**Definition:** The rate of a chemical reaction is the change in concentration of a reactant or product per unit time, expressed as:
**Rate of disappearance of reactant R:**
$$r_{av} = -\frac{\Delta[R]}{\Delta t}$$
**Rate of appearance of product P:**
$$r_{av} = +\frac{\Delta[P]}{\Delta t}$$
The negative sign for reactants is used because their concentration decreases, and we want rate to be positive. For products, concentration increases, so no negative sign is needed.
**Units of Rate:** mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹, mol L⁻¹ h⁻¹, atm s⁻¹ for gaseous reactions)
**Average Rate vs Instantaneous Rate:**
$$r_{inst} = -\frac{d[R]}{dt} = \frac{d[P]}{dt}$$
Mathematically obtained using calculus (derivative). Graphically found by drawing tangent to concentration vs time curve at a specific time point and finding slope of tangent.
**Important Point:** Average rate always decreases with time as reactant concentration decreases, but instantaneous rate at any moment gives accurate picture of how fast reaction proceeds at that moment.
**Expression for Reactions with Unequal Stoichiometric Coefficients:**
For reaction: **aA + bB → cC + dD**
$$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$$
Example: For **2HI(g) → H₂(g) + I₂(g)**
$$\text{Rate} = -\frac{1}{2}\frac{\Delta[HI]}{\Delta t} = \frac{\Delta[H_2]}{\Delta t} = \frac{\Delta[I_2]}{\Delta t}$$
This ensures rate of reaction is same regardless of which species we follow (elimination of stoichiometric ambiguity).
**Definition:** The **rate law** (or rate expression) is a mathematical expression showing how reaction rate depends on concentration of reactants at constant temperature.
**General Form:**
$$\text{Rate} = k[A]^x[B]^y$$
where:
**Differential Form:**
$$-\frac{d[R]}{dt} = k[A]^x[B]^y$$
**Critical Point:** Rate law CANNOT be predicted from balanced equation alone; it must be determined experimentally by studying how initial rate changes with initial concentrations.
**Example:** For reaction **2NO(g) + O₂(g) → 2NO₂(g)**
Initial rate data shows:
Therefore: **Rate = k[NO]²[O₂]**
Note: Exponents (2 and 1) differ from stoichiometric coefficients (2 and 1 in this case match, but not always). For **CHCl₃ + Cl₂ → CCl₄ + HCl**, experimentally determined rate = k[CHCl₃][Cl₂]^(1/2), where exponent ½ ≠ stoichiometric coefficient 1.
**Definition:** The **order of a reaction** is the sum of exponents of concentration terms in the rate law expression.
$$\text{Overall Order} = x + y$$
**Types Based on Order:**
**Zero Order Reaction:**
**First Order Reaction:**
**Second Order Reaction:**
**Fractional Order:**
**Units of Rate Constant:**
Rate constant units depend on overall order:
| Order | Rate Equation Form | Units of k |
|-------|-------------------|------------|
| Zero | Rate = k | mol L⁻¹ s⁻¹ |
| First | Rate = k[A] | s⁻¹ |
| Second | Rate = k[A]² or k[A][B] | L mol⁻¹ s⁻¹ |
| Third | Rate = k[A]³ etc. | L² mol⁻² s⁻¹ |
**Derivation of units:** Since Rate (mol L⁻¹ s⁻¹) = k × [concentration terms]^n
$$k = \frac{\text{Rate}}{[\text{concentration}]^n} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^n} = \text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}$$
**Molecularity:**
**Order:**
**Key Difference:** Molecularity applies to elementary steps; order applies to overall reaction. For complex reactions, they differ. Overall order = order of rate-determining step + contributions from pre-equilibrium steps.
