Chemical Kinetics - Class 12 Chemistry

Class 12 Chemistry | Chapter 3

Chemical Kinetics

Rate of Reaction • Rate Law • Order & Molecularity • Integrated Rate Equations • Arrhenius Equation

1. Rate of a Chemical Reaction

Chemical Kinetics: The branch of chemistry that deals with the study of reaction rates and their mechanisms.

The rate of a reaction is the change in concentration of a reactant or product per unit time.

Average Rate (r_av): Δ[R]/Δt or Δ[P]/Δt over a large time interval.
Instantaneous Rate (r_inst): -d[R]/dt or +d[P]/dt at any particular instant of time (slope of the tangent to the concentration-time curve).
Unit of rate: mol L⁻¹ s⁻¹ (or atm s⁻¹ for gaseous reactions).

1.1 Rate Expression from Stoichiometry

For a general reaction: aA + bB → cC + dD

Rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt

Note: Negative sign indicates decrease in concentration of reactants. Rate is always a positive quantity.

2. Rate Law and Rate Constant

Rate Law: The expression which relates the rate of reaction to concentration of reactants. It is an experimentally determined equation.

Rate = k [A]^x [B]^y

k: Rate constant or specific reaction rate. It is the rate of reaction when concentration of each reactant is unity (1 M).
Dependence of k: Depends on temperature and catalyst (NOT on initial concentration).

2.1 Order of a Reaction (n)

The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

Order = x + y.
Order can be 0, 1, 2, 3, or even a fraction or negative. It is an experimental value.

2.2 Units of Rate Constant (k)

Unit of rate (mol L⁻¹ s⁻¹) = k × (mol L⁻¹)ⁿ. Hence, Unit of k = (mol L⁻¹)^1-n s⁻¹

Order (n)	Reaction Type	Unit of k
0	Zero Order	mol L⁻¹ s⁻¹ (Same as rate unit)
1	First Order	s⁻¹
2	Second Order	mol⁻¹ L s⁻¹ (or L mol⁻¹ s⁻¹)

3. Molecularity of a Reaction

The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

It can be 1 (unimolecular), 2 (bimolecular), or 3 (termolecular).
It is a theoretical concept applicable only to elementary (single-step) reactions.
It can never be zero, fractional, or negative.
Molecularity greater than 3 is very rare (probability of more than 3 molecules colliding simultaneously is negligible).
For complex reactions, order is given by the slowest step (Rate Determining Step), but molecularity has no meaning for the overall complex reaction.

4. Integrated Rate Equations

Integrating the differential rate equations gives a relation between concentration at different times and rate constant.

4.1 Zero Order Reactions

Rate is independent of the concentration of reactants. (e.g., decomposition of gases on solid surfaces like Pt or Mo at high pressure, photochemical reactions).

[R] = −kt + [R]₀
Equation of a straight line (y = mx + c). Graph: [R] vs t gives a straight line with slope = −k and intercept = [R]₀.

Half-life (t_1/2): Time in which concentration of reactant is reduced to one half of its initial concentration.

t_1/2 = [R]₀ / 2k (t_1/2 is directly proportional to initial concentration)

4.2 First Order Reactions

Rate is proportional to the first power of concentration. (e.g., all natural and artificial radioactive decays, inversion of cane sugar).

k = (2.303 / t) log([R]₀ / [R])
Exponential form: [R] = [R]₀e^−kt

Graph: log[R] vs t gives a straight line with slope = −k/2.303 and intercept = log[R]₀.

Half-life (t_1/2):

t_1/2 = 0.693 / k (t_1/2 is INDEPENDENT of initial concentration)

4.3 Pseudo First Order Reactions

Reactions which are not truly of the first order but under certain conditions behave as first order reactions. This happens when one reactant is present in large excess (usually the solvent, e.g., water), so its concentration remains almost constant.

