JEE Chemistry Chemical Kinetics Notes Set 02

Rate of Reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time. For aA + bB → cC + dD:
\( \text{Rate} = -\dfrac1a\dfrac{d[A]}{dt} = -\dfrac1b\dfrac{d[B]}{dt} = \dfrac1c\dfrac{d[C]}{dt} = \dfrac1d\dfrac{d[D]}{dt} \)
The average rate is Δ[concentration]/Δt over a time interval and the instantaneous rate is the slope of the tangent to the concentration-time curve. The unit of rate is mol L⁻¹ s⁻¹ (for gases, atm s⁻¹).

Rate Law, Order and Molecularity

The rate law expresses rate in terms of concentrations, with exponents found by experiment: rate = k[A]^m[B]ⁿ. The order is m + n; it can be zero, fractional or negative, and is not given by the stoichiometric coefficients unless the reaction is elementary.
Molecularity is the number of species colliding in an elementary step. It is always a whole number (1, 2 or rarely 3) and is meaningless for a complex reaction.
Unit of k = (mol L⁻¹)¹⁻ⁿ s⁻¹: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹.
Pseudo first order: when one reactant is in large excess its concentration is practically constant, so the reaction obeys first order kinetics (acid hydrolysis of an ester, inversion of cane sugar).

Worked example 1. For 2A + B → products the data ([A], [B], rate) are (0.1, 0.1, 2 × 10⁻³), (0.2, 0.1, 8 × 10⁻³), (0.2, 0.2, 1.6 × 10⁻²). Find the rate law.
Answer:
Doubling [A] gives 4 times the rate, so order in A is 2.
Doubling [B] doubles the rate, so order in B is 1.
Rate = k[A]²[B]; order 3.

Reaction Mechanism

A complex reaction proceeds by a sequence of elementary steps. The slowest step is the rate-determining step and controls the rate law. Intermediates formed in fast pre-equilibria must be eliminated using the equilibrium constant.
Example: 2NO + Cl₂ → 2NOCl by NO + Cl₂ ⇌ NOCl₂ (fast) and NOCl₂ + NO → 2NOCl (slow). Rate = k[NOCl₂][NO] = kK[NO]²[Cl₂].

Integrated Rate Equations

First order: \( k = \dfrac{2.303}{t}\log\dfrac{a}{a-x} \), where a is the initial amount and a − x is the amount left. 75% completion takes two half lives and 87.5% takes three. Half life of any order n is proportional to a¹⁻ⁿ.
For a gaseous first order reaction in which the number of moles changes, use the total pressure to find the partial pressure of the reactant, e.g. for A(g) → 2B(g) + C(g): P_A = (3P₀ − P_t)/2.

Worked example 2. A first order reaction is 75% complete in 32 min. Find the rate constant.
Answer:
\( k = \dfrac{2.303}{32}\log 4 = 0.0433\ \text{min}^{-1} \)

Collision Theory and Activation Energy

Reaction occurs only through effective collisions: the colliding molecules must have energy at least equal to the threshold energy and the proper orientation. Rate = (collision frequency) × (steric factor) × e^−Ea/RT.
Activation energy E_a is the extra energy that reactants must acquire to reach the activated complex. For a reaction, ΔH = E_a(forward) − E_a(backward), so for an endothermic reaction E_a is greater than ΔH. A catalyst lowers E_a for both directions equally and does not change ΔH.

Effect of Temperature: Arrhenius Equation

\( k = Ae^{-E_a/RT} \), so \( \ln k = \ln A - \dfrac{E_a}{RT} \). A plot of ln k against 1/T is a straight line with slope −E_a/R.
Between two temperatures: \( \log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right) \). As T → ∞, k → A.
For many reactions the rate doubles for every 10 K rise in temperature (temperature coefficient about 2), because the fraction of molecules with energy greater than E_a increases sharply.

Worked example 3. The rate doubles from 293 K to 308 K. Find E_a.
Answer:
\( E_a = \dfrac{2.303\times8.314\times\log2\times293\times308}{15} = 3.47\times10^4\ \text{J mol}^{-1} \)