JEE Chemistry Chemical Kinetics Notes Set 03

Rate of Reaction

The rate of reaction is the change of concentration of a reactant or product per unit time. For \( aA + bB \rightarrow cC + dD \):

\[ \text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt} \]

1. Average rate is measured over a time interval; instantaneous rate is the slope of the concentration-time curve at one moment.
2. Unit of rate is concentration per unit time, for example mol L⁻¹ s⁻¹ (atm s⁻¹ for gases).

Rate Law, Order and Molecularity

1. The rate law expresses rate as a product of concentration powers, for example rate = k[A]^m[B]ⁿ. The powers m and n are found by experiment and the order is m + n.
2. Molecularity is the number of species that collide in an elementary step. It is a whole number (1, 2 or 3), while order can be zero, fractional or negative.
3. For an elementary reaction the order equals the molecularity, but for a complex reaction the order comes from the slowest step.
4. A pseudo first order reaction has a second order rate law, but one reactant is in such large excess that its concentration stays constant (for example hydrolysis of an ester in water).

To find an order from data, change one concentration at a time and see how the initial rate changes: if the concentration doubles and the rate becomes 2^m times, the order is m.

Units of Rate Constant

Order n	Unit of k
0	mol L⁻¹ s⁻¹
1	s⁻¹
2	L mol⁻¹ s⁻¹
n	(mol L⁻¹)1−n s⁻¹

Integrated Rate Equations and Half-Life

Order	Integrated equation	Half-life	Straight-line plot
Zero	\( [A] = [A]_0 - kt \)	\( \dfrac{[A]_0}{2k} \)	[A] against t
First	\( k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]} \)	\( \dfrac{0.693}{k} \) (independent of concentration)	log [A] against t
Second	\( \dfrac{1}{[A]} - \dfrac{1}{[A]_0} = kt \)	\( \dfrac{1}{k[A]_0} \)	1/[A] against t

1. For first order, equal fractions of reaction take equal times: 75% takes two half-lives, 87.5% takes three, 99.9% takes about ten.
2. In general \( t_{1/2} \propto a^{1-n} \).

Reaction Mechanism

1. A complex reaction proceeds through elementary steps. The slowest step is the rate-determining step and the rate law is written from it.
2. If the slow step involves an intermediate, express the intermediate through the preceding fast equilibrium so that only reactants appear in the final rate law.

Temperature Dependence and Arrhenius Equation

\[ k = A\,e^{-E_a/RT}, \qquad \log k = \log A - \frac{E_a}{2.303\,RT} \] \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right] \]

1. A graph of log k against 1/T is a straight line of slope −Ea/(2.303R) and intercept log A.
2. Temperature raises the fraction of molecules having energy above Ea, which is why rate rises sharply. Many reactions roughly double in rate for a 10°C rise.
3. A catalyst lowers Ea, so both forward and backward rates increase equally and the equilibrium position is unchanged.
4. For a reversible reaction, \( E_a(\text{forward}) - E_a(\text{reverse}) = \Delta H \).

Collision Theory

A reaction occurs only when molecules collide with energy equal to or greater than the threshold energy and with a suitable orientation. The rate depends on the collision frequency multiplied by the fraction of effective collisions.

Worked Examples

Example 1. A first order reaction is 75% complete in 40 min. Two half-lives equal 40 min, so \( t_{1/2} = 20 \) min and \( k = 0.693/20 = 0.0347 \) min⁻¹.

Example 2. Doubling [A] at constant [B] gives a 4-fold rate and doubling [B] at constant [A] gives a 2-fold rate. Rate = k[A]²[B], and the overall order is 3.

Example 3. For a zero order reaction with k = 0.02 mol L⁻¹ s⁻¹ and [A]₀ = 1 M, the reactant is used up completely in 1/0.02 = 50 s.