Acid catalyzed hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH. Rate = k'[CH₃COOC₂H₅][H₂O] = k[CH₃COOC₂H₅]. (Molecularity=2, Order=1).
Inversion of cane sugar: C₁₂H₂₂O₁₁ + H₂O (excess) → Glucose + Fructose.

5. Temperature Dependence of the Rate of a Reaction

For a chemical reaction, the rate constant (k) typically doubles for every 10°C rise in temperature. This ratio is called the Temperature Coefficient (k_t+10 / k_t ≈ 2 to 3).

5.1 Arrhenius Equation

Provides a quantitative relationship between temperature and rate constant.

k = A e^{−E_a / RT}

A: Arrhenius factor, Frequency factor, or pre-exponential factor. Related to collision frequency.
E_a: Activation energy (J mol⁻¹). The minimum extra energy required by reactant molecules to form the activated complex (transition state).
e^−E_a/RT: Represents the fraction of molecules with kinetic energy greater than or equal to E_a.

5.2 Logarithmic Form

ln k = −(E_a / RT) + ln A
log k = log A - E_a / (2.303 RT)

Graph: log k vs (1/T) gives a straight line with slope = −E_a / 2.303R and y-intercept = log A.

5.3 Comparing Two Temperatures

If k₁ and k₂ are rate constants at temperatures T₁ and T₂ respectively:

log (k₂ / k₁) = (E_a / 2.303 R) × [ (T₂ - T₁) / (T₁ T₂) ]

5.4 Effect of Catalyst

A catalyst increases the rate of both forward and backward reactions by providing an alternate path or mechanism with a lower activation energy (E_a decreases). It does not alter ΔG (Gibbs free energy) or K_c (equilibrium constant).

6. Collision Theory of Chemical Reactions

Based on kinetic theory of gases. Reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other.

Rate = Z_AB e^{−E_a / RT}

Where Z_AB is the collision frequency (number of collisions per second per unit volume).

For complex molecules, collisions with sufficient energy do not always lead to a reaction. Proper orientation is also required.

Rate = P · Z_AB · e^{−E_a / RT}

P (Probability or Steric factor): Accounts for effective collisions having proper orientation.
Criteria for effective collision: (1) Energy barrier (E ≥ E_a or Threshold Energy), (2) Orientation barrier (Proper spatial collision mode).

⚠️ NEET TIP: Threshold Energy = Activation Energy (E_a) + Average kinetic energy of reactant molecules. Reactants must cross the threshold energy barrier to convert to products.

🎓 NEET Previous Year Questions

Q1. [NEET 2022] The given graph is a representation of kinetics of a reaction:
(y-axis: log k, x-axis: 1/T). The y and x axes for zero and first order reactions, respectively are: (Wait, question actually asked what slope represents). Slope of the line in the graph of log k vs 1/T is:

Answer From Arrhenius Eq: log k = log A - E_a/(2.303 RT). Comparing with y = mx + c, slope m = -E_a / 2.303 R.

Q2. [NEET 2021] For a reaction A → B, enthalpy of reaction is -4.2 kJ mol⁻¹ and enthalpy of activation is 9.6 kJ mol⁻¹. The correct potential energy profile for the reaction is shown in option: (Qualitative).

Answer ΔH is negative (-4.2), so the reaction is exothermic. In an exothermic reaction, the potential energy of products (B) is LOWER than the potential energy of reactants (A). Activation energy (Hump) is +9.6 above A.

Q3. [NEET 2020] Mechanism of a hypothetical reaction X2 + Y2 → 2XY is given below:
(i) X2 ⇌ X + X (fast)
(ii) X + Y2 → XY + Y (slow)
(iii) X + Y → XY (fast)
The overall order of the reaction will be:

Answer Rate depends on slow step: Rate = k[X][Y2]. From fast equilibrium (i): Keq = [X]² / [X2] ⇒ [X] = K' [X2]^1/2. Substitute in rate law: Rate = k·K' [X2]^1/2 [Y2]¹. Overall order = 1/2 + 1 = 1.5.

