JEE Chemistry Chemical Kinetics Notes Set 05

Rate of Reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time. For \( aA + bB \rightarrow cC + dD \): rate = \( -\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt} \). Unit: mol L\(^{-1}\) s\(^{-1}\).
Factors: concentration, temperature, catalyst, nature of reactants, surface area of solids and (for photochemical reactions) light intensity.

Rate Law, Order and Molecularity

Rate law: rate = \( k[A]^m[B]^n \) found only by experiment. The order is \( m+n \); it may be zero, fractional or a whole number. The rate constant k is independent of concentration but depends on temperature.
Molecularity is the number of molecules colliding in an elementary step; it is always a whole number (1, 2 or rarely 3) and has no meaning for a complex reaction. A complex reaction proceeds through several steps and the slowest step (rate determining step) fixes the rate law.
Pseudo first order reactions: when one reactant is in large excess its concentration stays constant, e.g. acid hydrolysis of ethyl acetate and inversion of cane sugar.

Integrated Rate Equations

Zero order: \( [A]_0-[A]=kt \), \( t_{1/2}=\dfrac{[A]_0}{2k} \), unit of k: mol L\(^{-1}\) s\(^{-1}\).
First order: \( k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]} \), \( t_{1/2}=\dfrac{0.693}{k} \) (independent of initial concentration), unit of k: s\(^{-1}\). Time for 99.9% completion is about 10 half-lives; for 90% it is 3.3 half-lives.
Second order (one reactant): \( \dfrac{1}{[A]}-\dfrac{1}{[A]_0}=kt \), \( t_{1/2}=\dfrac{1}{k[A]_0} \), unit of k: L mol\(^{-1}\) s\(^{-1}\).
In general \( t_{1/2} \propto [A]_0^{\,1-n} \) and the unit of k is (mol/L)\(^{1-n}\) s\(^{-1}\).
After n half-lives the fraction remaining is \( (1/2)^n \).

Temperature Dependence and Catalysis

Temperature coefficient: ratio of the rate constants at T + 10 and T, usually between 2 and 3.
Arrhenius equation: \( k = Ae^{-E_a/RT} \) or \( \log k = \log A - \dfrac{E_a}{2.303RT} \). A plot of log k against 1/T is a straight line with slope \( -E_a/2.303R \). For two temperatures: \( \log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \).
Activation energy is the minimum extra energy that reactant molecules must acquire to react. A catalyst provides a new path of lower activation energy; it speeds up the forward and backward reactions equally and does not change \( \Delta H \), K or the extent of reaction.
Collision theory: only collisions with energy at least \( E_a \) and proper orientation lead to products; the fraction of effective collisions is \( e^{-E_a/RT} \).

Worked Examples

Example 1. A first order reaction has \( k = 2.303\times10^{-3}\ s^{-1} \). How long does it take for the concentration to fall to one-tenth?
Answer:
\( t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]} = \dfrac{2.303}{2.303\times10^{-3}}\log10 = 1000 \) s.

Example 2. The rate of a reaction doubles when the concentration of A is increased four times. Find the order in A.
Answer:
\( 4^n = 2 \), so \( n = \tfrac12 \).