CBSE Class 12 Chemistry Chemical Kinetics Notes Set 07

These Class 12 CBSE Chemistry notes for Unit 3 Chemical Kinetics explain how the rate of a reaction is defined and measured, how the rate law, order and molecularity are found, and how the integrated equations of zero and first order reactions give the rate constant and half-life. They also cover collision theory, activation energy, potential energy diagrams and the Arrhenius equation, and end with nine solved numerical examples.

Important Notes and Formulas

Rate of a Chemical Reaction

1. Chemical kinetics is the branch of chemistry that studies the rates of reactions, the factors that affect them and the steps by which they take place. Reactions vary from very fast (ionic reactions such as AgNO₃ + NaCl) to very slow (rusting of iron) and moderate (hydrolysis of an ester).
2. The rate of a reaction is the change in the concentration of a reactant or a product per unit time. It is always positive. For R → P: \( \text{rate} = -\frac{\Delta[R]}{\Delta t} = +\frac{\Delta[P]}{\Delta t} \).
3. The unit of rate is mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹, mol L⁻¹ h⁻¹). For gases the unit may be atm s⁻¹.
4. The average rate is measured over a finite interval of time and equals the slope of the chord joining two points on the concentration–time curve.
5. The instantaneous rate is the rate at one instant, \( -\frac{d[R]}{dt} \), and equals the slope of the tangent to the curve at that instant. It is usually highest at the start and falls as the reactants are used up.

For the general reaction aA + bB → cC + dD, the rate of reaction is the same for every species once each term is divided by its stoichiometric coefficient:

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

Factors Affecting the Rate of a Reaction

1. Concentration: the rate generally increases when the concentration of a reactant increases, as the relation is expressed by the rate law.
2. Temperature: the rate (and the rate constant) increases with temperature. For many reactions the rate roughly doubles for every 10 °C rise.
3. Catalyst: a catalyst increases the rate by providing an alternative path of lower activation energy. It does not change ΔH or the equilibrium composition.
4. Nature and physical state of the reactants: reactions of ions in solution are almost instantaneous, while those involving bond breaking in molecules are slower. A solid reacts faster when it is finely divided because its surface area is larger.
5. Radiation: some reactions, such as H₂ + Cl₂ → 2HCl, are very slow in the dark but fast in light (photochemical reactions).

Rate Law, Rate Constant and Order of a Reaction

1. The rate law (rate expression) is the expression of the rate in terms of the concentrations of the reactants, each raised to a power found by experiment. For aA + bB → products: rate = k[A]ˣ[B]ʸ. The powers x and y need not equal the stoichiometric coefficients a and b.
2. The proportionality constant k is the rate constant (specific reaction rate). When every concentration is 1 mol L⁻¹, rate = k. A large k means a fast reaction. k depends on temperature and the catalyst but not on the concentrations.
3. The order of a reaction is the sum of the powers of the concentration terms in the rate law (x + y here). It can be zero, a whole number or a fraction, and it is found only by experiment.
4. Examples of experimental rate laws: CHCl₃ + Cl₂ → CCl₄ + HCl, rate = k[CHCl₃][Cl₂]^½ (order 3/2); CH₃CHO → CH₄ + CO, rate = k[CH₃CHO]^3/2 (order 3/2); 2NO + O₂ → 2NO₂, rate = k[NO]²[O₂] (order 3).

Units of the Rate Constant

For rate = k[A]ⁿ, \( k = \dfrac{\text{rate}}{[A]^n} \), so the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹.

