Introduction to Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the study of reaction rates and their mechanisms. It helps us understand how chemical reactions occur and what factors influence their speed.
Why Study Chemical Kinetics?
- To understand the speed of chemical reactions
- To determine the factors affecting reaction rates
- To predict how long a reaction will take to complete
- To design optimal conditions for industrial processes
- To understand reaction mechanisms
Note: Thermodynamics tells us whether a reaction is feasible, while kinetics tells us how fast it will occur.
Rate of a Reaction
Definition
The rate of a reaction is defined as the change in concentration of a reactant or product per unit time.
For a reaction: R → P
Rate of disappearance of R = -Δ[R]/Δt
Rate of appearance of P = +Δ[P]/Δt
Average Rate
The average rate of a reaction is the change in concentration over a finite time interval.
rav = -Δ[R]/Δt = +Δ[P]/Δt
Instantaneous Rate
The instantaneous rate is the rate at a particular instant of time, obtained when Δt approaches zero.
rinst = -d[R]/dt = +d[P]/dt
Fig: Concentration vs Time graph showing instantaneous rate as the slope of tangent
NEET/JEE Tip: For reactions with different stoichiometric coefficients, divide the rate by the coefficient to get the unique rate of reaction.
Example: For the reaction 2HI(g) → H₂(g) + I₂(g), express the rate of reaction.
Solution: Rate = -½ d[HI]/dt = d[H₂]/dt = d[I₂]/dt
Factors Influencing Rate of Reaction
- Concentration of reactants: Generally, rate increases with increasing concentration
- Temperature: Rate increases with increasing temperature
- Catalyst: Increases rate without being consumed
- Surface area: For heterogeneous reactions, rate increases with surface area
- Pressure: For gaseous reactions, rate increases with pressure
Rate Law and Rate Constant
Rate Law Expression
For a general reaction: aA + bB → cC + dD
Rate = k [A]x [B]y
where:
- k = rate constant
- x = order with respect to A
- y = order with respect to B
- (x + y) = overall order of reaction
Important: Rate law cannot be predicted from the stoichiometric equation; it must be determined experimentally.
Rate Constant (k)
The rate constant is a proportionality constant in the rate law. Its units depend on the overall order of the reaction.
| Order | Units of k |
|---|---|
| Zero | mol L⁻¹ s⁻¹ |
| First | s⁻¹ |
| Second | L mol⁻¹ s⁻¹ |
| Third | L² mol⁻² s⁻¹ |
NEET/JEE Tip: You can determine the order of reaction from the units of the rate constant.
Order and Molecularity of Reactions
Order of Reaction
The sum of powers of concentration terms in the rate law expression.
Characteristics of Order:
- Experimental quantity
- Can be zero, fractional, or integer
- Applicable to both elementary and complex reactions
Molecularity of Reaction
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction.
Characteristics of Molecularity:
- Theoretical concept
- Always a whole number (1, 2, or 3)
- Applicable only to elementary reactions
| Property | Order | Molecularity |
|---|---|---|
| Definition | Sum of powers of concentration terms in rate law | Number of reacting species in an elementary step |
| Value | Can be 0, 1, 2, 3 or fraction | Always 1, 2, or 3 |
| Applicability | Elementary and complex reactions | Only elementary reactions |
| Determination | Experimental | Theoretical (from reaction mechanism) |
Question: For a reaction, A + B → Product; the rate law is given by r = k [A]1/2 [B]2. What is the order of the reaction?
Answer: Order = 1/2 + 2 = 2.5 (half order)
Integrated Rate Equations
Zero Order Reactions
Reactions where rate is independent of concentration of reactants.
Rate = -d[R]/dt = k
[R] = -kt + [R]0
t1/2 = [R]0/2k
Fig: For zero order reaction, [R] vs t is a straight line with slope = -k
First Order Reactions
Reactions where rate depends on the first power of concentration of one reactant.
Rate = -d[R]/dt = k[R]
ln[R] = -kt + ln[R]0
or k = (2.303/t) log([R]0/[R])
t1/2 = 0.693/k
Fig: For first order reaction, ln[R] vs t is a straight line with slope = -k
Important: For first order reactions, half-life is independent of initial concentration.
Pseudo First Order Reactions
Reactions that are actually of higher order but behave as first order when one reactant is in excess.