**Elementary Reaction:**
**Complex (Multi-step) Reaction:**
**Example:** Hydrogen peroxide decomposition (catalyzed by I⁻):
**Overall:** 2H₂O₂ → 2H₂O + O₂
**Mechanism (2 elementary steps):**
1. H₂O₂ + I⁻ → H₂O + IO⁻ (slow step = rate-determining)
2. H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast step)
**Intermediate:** IO⁻ (formed in step 1, consumed in step 2)
**Observed Rate Law:** Rate = k[H₂O₂][I⁻]
This can be derived from mechanism:
**Rate-Determining Step (RDS):**
**Determining Rate Law from Mechanism:**
If rate law involves species in fast pre-equilibrium steps, use equilibrium constant to eliminate intermediate:
Example: Reaction with fast pre-equilibrium:
1. A + B ⇌ C (fast, equilibrium)
2. C + D → E (slow)
For step 1: K = [C]/([A][B]), so [C] = K[A][B]
Rate = k₂[C][D] = k₂K[A][B][D] = k_overall[A][B][D]
Observed rate law includes all reactants affecting slow step via equilibrium.
**For Zero Order Reaction:**
Rate = k (independent of concentration)
Integrated form (by integration):
$$[A] = [A]_0 - kt$$
Half-life: $$t_{1/2} = \frac{[A]_0}{2k}$$
Linear plot: [A] vs t gives straight line with slope = -k
**For First Order Reaction:**
Rate = k[A]
Integrated form (differential equation solution):
$$\ln[A] = \ln[A]_0 - kt$$
or
$$\ln\frac{[A]_0}{[A]} = kt$$
**Half-life formula (key result):**
$$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$$
Important: **First order half-life is independent of initial concentration** — each half-life period reduces concentration by half, regardless of starting amount.
Linear plot: ln[A] vs t gives straight line with slope = -k
**Determining Reaction Order Experimentally:**
1. **Graphical Method:** Plot concentration data in different forms:
2. **Half-life Method:** If each successive half-life doubles (for zero order) or remains constant (for first order), determine order
3. **Initial Rate Method:** Vary one reactant concentration while keeping others constant; compare how rate changes
**Arrhenius Equation:**
$$k = Ae^{-E_a/RT}$$
or logarithmic form:
$$\ln k = \ln A - \frac{E_a}{RT}$$
where:
**Important Points:**
**Comparing Two Temperatures:**
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
Or rearranged:
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{T_2 - T_1}{T_1T_2}\right)$$
**Determining E_a:**
From two rate constants at different temperatures or from slope of ln k vs 1/T plot (slope = -E_a/R)
**Basic Postulates:**
1. **Collision Requirement:** For reaction to occur, reactant molecules must collide
2. **Energy Requirement:** Colliding molecules must possess minimum energy (activation energy E_a)
3. **Orientation Requirement:** Molecules must be properly oriented at moment of collision for bonds to form/break
**Reaction Rate Expression from Collision Theory:**
$$k = Az e^{-E_a/RT}$$
where:
Smaller steric factor p means fewer successful collisions despite proper energy.
**Example:** For 2HI → H₂ + I₂, collision theory predicts rate ∝ [HI]². Experimental rate law = k[HI]² confirms bimolecular mechanism. Activation energy from Arrhenius equation gives insight into reaction pathway.
**Energy Diagram:**
**Catalyst Effect:**
**Observation:** As reactants are consumed, rate decreases because:
**Example:** Hydrolysis of butyl chloride C₄H₉Cl + H₂O → C₄H₉OH + HCl
Initial data at higher concentrations: rate ≈ 1.90 × 10⁻⁴ mol L⁻¹ s⁻¹
Later data at lower concentrations: rate ≈ 0.4 × 10⁻⁴ mol L⁻¹ s⁻¹
This continuous decrease is expected for any order > 0.
