JEE Chemistry Chemical Kinetics Notes Set 01

Rate of a Reaction

Chemical kinetics studies how fast reactions go, what path (mechanism) they follow, and what changes the speed. Reactions can be almost instantaneous (ionic precipitation), very slow (rusting) or of moderate speed (the ones that can be measured in a laboratory).

Average rate is the change in concentration over a finite time interval, \( r_{av} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t} \). Instantaneous rate is the same quantity for an infinitesimally small interval, i.e. the slope of the tangent to the concentration-time curve at that moment.

For a general reaction \( aA + bB \to cC + dD \), the rate of reaction is \[ r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}. \] The unit is mol L⁻¹ s⁻¹ (for gases it can be atm s⁻¹).

Rate Law, Order and Molecularity

The rate law is an experimental expression connecting the rate with the concentrations of the reactants, e.g. \( r = k[A]^m[B]^n \). The exponents m and n are found only by experiment and need not match the stoichiometric coefficients. The overall order is m + n and can be zero, a fraction or negative. The constant k is the specific rate constant: it equals the rate when every concentration is 1 mol L⁻¹.

The unit of k for order n is \( (\text{mol L}^{-1})^{1-n}\,\text{s}^{-1} \): s⁻¹ for first order, mol L⁻¹ s⁻¹ for zero order and L mol⁻¹ s⁻¹ for second order.

Molecularity is the number of species that collide in a single elementary step. It is a whole number (1, 2 or rarely 3), is defined only for elementary steps and can never be zero or fractional. Order and molecularity agree only for elementary reactions. Reactions of molecularity above three are practically unknown because the collision of four or more particles together is extremely improbable.

Mechanism and the Rate-Determining Step

Most reactions occur in several elementary steps. The slowest step controls the overall rate, so the rate law is written from the slow step. If the slow step contains an intermediate, replace it using the preceding fast equilibrium. Example: for \( 2NO + Cl_2 \to 2NOCl \) via fast \( NO + Cl_2 \rightleftharpoons NOCl_2 \) and slow \( NOCl_2 + NO \to 2NOCl \), the rate is \( k[NO]^2[Cl_2] \).

Pseudo First-Order Reactions

When one reactant is present in huge excess its concentration hardly changes, and the observed order drops. Hydrolysis of an ester or of cane sugar in dilute acid is bimolecular but follows first-order kinetics because water is in large excess. The experimental constant is \( k' = k[H_2O] \).

Integrated Rate Equations

Zero order: \( [A] = [A]_0 - kt \); a plot of [A] against t is a straight line of slope −k; \( t_{1/2} = \frac{[A]_0}{2k} \). The rate does not change with concentration. Such reactions occur on metal surfaces when the surface is saturated (e.g. decomposition of ammonia on hot platinum).

First order: \( k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} \) or \( [A] = [A]_0 e^{-kt} \); a plot of \( \log[A] \) against t is a straight line of slope \( -\frac{k}{2.303} \). The half-life \( t_{1/2} = \frac{0.693}{k} \) does not depend on the starting concentration, so equal fractions of the reactant are used up in equal times. Also \( t_{75\%} = 2t_{1/2} \) and \( t_{99.9\%} \approx 10\,t_{1/2} \).

Second order (one reactant): \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \); a plot of 1/[A] against t is linear; \( t_{1/2} = \frac{1}{k[A]_0} \). In general \( t_{1/2} \propto [A]_0^{\,1-n} \), which is a quick way to find the order from half-life data.

Collision Theory and Activation Energy

For a reaction to occur, molecules must collide with at least the threshold energy and with a proper orientation. The rate is the collision frequency multiplied by the fraction of effective collisions. The minimum extra energy that reactants need above their average energy is the activation energy Ea. For a reversible reaction \( \Delta H = E_{a(f)} - E_{a(r)} \); an endothermic reaction has Ea(f) greater than ΔH.

Temperature Dependence: Arrhenius Equation

\( k = Ae^{-E_a/RT} \) so that \( \log k = \log A - \frac{E_a}{2.303RT} \). A plot of log k against 1/T is a straight line with slope \( -\frac{E_a}{2.303R} \) and intercept log A. For two temperatures, \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right). \] As T tends to infinity, k tends to A. The temperature coefficient (ratio of k for a 10 K rise) is usually 2 to 3.

A catalyst provides an alternative path with lower activation energy. It increases the rate of both the forward and reverse reactions equally, so it does not change ΔH or the equilibrium position, only the time taken to reach it.

Solved Examples

Example 1. A first-order reaction is 75% complete in 32 min. Find its rate constant and the time for 90% completion.
Answer:
75% completion is two half-lives, so \( t_{1/2} = 16 \) min.
\( k = \frac{0.693}{16} = 0.0433\ \text{min}^{-1} \).
\( t_{90\%} = \frac{2.303}{k}\log 10 = \frac{2.303}{0.0433} = 53.2 \) min.

Example 2. The rate of a reaction triples when the temperature rises from 300 K to 310 K. Find the activation energy (R = 8.314 J K⁻¹ mol⁻¹).
Answer:
\( \log 3 = \frac{E_a}{2.303\times8.314}\left(\frac{1}{300}-\frac{1}{310}\right) \).
\( \frac{1}{300}-\frac{1}{310} = 1.075\times10^{-4} \) K⁻¹ and \( \log 3 = 0.4771 \).
\( E_a = \frac{2.303\times8.314\times0.4771}{1.075\times10^{-4}} \approx 85.0 \) kJ mol⁻¹.

Example 3. The rate law of a reaction is rate = k[A]²[B]. By what factor does the rate change if [A] is halved and [B] is tripled?
Answer:
Factor = (½)² × 3 = ¾.
The rate becomes 0.75 times the original.