Examples:
- Hydrolysis of ester in excess water: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
- Inversion of cane sugar: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
Example: A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
Solution: t = (2.303/k) log([R]0/[R]) = (2.303/1.15×10⁻³) log(5/3) = 444 s
Temperature Dependence of Reaction Rate
Arrhenius Equation
k = A e-Ea/RT
where:
- k = rate constant
- A = Arrhenius factor or frequency factor
- Ea = activation energy
- R = gas constant
- T = temperature in Kelvin
Logarithmic Form of Arrhenius Equation
ln k = ln A - Ea/RT
or log k = log A - Ea/(2.303RT)
Fig: Plot of ln k vs 1/T gives a straight line with slope = -Ea/R
Two-Point Form of Arrhenius Equation
log(k₂/k₁) = Ea/(2.303R) × (1/T₁ - 1/T₂)
NEET/JEE Tip: For most reactions, rate doubles for every 10°C rise in temperature.
Example: The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.
Solution: Using log(k₂/k₁) = Ea/(2.303R) × (1/T₁ - 1/T₂)
log 2 = Ea/(2.303×8.314) × (1/298 - 1/308)
Ea = 52.897 kJ mol⁻¹
Collision Theory of Chemical Reactions
According to collision theory, for a reaction to occur:
- Molecules must collide
- Collisions must have sufficient energy (≥ Ea)
- Collisions must have proper orientation
Rate = PZ e-Ea/RT
where:
- P = probability or steric factor
- Z = collision frequency
- e-Ea/RT = fraction of molecules with energy ≥ Ea
Fig: Maxwell-Boltzmann distribution of molecular energies
Effect of Catalyst
A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change.
Characteristics of Catalysts:
- Lowers activation energy by providing an alternative path
- Does not change ΔG of the reaction
- Does not alter the equilibrium constant
- Catalyses both forward and backward reactions equally
- Small amount can catalyse large amount of reactants
Fig: Energy profile diagram showing effect of catalyst
Important Formulas Summary
| Concept | Formula |
|---|---|
| Average Rate | rav = -Δ[R]/Δt = +Δ[P]/Δt |
| Instantaneous Rate | rinst = -d[R]/dt = +d[P]/dt |
| Rate Law | Rate = k [A]x [B]y |
| Zero Order Integrated | [R] = -kt + [R]0 |
| Zero Order Half-life | t1/2 = [R]0/2k |
| First Order Integrated | ln[R] = -kt + ln[R]0 or k = (2.303/t) log([R]0/[R]) |
| First Order Half-life | t1/2 = 0.693/k |
| Arrhenius Equation | k = A e-Ea/RT |
| Two-point Form | log(k₂/k₁) = Ea/(2.303R) × (1/T₁ - 1/T₂) |
Important Questions for NEET/JEE
1. Define order of reaction. How is it different from molecularity?
Order of reaction is the sum of powers of concentration terms in the rate law expression. It is an experimental quantity and can be zero, fractional or integer. Molecularity is the number of reacting species in an elementary reaction and is always a whole number (1, 2, or 3).
2. A reaction is second order with respect to a reactant. How is the rate affected if concentration is doubled?
For a second order reaction, Rate = k[A]². If [A] is doubled, rate becomes 4 times.
3. The half-life of a reaction is 100 seconds. Calculate the time required for 75% completion for a first order reaction.
For first order reaction, t1/2 = 0.693/k = 100 s ⇒ k = 0.00693 s⁻¹
For 75% completion, [R] = 0.25[R]0
t = (2.303/k) log([R]0/[R]) = (2.303/0.00693) log(4) = 200 s
4. What is the effect of catalyst on activation energy and rate constant?
Catalyst lowers the activation energy by providing an alternative path. According to Arrhenius equation, lower Ea leads to higher rate constant, thus increasing the rate of reaction.
5. Show that for a first order reaction, time required for 99% completion is twice the time required for 90% completion.
For first order reaction: t = (2.303/k) log([R]0/[R])
For 90% completion: [R] = 0.1[R]0 ⇒ t90% = (2.303/k) log(10)
For 99% completion: [R] = 0.01[R]0 ⇒ t99% = (2.303/k) log(100) = (2.303/k) × 2 log(10)
∴ t99% = 2 × t90%