Rate Laws – Advanced University‑Level Study Notes
(MIT‑style) – Chemistry, IIT‑JEE (Main + Advanced)
Goal: Derive rate laws from first principles, explore the underlying theory, examine edge cases, and solve a challenging problem.
1. Introduction
Kinetics studies how fast reactions occur and why reaction rates differ.
The rate law gives the instantaneous reaction rate (r) in terms of reactant concentrations ([A_i]) and a rate constant (k):
[ r = k \prod_i [A_i]^{\nu_i} ]
where (\nu_i) are the reaction orders (not necessarily equal to stoichiometric coefficients).
Key questions:
- How does one derive a rate law from a given mechanism?
- What governs the temperature dependence of (k)?
- When do deviations (fractional orders, diffusion control, surface reactions) arise?
2. Elementary Reactions & the Law of Mass Action
2.1 Definition
An elementary step is a single microscopic event (a single collision) that occurs in one quantum of time.
For an elementary reaction
[ aA + bB ;\xrightarrow{k}; \text{products} ]
the rate law follows directly from the law of mass action:
[ r = k [A]^a [B]^b ]
The exponent of each species equals its stoichiometric coefficient in that step.
2.2 Why the law holds
Assumptions:
- Reactants collide randomly in an ideal gas or dilute solution.
- Each collision that meets the necessary orientation and energy yields a successful reaction.
- No intermediate species have significant steady‑state concentration.
Under these assumptions, the number of successful collisions per unit time is proportional to the product of the reactant concentrations.
3. Mechanistic Rate Laws
Real reactions often involve multiple elementary steps.
A full mechanism:
[ \begin{aligned} &\text{Step 1: } aA + bB \xrightleftharpoons[k_{-1}]{k_1} C \ &\text{Step 2: } C \xrightarrow{k_2} \text{Products} \end{aligned} ]
3.1 Rate‑Determining Step (RDS) Approximation
If one step is much slower than the rest, the overall rate is governed by that step.
Let step 2 be rate‑determining:
[ r \approx k_2 [C] ]
But ([C]) is not known directly. Use steady‑state approximation (SSA) or pre‑equilibrium (PEA).
3.2 Steady‑State Approximation (SSA)
Assume the intermediate (C) is in a quasi‑steady state:
[ \frac{d[C]}{dt} = k_1 [A]^a [B]^b - k_{-1}[C] - k_2[C] \approx 0 ]
Solve for ([C]):
[ [C] = \frac{k_1 [A]^a [B]^b}{k_{-1} + k_2} ]
Hence
[ r = k_2 [C] = \frac{k_1 k_2}{k_{-1}+k_2} [A]^a [B]^b ]
Thus the overall reaction order equals the sum of the elementary orders of the RDS if the intermediate is in SSA.
3.3 Pre‑Equilibrium Approximation (PEA)
If step 1 is fast and reaches equilibrium quickly:
[ K_{\text{eq}} = \frac{k_1}{k_{-1}} = \frac{[C]}{[A]^a [B]^b} ;;\Rightarrow;; [C] = K_{\text{eq}} [A]^a [B]^b ]
Then
[ r = k_2 [C] = k_2 K_{\text{eq}} [A]^a [B]^b ]
Again the overall order equals the sum of the elementary orders of the RDS.
4. Transition State Theory (TST)
TST provides a microscopic picture of the activated complex (^\ddagger).
4.1 Derivation of Eyring Equation
Consider the equilibrium:
[ A + B ;\rightleftharpoons; ^\ddagger ]
The equilibrium constant (K^\ddagger) is
[ K^\ddagger = \frac{[^\ddagger]}{[A][B]} = \frac{k_B T}{h}, e^{-\Delta G^\ddagger / RT} ]
The rate is the flux over the transition state:
[ r = \frac{k_B T}{h} [^\ddagger] = k_B T , K^\ddagger , [A][B] ]
Thus
[ k = \frac{k_B T}{h} e^{-\Delta G^\ddagger / RT} ]
or
[ k = \frac{k_B T}{h} e^{-\Delta H^\ddagger / RT} e^{\Delta S^\ddagger / R} ]
This is the Eyring equation.
4.2 Connection to Arrhenius Equation
Using (\Delta G^\ddagger = \Delta H^\ddagger - T \