Molecularity and Elementary Steps

1. An elementary reaction takes place in a single step. A complex reaction takes place in a sequence of elementary steps.
2. Molecularity is the number of reacting species (atoms, ions or molecules) that collide simultaneously in an elementary step. It is 1 (unimolecular, e.g. NH₄NO₂ → N₂ + 2H₂O), 2 (bimolecular, e.g. 2HI → H₂ + I₂) or 3 (termolecular, e.g. 2NO + O₂ → 2NO₂). Reactions of molecularity above 3 are very rare because the chance of four species colliding together is extremely small.
3. The rate of a complex reaction is controlled by its slowest step, the rate determining step. Example: in H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂ the observed rate is k[H₂O₂][I⁻]. This agrees with a mechanism whose slow first step is H₂O₂ + I⁻ → H₂O + IO⁻, followed by fast steps. The reaction is of first order in each of H₂O₂ and I⁻ although the overall equation has more species.
4. For an elementary reaction the order equals the molecularity, but for a complex reaction the order is decided by experiment (the slow step).

Pseudo First Order Reactions

A reaction which is not of first order but behaves like one because a reactant is present in large excess is a pseudo first order reaction. In the acid hydrolysis of ethyl acetate, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, the rate is k[CH₃COOC₂H₅][H₂O]. Water is in large excess, so [H₂O] is practically constant and rate = k′[CH₃COOC₂H₅] with k′ = k[H₂O]. The inversion of cane sugar is another example.

Determining the Order of a Reaction

1. Method of initial rates: measure the initial rate in several experiments, changing the concentration of one reactant at a time. If the concentration is doubled and the rate becomes 2ⁿ times, the order in that reactant is n.
2. Integrated rate equation (graphical) method: plot [R] against t (straight line for zero order) or ln [R] against t (straight line for first order); the plot that is a straight line gives the order.
3. Half-life method: a constant half-life indicates first order, and a half-life proportional to the initial concentration indicates zero order.

Integrated Rate Equation of a Zero Order Reaction

1. For R → P with rate = k[R]⁰ = k: \( -\frac{d[R]}{dt} = k \). Integrating and using [R] = [R]₀ at t = 0 gives [R] = [R]₀ − kt, so \( k = \frac{[R]_0 - [R]}{t} \).
2. A plot of [R] against t is a straight line with slope −k and intercept [R]₀. The rate is constant and independent of [R].
3. Half-life: \( t_{1/2} = \frac{[R]_0}{2k} \), which is directly proportional to the initial concentration. The reaction is complete at t = [R]₀/k.
4. Zero order reactions are uncommon. They occur when the rate is limited by something other than the concentration, e.g. the decomposition of NH₃ on a hot platinum surface (2NH₃ → N₂ + 3H₂) or the decomposition of HI on gold.

Integrated Rate Equation of a First Order Reaction

1. For R → P with rate = k[R]: \( \frac{d[R]}{[R]} = -k\,dt \). Integrating and using [R] = [R]₀ at t = 0 gives ln [R] = ln [R]₀ − kt.
2. Rate constant: \( k = \frac{1}{t}\ln\frac{[R]_0}{[R]} = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \). In the form [R] = [R]₀e^−kt, the concentration falls exponentially.
3. A plot of ln [R] against t is a straight line of slope −k, and a plot of log [R] against t has slope −k/2.303.
4. Half-life: \( t_{1/2} = \frac{0.693}{k} \), which is independent of the initial concentration. Every succeeding half-life is the same: after n half-lives the fraction left is (½)ⁿ.
5. The time for any given fraction of completion is independent of the initial concentration. For 75% completion t = 2t½ and for 99.9% completion t ≈ 10t½.
6. If [R]₁ and [R]₂ are the concentrations at times t₁ and t₂, \( k = \frac{2.303}{t_2 - t_1}\log\frac{[R]_1}{[R]_2} \).
7. Examples of first order reactions: decomposition of N₂O₅ and H₂O₂, and all radioactive decays.

