Unit 3: Chemical Kinetics
Introduction
Chemical kinetics is the branch of chemistry that studies the rate of chemical reactions and the factors that influence these rates. While thermodynamics tells us whether a reaction will occur, kinetics tells us how fast it will occur.
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Rate of a Reaction
The rate of reaction is the change in concentration of a reactant or product per unit time.
For a reaction A → B:
Rate = -d[A]/dt = +d[B]/dt
For aA + bB → cC + dD:
Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt
Units of rate: mol L-1 s-1 (or mol L-1 min-1)
Average rate = change in concentration / time interval
Instantaneous rate = slope of tangent to concentration-time curve at that point.
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Rate Law and Order of Reaction
The rate law expresses rate as a function of concentration:
Rate = k[A]m [B]n
- k = rate constant (specific rate constant)
- m = order with respect to A; n = order with respect to B
- Overall order = m + n
- Order is determined experimentally, not from stoichiometry.
Units of k: For an nth-order reaction: k has units of mol1-n Ln-1 s-1
For zero order: k has units mol L-1 s-1
For first order: k has units s-1
For second order: k has units L mol-1 s-1
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Integrated Rate Laws
Zero Order Reaction:
[A] = [A]0 - kt
t1/2 = [A]0 / 2k
First Order Reaction:
ln[A] = ln[A]0 - kt
OR [A] = [A]0 x e-kt
t1/2 = 0.693/k (independent of initial concentration — a key distinguishing feature!)
k = 2.303/t x log([A]0/[A])
Second Order Reaction:
1/[A] = 1/[A]0 + kt
t1/2 = 1/(k[A]0)
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Pseudo-First Order Reactions
When one reactant is in large excess (e.g., water in hydrolysis), its concentration remains nearly constant. The reaction appears to be first order in the other reactant. Example: hydrolysis of ethyl acetate in presence of excess water.
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Collision Theory and Activation Energy
- 1.Collision theory: Reactions occur when molecules collide with:
- 2.Sufficient energy (≥ activation energy, Ea)
- 3.Correct orientation
Activation energy (Ea): Minimum extra energy needed by reactants to overcome the energy barrier and form products.
Arrhenius Equation:
k = A e-Ea/RT
ln k = ln A - Ea/(RT)
In logarithmic form:
log k = log A - Ea/(2.303 RT)
For two temperatures T1 and T2:
log(k2/k1) = (Ea/2.303R) x (1/T1 - 1/T2)
where A = frequency factor (pre-exponential factor), R = 8.314 J mol-1 K-1.
Effect of Temperature: Rate roughly doubles for every 10°C rise in temperature (temperature coefficient ≈ 2).
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Factors Affecting Rate of Reaction
- 1.Concentration: Higher concentration → more collisions → faster rate
- 2.Temperature: Higher temperature → higher kinetic energy → more effective collisions
- 3.Catalyst: Provides an alternative reaction pathway with lower Ea; not consumed in the reaction
- 4.Surface area: More surface area (finer particles) → more collisions for solid reactants
- 5.Nature of reactants: Ionic reactions are faster; reactions requiring breaking of many strong bonds are slower
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Worked Examples
For 2N2O5 → 4NO2 + O2, if the rate of decrease of N2O5 is 0.04 mol L-1 s-1, find the rate of formation of NO2.
Rate = -(1/2) d[N2O5]/dt = +(1/4) d[NO2]/dt
-(1/2)(-0.04) = (1/4) d[NO2]/dt
0.02 = (1/4) d[NO2]/dt
d[NO2]/dt = 0.08 mol L-1 s-1
Rate = k[A]2[B]. If [A] is doubled and [B] is halved, how does the rate change?
New rate = k[2A]2[B/2] = k x 4[A]2 x [B]/2 = 2 x original rate (rate doubles)
For a first-order reaction, t1/2 = 693 s. Calculate k.
k = 0.693/t1/2 = 0.693/693 = 1 x 10-3 s-1
75% of a first-order reaction completes in 60 min. Find t1/2.
75% complete means 25% remains: [A] = [A]0/4
ln([A]0/[A]) = kt => ln(4) = k x 60
k = 1.386/60 = 0.02310 min-1
t1/2 = 0.693/0.02310 = 30 min
(Alternatively: 75% completion in 60 min = 2 half-lives, so t1/2 = 30 min)
Calculate Ea if k doubles from 300 K to 310 K. (R = 8.314 J mol-1 K-1)
log(k2/k1) = Ea/2.303R x (1/T1 - 1/T2)
log 2 = Ea / (2.303 x 8.314) x (1/300 - 1/310)
0.3010 = Ea/19.147 x (310-300)/(300 x 310)
0.3010 = Ea/19.147 x 10/93000
Ea = 0.3010 x 19.147 x 93000 / 10 = 53598 J/mol ≈ 53.6 kJ/mol
A first-order reaction has k = 1.5 x 10-3 s-1. How long does it take for 60% completion?
k x t = 2.303 log([A]0/[A]) = 2.303 log(100/40)
t = (2.303/k) x log(2.5) = (2.303/1.5 x 10-3) x 0.3979
t = 1535.3 x 0.3979 = 611 s
A reaction shows t1/2 independent of initial concentration. What is the order, and what is the unit of k?
t1/2 = 0.693/k (independent of [A]0) => first order reaction. Unit of k = s-1 (or min-1).
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Common mistakes
> - Assuming order equals stoichiometric coefficient. Order is always determined experimentally.
> - Using wrong units for k: always check units match the order of reaction.
> - Forgetting the stoichiometric factor when relating rates of different species in the same reaction.
> - Confusing activation energy with enthalpy of reaction. Ea is always positive (energy needed to cross the barrier); delta H can be negative (exothermic reaction).
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Summary
Chemical kinetics studies reaction rates, rate laws, and the factors affecting them. The rate law (Rate = k[A]m[B]n) is experimentally determined. Integrated rate laws give concentration-time relationships. First-order reactions have concentration-independent half-lives. The Arrhenius equation relates rate constant to temperature and activation energy. Catalysts lower activation energy. Understanding kinetics is crucial in industrial chemistry, pharmacology, and environmental science.