---
| Concept | Formula | Key Points |
|---------|---------|-----------|
| Average Rate | r_av = -Δ[R]/Δt | Over finite interval |
| Instantaneous Rate | r_inst = d[R]/dt | At specific moment; graphical tangent |
| Rate Law | Rate = k[A]^x[B]^y | Must be determined experimentally |
| Order | n = x + y | Sum of exponents; can be fractional |
| Zero Order Half-life | t₁/₂ = [A]₀/2k | Depends on initial concentration |
| First Order Half-life | t₁/₂ = 0.693/k | Independent of initial concentration |
| Arrhenius Equation | k = Ae^(-E_a/RT) | Temperature dependence |
| Two-Temperature Arrhenius | ln(k₂/k₁) = (E_a/R)(1/T₁ - 1/T₂) | Calculating E_a |
| Stoichiometric Rate | Rate = -(1/a)Δ[A]/Δt = (1/c)Δ[C]/Δt | Coefficients normalize rates |
Q1. For the reaction: N₂O₅ → N₂O₄ + ½O₂, if the rate of disappearance of N₂O₅ is 1.4 × 10⁻⁵ mol L⁻¹ s⁻¹, what is the rate of appearance of O₂?
Answer: A — From stoichiometry, 1 mole N₂O₅ produces 0.5 mole O₂, so rate of O₂ = 0.5 × 1.4 × 10⁻⁵ = 0.7 × 10⁻⁵ mol L⁻¹ s⁻¹.
Q2. Which statement correctly distinguishes thermodynamics from chemical kinetics?
Answer: B — Thermodynamics answers 'will it happen?' (ΔG < 0); kinetics answers 'how fast will it happen?' — they are complementary.
Q3. From the data below for reaction A → B, calculate average rate between t = 50s and t = 100s: t(s): 0, 50, 100, 150 [A](mol L⁻¹): 1.00, 0.80, 0.64, 0.51
Answer: B — rav = Δ[A]/Δt = (0.80 – 0.64)/(100 – 50) = 0.16/50 = 0.0032 = 3.2 × 10⁻³ mol L⁻¹ s⁻¹.
Q4. Why is a negative sign used in the expression –Δ[R]/Δt for the rate of disappearance of reactant R?
Answer: B — Reactant concentration decreases, so Δ[R] is negative; multiplying by –1 ensures the reported rate is positive, matching convention that reaction rates are positive quantities.
Q5. Which of the following is NOT a correct statement about instantaneous rate?
Answer: B — Instantaneous rate equals average rate only by coincidence; average rate is constant over its interval, while instantaneous rate varies at each moment.
Q6. For a reaction at 25°C, the concentration of reactant changes from 0.500 mol L⁻¹ to 0.250 mol L⁻¹ in 200 seconds. What is the average rate of reaction?
Answer: A — rav = Δ[C]/Δt = (0.500 – 0.250)/200 = 0.250/200 = 1.25 × 10⁻³ mol L⁻¹ s⁻¹.
Q7. The concentration-time graph for a reaction shows that the average rate decreases with successive time intervals. What does this indicate about the reaction?
Answer: B — As reactants are consumed, their concentration decreases, making Δ[R] smaller in each successive interval; thus average rate = Δ[R]/Δt also decreases.
Q8. Consider two reactions: (I) Diamond → Graphite (thermodynamically favourable but occurs imperceptibly slowly) and (II) AgNO₃ + NaCl → AgCl (occurs instantaneously). Both statements below are given: Statement 1: Reaction (I) has negative ΔG but very high activation energy; Reaction (II) has negative ΔG and low activation energy. Statement 2: Kinetics determines the observable speed regardless of thermodynamic feasibility. (A) Both statements are correct (B) Statement 1 is correct, Statement 2 is incorrect (C) Statement 1 is incorrect, Statement 2 is correct (D) Both statements are incorrect
Answer: A — Statement 1: Correct — high Ea makes reaction slow despite favourable ΔG; low Ea makes reaction fast. Statement 2: Correct — kinetics explains observable rates; thermodynamics alone cannot.
Q9. If the instantaneous rate of a reaction at t = 300s is calculated from the tangent to a [C] vs t curve, and the tangent passes through points (250s, 0.80 mol L⁻¹) and (350s, 0.60 mol L⁻¹), what is the instantaneous rate at 300s?