Q4. [NEET 2019] If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

Answer t = (2.303/k) log( [A]o / [A] ). For 99% completion, [A] = 100 - 99 = 1%.
t = (2.303/k) log(100/1) = (2.303/k) log(10²) = 2(2.303/k) = 4.606/k.

Q5. [NEET 2018] Which of the following statements is not correct about order of a reaction?

Answer Wrong statement often given: "Order of reaction cannot be fractional". Correct fact: Order can be zero, fractional or integer, and it is an experimental quantity. (Molecularity cannot be zero or fractional).

💡 Rapid Revision

Unit of rate constant (k): 0th order (mol/L/s); 1st order (1/s); 2nd order (L/mol/s).
For 1st order: Half life is independent of initial concentration (t_1/2 = 0.693/k).
For 0th order: Half life is directly proportional to initial concentration.
Catalyst lowers E_a but does not change ΔH or equilibrium point.
Rate law is ALWAYS determined experimentally. It cannot be predicted merely by looking at the balanced equation (except for elementary reactions).

Numericals - Chemical Kinetics - Class 12 Chemistry

CLASS 12 CHEMISTRY | NCERT SOLUTIONS

Chapter 3 — Chemical Kinetics

22 Solved Numericals — Step-by-Step Breakdown

Key Formulas: Rate = k[A]ˢ[B]ˣ | k = (2.303/t)log([R]o/[R]) | t₁⁄₂ = 0.693/k | log(k₂/k₁) = (Ea / 2.303R)[(T₂-T₁)/(T₁T₂)]

📝 Rate of Reaction & Order (Q1 – Q7)

3 MarksQ1. For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

✓ Solution
1. Rate in minutes:
Average rate (r_av) = -Δ[R] / Δt = - (0.02 - 0.03) / 25
r_av = - (-0.01) / 25 = 1/2500 = 4 × 10⁻⁴ M min⁻¹.
2. Rate in seconds:
Δt = 25 min = 25 × 60 = 1500 s.

r_av = - (-0.01) / 1500 = 1 / 150000 = 6.66 × 10⁻⁶ M s⁻¹.

3 MarksQ2. For a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval.

✓ Solution
For the reaction 2A → Products, rate expression is:
Rate = - (1/2) Δ[A] / Δt
Δ[A] = 0.4 - 0.5 = -0.1 mol L⁻¹.
Δt = 10 min.

Rate = - (1/2) × (-0.1 / 10) = 0.1 / 20 = 0.005 mol L⁻¹ min⁻¹.

3 MarksQ3. For a reaction, A + B → Product; the rate law is given by, r = k[A]½ [B]². What is the order of the reaction?

✓ Solution
The overall order of a reaction is the sum of the powers to which the concentration terms are raised in the rate law expression.
Order wrt A = 1/2. Order wrt B = 2.
Overall order = 1/2 + 2 = 5/2 = 2.5.

The order of the reaction is 2.5 (Fractional order is possible).

3 MarksQ4. Identify the reaction order from each of the following rate constants: (i) k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹ (ii) k = 3 × 10⁻⁴ s⁻¹

✓ Solution
The unit of rate constant (k) depends on the order (n) as: (mol L⁻¹)¹⁻ⁿ s⁻¹.
(i) Unit is L mol⁻¹ s⁻¹. This implies 1 - n = -1 ⇒ n = 2.
So, it is a Second order reaction.
(ii) Unit is s⁻¹. This implies 1 - n = 0 ⇒ n = 1.

So, it is a First order reaction.

3 MarksQ5. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

✓ Solution
Rate law: r = k[A]².
(i) If [A] is doubled: [A]' = 2A.
r' = k(2A)² = 4k[A]² = 4r. The rate increases 4 times.
(ii) If [A] is halved: [A]' = A/2.

r' = k(A/2)² = (1/4) k[A]² = r/4. The rate becomes 1/4th (one fourth).