Collision Theory and Activation Energy

1. According to collision theory, the reactant molecules must collide for a reaction to occur. The collision frequency (Z) of gases is very high (about 10²⁵ to 10²⁸ collisions per second per cm³), yet only a small fraction of collisions lead to products. These are the effective collisions.
2. A collision is effective only if the molecules (i) have a minimum energy, equal to or greater than the threshold energy, and (ii) collide with the proper orientation.
3. The activation energy (E_a) is the minimum extra energy that the reactant molecules must acquire to form the activated complex (transition state): E_a = threshold energy − average energy of the reactants. The activated complex is an unstable species at the top of the energy barrier which breaks to give products.
4. The rate constant can be written as k = PZe^−E_a/RT, where P is the steric (probability) factor for the orientation of colliding molecules and e^−E_a/RT is the fraction of molecules with energy ≥ E_a.
5. In a potential energy diagram, ΔH = E_a(forward) − E_a(reverse). For an exothermic reaction ΔH is negative and the reverse barrier is higher; for an endothermic reaction ΔH is positive and the reverse barrier is lower.

1. A catalyst provides an alternative path of lower activation energy. It lowers the barrier for the forward and the reverse reaction by the same amount, increases the rate, but leaves ΔH and the equilibrium constant unchanged.
2. On raising the temperature the kinetic energy of the molecules increases and the energy distribution curve shifts and flattens. The fraction of molecules with energy ≥ E_a (the shaded area) increases sharply, so the rate increases.

Temperature Dependence: the Arrhenius Equation

1. The Arrhenius equation is k = Ae^−E_a/RT, where A is the Arrhenius (pre-exponential or frequency) factor, E_a is the activation energy, R = 8.314 J K⁻¹ mol⁻¹ and T is the temperature in kelvin. A is related to the collision frequency and orientation factor.
2. Taking logarithms: ln k = ln A − E_a/RT, or log k = log A − E_a/(2.303RT). A plot of ln k against 1/T is a straight line with slope −E_a/R and intercept ln A, so E_a can be found from the slope.
3. If k₁ and k₂ are the rate constants at temperatures T₁ and T₂: \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \). The energy E_a comes out in J mol⁻¹ when R = 8.314 J K⁻¹ mol⁻¹.
4. k increases with temperature, and for a reaction with E_a of about 50 kJ mol⁻¹ it roughly doubles for a 10 K rise near room temperature. A reaction with higher E_a is more sensitive to temperature. A catalyst lowers E_a and therefore increases k at the same temperature.

Solved Examples

Example 1. For the reaction 4NH₃ + 5O₂ → 4NO + 6H₂O, the rate of formation of NO is 3.6 × 10⁻³ mol L⁻¹ s⁻¹. Calculate the rate of disappearance of NH₃ and O₂, the rate of formation of H₂O and the rate of reaction.
Answer:
Rate of reaction = \( -\frac{1}{4}\frac{d[NH_3]}{dt} = -\frac{1}{5}\frac{d[O_2]}{dt} = \frac{1}{4}\frac{d[NO]}{dt} = \frac{1}{6}\frac{d[H_2O]}{dt} \)
Rate of reaction = ¼ × 3.6 × 10⁻³ = 9.0 × 10⁻⁴ mol L⁻¹ s⁻¹.
−d[NH₃]/dt = d[NO]/dt = 3.6 × 10⁻³ mol L⁻¹ s⁻¹.
−d[O₂]/dt = 5 × 9.0 × 10⁻⁴ = 4.5 × 10⁻³ mol L⁻¹ s⁻¹.
d[H₂O]/dt = 6 × 9.0 × 10⁻⁴ = 5.4 × 10⁻³ mol L⁻¹ s⁻¹.

Example 2. The initial rates of the reaction A + B → products were measured for three runs: ([A], [B], rate) = (0.10 M, 0.10 M, 5.0 × 10⁻⁴ M s⁻¹), (0.20 M, 0.10 M, 1.0 × 10⁻³ M s⁻¹) and (0.10 M, 0.30 M, 4.5 × 10⁻³ M s⁻¹). Find the order with respect to A and B, the overall order and the rate constant.
Answer:
Runs 1 and 2: [A] doubles with [B] constant and the rate doubles, so 2ˣ = 2 and the order in A is 1.
Runs 1 and 3: [B] triples with [A] constant and the rate becomes 4.5 × 10⁻³ ÷ 5.0 × 10⁻⁴ = 9 times, so 3ʸ = 9 and the order in B is 2.
Rate = k[A][B]²; overall order = 3.
k = 5.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 0.50 mol⁻² L² s⁻¹.