Answer: A — Slope of tangent = Δ[C]/Δt = (0.60 – 0.80)/(350 – 250) = –0.20/100 = –0.002; instantaneous rate = |–0.002| = 0.002 mol L⁻¹ s⁻¹.
Q10. Which of the following correctly ranks the three time intervals in Example 3.1 (C₄H₉Cl hydrolysis) in order of decreasing average rate?
Answer: A — From Table 3.1: 0–50s has rate 1.90 × 10⁻⁴, 50–100s has 1.70 × 10⁻⁴, and 300–400s has 1.10 × 10⁻⁴ mol L⁻¹s⁻¹; early intervals show higher rates as reactant concentration is higher.
What is the fundamental difference between thermodynamics and chemical kinetics?
Thermodynamics predicts whether a reaction is feasible (ΔG < 0), while kinetics determines how fast that reaction actually proceeds.
Define average rate of a reaction with proper sign convention.
Average rate = –Δ[R]/Δt for reactants or +Δ[P]/Δt for products, where negative sign ensures rate is expressed as a positive quantity.
What is instantaneous rate and how is it obtained graphically?
Instantaneous rate is the reaction rate at a specific moment, obtained by drawing a tangent to the concentration-time curve at that instant and calculating its slope.
Why does the average rate of a reaction decrease over time?
As reactant concentration decreases with time, the change in concentration over successive time intervals becomes smaller, making the average rate lower.
What are the units of reaction rate when concentration is in mol L⁻¹ and time in seconds?
Units are mol L⁻¹ s⁻¹, derived from [concentration]/[time] = (mol L⁻¹)/(s).
For the reaction R → P with equal stoichiometry, how do the rates of disappearance and appearance compare?
The magnitude of the rate of disappearance of R equals the rate of appearance of P, though expressed with opposite signs in equations.
Why is it impossible to predict reaction rate at a specific moment using only average rate?
Average rate is constant over its defined time interval and does not reflect changes in concentration at individual moments; instantaneous rate is needed for that.
What does the slope of a tangent line on a [concentration] vs time graph represent?
The slope represents the instantaneous rate of change of concentration at the exact point where the tangent touches the curve.
State the relationship between kinetic studies and practical applications like food preservation or fuel combustion control.
Kinetics helps determine how to alter reaction conditions (concentration, temperature, catalyst) to speed up or slow down reactions for practical purposes.
Why must we use a negative sign when expressing the rate of disappearance of a reactant?
Reactant concentration decreases (negative change), so the negative sign converts the negative Δ[R] into a positive reaction rate value.
Define average rate and instantaneous rate of a reaction. Write the mathematical expression for instantaneous rate. [2 marks]
State that average rate = Δ[C]/Δt over a finite interval Δt, and instantaneous rate = d[C]/dt as Δt → 0 (limit of derivative). Mention that instantaneous rate is obtained from tangent slope.
For the reaction 2A → B, the concentration of A decreases from 0.80 mol L⁻¹ to 0.60 mol L⁻¹ in 10 minutes. Calculate (i) the average rate of disappearance of A, and (ii) the average rate of appearance of B. Show all working steps. [5 marks]
Use rav = –Δ[A]/Δt for disappearance (ensure negative sign); apply stoichiometry (2 moles A produce 1 mole B) so rate of B formation = 0.5 × rate of A disappearance. Convert time to seconds if asked for mol L⁻¹ s⁻¹.
Explain why kinetic studies are as important as thermodynamic studies for understanding chemical reactions. Support your answer with a real example, showing how a reaction can be thermodynamically favourable but kinetically too slow to observe. Discuss how chemists use kinetic knowledge in practical applications (food preservation, drug design, industrial catalysts). [6 marks]
Contrast ΔG < 0 (feasible, thermodynamics) with rate of reaction (kinetics). Use diamond-to-graphite example: ΔG < 0 but extremely slow. Explain activation energy barrier; discuss catalyst role; mention concentration, temperature effects on observable speed; link to real-world applications like decomposition prevention or reaction acceleration.
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