4 MarksQ6. Given the reaction: 2NO(g) + O₂(g) → 2NO₂(g). The rate law is Rate = k[NO]²[O₂]. What happens to the rate if the volume of the reaction vessel is reduced to 1/3 of its original volume?

✓ Solution
Concentration = moles / Volume (c = n/V).
If volume V is reduced to V/3, then the concentration of ALL gases will become 3 times their original values.
[NO]' = 3[NO] ; [O₂]' = 3[O₂].
New Rate, r' = k [3NO]² [3O₂]
r' = k(9 [NO]²) (3 [O₂]) = 27 k[NO]²[O₂]
r' = 27 × Original Rate.

The rate increases by a factor of 27.

3 MarksQ7. In a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Determine the rate law.

✓ Solution
Let Rate = k[A]ˢ[B]ˣ.
1. When [B] is doubled, rate doubles: 2(Rate) = k[A]ˢ[2B]ˣ = 2ˣ k[A]ˢ[B]ˣ.
So, 2ˣ = 2¹ ⇒ y = 1.
2. When both are doubled, rate becomes 8 times: 8(Rate) = k[2A]ˢ[2B]¹ = 2ˢ × 2 × k[A]ˢ[B].
So, 2ˢ⁺¹ = 8 = 2³ ⇒ x + 1 = 3 ⇒ x = 2.

Rate Law: Rate = k[A]²[B]¹

📈 First Order Equations & Half-Life (Q8 – Q15)

3 MarksQ8. A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

✓ Solution
For first order: k = (2.303 / t) log([R]₀ / [R])
1.15 × 10⁻³ = (2.303 / t) log(5 / 3)
t = (2.303 / 1.15 × 10⁻³) × [log 5 - log 3]
t = 2002.6 × [0.6990 - 0.4771]
t = 2002.6 × 0.2219

t = 444.38 s

4 MarksQ9. Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t₁⁄₂) of the reaction.

✓ Solution
1. Half life: t_1/2 = 0.693 / k = (2.303 × 0.3010) / k = 2.303 log(2) / k.
2. Time for 99.9%: Let [R]₀ = 100.
Amount reacted = 99.9. Amount left, [R] = 100 - 99.9 = 0.1.
t_99.9% = (2.303 / k) log(100 / 0.1) = (2.303 / k) log(1000) = (2.303 / k) × 3.
3. Ratio:
t_99.9% / t_1/2 = [(2.303 × 3) / k] / [(2.303 × 0.3010) / k]
Ratio = 3 / 0.3010 = 9.96 ≈ 10.

Therefore, t_99.9% ≈ 10 × t_1/2 (Proved).

3 MarksQ10. Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

✓ Solution
Given: t_1/2 = 60 min.
For a first order reaction: k = 0.693 / t_1/2
k = 0.693 / 60 min = 0.01155 min⁻¹.
Or in seconds: t_1/2 = 60 × 60 = 3600 s.

k = 0.693 / 3600 s = 1.925 × 10⁻⁴ s⁻¹

3 MarksQ11. A first order reaction takes 40 min for 30% decomposition. Calculate t½.

✓ Solution
Let [R]₀ = 100. Since 30% decomposed, [R] left = 100 - 30 = 70. Time t = 40 min.
k = (2.303 / t) log([R]₀ / [R])
k = (2.303 / 40) log(100 / 70) = (2.303 / 40) log(10/7)
k = (2.303 / 40) [log 10 - log 7] = (2.303 / 40) [1 - 0.8451]
k = (2.303 / 40) × 0.1549 = 0.3567 / 40 = 8.918 × 10⁻³ min⁻¹.

t_1/2 = 0.693 / k = 0.693 / (8.918 × 10⁻³) = 77.7 min

3 MarksQ12. What will be the half-life of a zero order reaction if the initial concentration of reactant is 0.1 M and rate constant is 1.0 × 10⁻³ mol L⁻¹ min⁻¹?

