate increase for reactions with higher activation energy. The reaction with Eₐ = 120 kJ/mol is about 4.8 times faster, while the one with Eₐ = 60 kJ/mol is only about 2.2 times faster. This is because Eₐ appears in the numerator of the Arrhenius exponent — higher Eₐ means greater sensitivity to temperature changes.
Q20. The decomposition of N₂O₅ is first-order: $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$. At a certain temperature, k = 5.0 × 10⁻⁴ s⁻¹.
(a) If the initial concentration of N₂O₅ is 0.500 M, what is [N₂O₅] after 1000 seconds?
(b) What is the rate of formation of NO₂ at that moment?
(c) What is the rate of formation of O₂ at that moment?
(a) If the initial concentration of N₂O₅ is 0.500 M, what is [N₂O₅] after 1000 seconds?
(b) What is the rate of formation of NO₂ at that moment?
(c) What is the rate of formation of O₂ at that moment?
Answer:
(a) $$\ln[\text{N}_2\text{O}_5]_t = \ln(0.500) – (5.0 \times 10^{-4})(1000)$$
$$= -0.693 – 0.500 = -1.193$$
$$[\text{N}_2\text{O}_5]_t = e^{-1.193} = 0.304 \text{ M}$$
(b) First find the rate of reaction:
$$\text{Rate} = k[\text{N}_2\text{O}_5] = (5.0 \times 10^{-4})(0.304) = 1.52 \times 10^{-4} \text{ M/s}$$
$$\frac{\Delta[\text{NO}_2]}{\Delta t} = 4 \times \text{Rate} = 4 \times 1.52 \times 10^{-4} = 6.08 \times 10^{-4} \text{ M/s}$$
(c) $$\frac{\Delta[\text{O}_2]}{\Delta t} = 1 \times \text{Rate} = 1.52 \times 10^{-4} \text{ M/s}$$
(a) $$\ln[\text{N}_2\text{O}_5]_t = \ln(0.500) – (5.0 \times 10^{-4})(1000)$$
$$= -0.693 – 0.500 = -1.193$$
$$[\text{N}_2\text{O}_5]_t = e^{-1.193} = 0.304 \text{ M}$$
(b) First find the rate of reaction:
$$\text{Rate} = k[\text{N}_2\text{O}_5] = (5.0 \times 10^{-4})(0.304) = 1.52 \times 10^{-4} \text{ M/s}$$
$$\frac{\Delta[\text{NO}_2]}{\Delta t} = 4 \times \text{Rate} = 4 \times 1.52 \times 10^{-4} = 6.08 \times 10^{-4} \text{ M/s}$$
(c) $$\frac{\Delta[\text{O}_2]}{\Delta t} = 1 \times \text{Rate} = 1.52 \times 10^{-4} \text{ M/s}$$