Example 3. A zero order reaction has a rate constant of 2.0 × 10⁻³ mol L⁻¹ s⁻¹ and an initial concentration of 0.10 mol L⁻¹. Find the concentration after 20 s, the half-life and the time for complete reaction.
Answer:
[R] = [R]₀ − kt = 0.10 − 2.0 × 10⁻³ × 20 = 0.060 mol L⁻¹.
t½ = [R]₀/2k = 0.10 ÷ (2 × 2.0 × 10⁻³) = 25 s.
Complete reaction: t = [R]₀/k = 0.10 ÷ 2.0 × 10⁻³ = 50 s.

Example 4. In a first order reaction the concentration of the reactant falls from 0.50 mol L⁻¹ to 0.20 mol L⁻¹ in 30 minutes. Calculate the rate constant, the half-life and the time taken for 90% completion.
Answer:
k = (2.303/t) log ([R]₀/[R]) = (2.303 ÷ 30) × log 2.5 = (2.303 ÷ 30) × 0.3979 = 0.0305 min⁻¹.
t½ = 0.693 ÷ 0.0305 = 22.7 min.
For 90% completion [R]₀/[R] = 10, so t = 2.303 ÷ 0.0305 × log 10 = 75.4 min.

Example 5. The rate constant of a first order decomposition is 5.0 × 10⁻³ s⁻¹. Calculate the time for 99% completion and compare it with the half-life.
Answer:
For 99% completion, [R]₀/[R] = 100, so t = (2.303/k) log 100 = 2 × 2.303 ÷ 5.0 × 10⁻³ = 921 s.
t½ = 0.693 ÷ 5.0 × 10⁻³ = 138.6 s, so 99% completion takes about 6.6 half-lives (921 ÷ 138.6 ≈ 6.6).

Example 6. The rate constant of a reaction is 3.0 × 10⁻⁴ s⁻¹ at 300 K and 1.2 × 10⁻³ s⁻¹ at 320 K. Calculate the activation energy (R = 8.314 J K⁻¹ mol⁻¹).
Answer:
k₂/k₁ = 1.2 × 10⁻³ ÷ 3.0 × 10⁻⁴ = 4, so log 4 = 0.6021.
log (k₂/k₁) = E_a/(2.303R) × (T₂ − T₁)/(T₁T₂) gives 0.6021 = E_a ÷ (2.303 × 8.314) × (20 ÷ (300 × 320)).
E_a = 0.6021 × 2.303 × 8.314 × 300 × 320 ÷ 20 = 5.53 × 10⁴ J mol⁻¹ = 55.3 kJ mol⁻¹.

Example 7. Calculate the fraction of molecules having energy equal to or greater than the activation energy at 500 K for a reaction with E_a = 100 kJ mol⁻¹.
Answer:
Fraction = e^−E_a/RT = e^{−100000/(8.314 × 500)} = e^−24.06.
Fraction = 3.6 × 10⁻¹¹. Only about 4 in every 10¹¹ molecules have enough energy, which explains why the reaction is slow.

Example 8. For the endothermic reaction A → B, ΔH = +30 kJ mol⁻¹ and the activation energy of the forward reaction is 75 kJ mol⁻¹. What is the activation energy of the reverse reaction?
Answer:
ΔH = E_a(forward) − E_a(reverse), so 30 = 75 − E_a(reverse).
E_a(reverse) = 75 − 30 = 45 kJ mol⁻¹. The reverse barrier is lower because the products lie at a higher energy than the reactants.

Example 9. A catalyst lowers the activation energy of a reaction from 75 kJ mol⁻¹ to 50 kJ mol⁻¹ at 300 K. By what factor does the rate constant change if the pre-exponential factor A is the same?
Answer:
k_cat/k = e^{(75000 − 50000)/(8.314 × 300)} = e^10.02.
The ratio is about 2.3 × 10⁴, so the catalysed reaction is more than twenty thousand times faster.