✓ Solution
For a ZERO order reaction, half life t_1/2 depends on initial concentration.
Formula: t_1/2 = [R]₀ / 2k
t_1/2 = 0.1 / (2 × 1.0 × 10⁻³) = 0.1 / 0.002

t_1/2 = 50 minutes

3 MarksQ13. In a first order reaction, the concentration of the reactant is reduced from 0.6 M to 0.2 M in 5 minutes. Calculate the rate constant (k).

✓ Solution
[R]₀ = 0.6 M; [R] = 0.2 M; t = 5 min.
k = (2.303 / t) log([R]₀ / [R])
k = (2.303 / 5) log(0.6 / 0.2) = (2.303 / 5) log(3)
k = 0.4606 × 0.4771

k = 0.2198 min⁻¹

4 MarksQ14. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained:
t=0, P = 35.0 mm Hg; t=360, P = 54.0 mm Hg. Calculate rate constant. (Decomposition is A(g) → B(g) + C(g)).

✓ Solution
Reaction: A(g) → B(g) + C(g)
At t=0: P₀. At t=t: P₀ - p, p, p. Total P_t = P₀ - p + p + p = P₀ + p.
Pressure of A at time t (p_A) = P₀ - p = P₀ - (P_t - P₀) = 2P₀ - P_t.
P₀ = 35. P_t = 54 at t = 360 s.
p_A = 2(35) - 54 = 70 - 54 = 16 mm Hg.
k = (2.303 / t) log(P₀ / p_A)
k = (2.303 / 360) log(35 / 16) = 0.00639 × log(2.187) = 0.00639 × 0.339

k = 2.16 × 10⁻³ s⁻¹

3 MarksQ15. The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

✓ Solution
Radioactive decay is always First Order.
k = 0.693 / t_1/2 = 0.693 / 5730 = 1.209 × 10⁻⁴ yr⁻¹.
[R]₀ = 100; [R] = 80.
t = (2.303 / k) log(100 / 80) = (2.303 / 1.209 × 10⁻⁴) log(1.25)
t = 19048.8 × 0.0969

Age (t) = 1845 years

🔥 Arrhenius Equation & Activation Energy (Q16 – Q22)

4 MarksQ16. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

✓ Solution
T₁ = 293 K; T₂ = 313 K. Rate quadruples ⇒ k₂ / k₁ = 4. R = 8.314 J K⁻¹ mol⁻¹.
log(k₂ / k₁) = (E_a / 2.303R) [(T₂ - T₁) / (T₁ T₂)]
log(4) = (E_a / 2.303 × 8.314) × [(313 - 293) / (293 × 313)]
0.6020 = (E_a / 19.147) × [20 / 91709]
E_a = (0.6020 × 19.147 × 91709) / 20 = 1057075.8 / 20

E_a = 52853 J mol⁻¹ = 52.85 kJ mol⁻¹

3 MarksQ17. The activation energy for the reaction 2HI(g) → H₂ + I₂ is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

✓ Solution
Fraction of molecules = e^−Ea/RT.
Let x = e^−Ea/RT. Taking ln on both sides:
ln x = -E_a / RT. Thus, log x = -E_a / (2.303 RT).
E_a = 209.5 kJ = 209500 J. T = 581 K.
log x = -209500 / (2.303 × 8.314 × 581) = -209500 / 11124.6 = -18.832
x = Antilog(-18.832) = Antilog(-19 + 0.168).

Fraction x = 1.47 × 10⁻¹⁹

4 MarksQ18. The decomposition of a hydrocarbon follows the equation: k = (4.5 × 10¹¹ s⁻¹) e^-28000K/T. Calculate E_a.

✓ Solution
Compare the given equation with Arrhenius equation: k = A e^-Ea/RT.
Therefore, -E_a / RT = -28000 / T
E_a / R = 28000
E_a = 28000 × R = 28000 × 8.314 J mol⁻¹
E_a = 232792 J mol⁻¹

E_a = 232.79 kJ mol⁻¹

4 MarksQ19. At what temperature will the rate of a reaction be exactly double its rate at 300 K? (Activation energy = +50 kJ mol⁻¹).

✓ Solution
T₁ = 300 K. k₂/k₁ = 2. E_a = 50000 J mol⁻¹.
log 2 = (50000 / 19.147) [ (T₂ - 300) / (300 × T₂) ]
0.3010 = 2611.37 [ (T₂ - 300) / 300 T₂ ]
0.3010 / 2611.37 = (T₂ - 300) / 300 T₂
1.152 × 10⁻⁴ = 1/300 - 1/T₂
1/T₂ = 1/300 - 0.0001152 = 0.003333 - 0.0001152 = 0.0032178 K⁻¹

T₂ = 1 / 0.0032178 = 310.7 K

3 MarksQ20. In a reaction, E_a = 0. What will be the value of rate constant if temperature approaches infinity?

✓ Solution
Arrhenius equation: k = A e^-Ea/RT.
If E_a = 0, then the term e⁰ = 1.
Therefore, k = A (Frequency factor).
Even if T approaches infinity, if E_a=0 or if T→∞ for any E_a, the term -Ea/RT approaches 0, and e⁰=1.

Rate constant (k) will be equal to Arrhenius constant A.

4 MarksQ21. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

✓ Solution
1. Time for 99%: [R] = 100 - 99 = 1.
t₉₉ = (2.303/k) log(100/1) = (2.303/k) log(10²) = (2.303/k) × 2.
2. Time for 90%: [R] = 100 - 90 = 10.
t₉₀ = (2.303/k) log(100/10) = (2.303/k) log(10) = (2.303/k) × 1.
3. Compare:
t₉₉ = 2 × (2.303/k) = 2 × t₉₀.

Hence, t_99% = 2 × t_90%. (Proved).

3 MarksQ22. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of reaction?

✓ Solution
50% completion means 1 half-life (t_1/2) = 2 hours.
75% completion means 25% reactant is left. This corresponds to 2 half-lives (100% → 50% → 25%).
So, 2 × t_1/2 = 2 × 2 = 4 hours.
This exact relationship (t_75% = 2 × t_1/2) is ONLY true for a First Order reaction, because its half-life is independent of the initial concentration.

The order of the reaction is 1 (First order).

✍ Score Guide — 22 Questions
All questions from NCERT aligned with CBSE marking scheme.

High-Yield Facts & Formulas: Chemical Kinetics

Chemical Kinetics Definition
The branch of chemistry which deals with the study of reaction rates and their mechanisms.

Average Rate of Reaction
Rate = -([R]₂ - [R]₁) / (t₂ - t₁).

Instantaneous Rate of Reaction
Rate = -d[R]/dt = d[P]/dt.

Rate Law
An expression in which reaction rate is given in terms of molar concentration of reactants.

Rate Constant (k)
The rate of reaction when the concentration of each reactant is unity.

Order of Reaction
The sum of powers of the concentration of reactants in the rate law expression.

Molecularity of Reaction
The number of reacting species taking part in an elementary reaction.

Zero Order Reaction Rate Unit
mol L^-1 s^-1.

First Order Reaction Rate Unit
s^-1.

Second Order Reaction Rate Unit
L mol^-1 s^-1.

Zero Order Integrated Rate Law
[R] = -kt + [R]₀

First Order Integrated Rate Law
k = (2.303/t) log ([R]₀/[R])

Zero Order Half-life
t_1/2 = [R]₀ / 2k

First Order Half-life
t_1/2 = 0.693 / k

Activation Energy (E_a)
Minimum extra energy required by a reactant molecule to get converted into product.

Arrhenius Equation
k = A e^-Ea/RT

Effect of Catalyst
Increases the rate of reaction by providing a path of lower activation energy.

Collision Frequency (Z)
Number of collisions per second per unit volume.

Pseudo First Order Reaction
Reactions that follow first order kinetics but are of higher order chemically.

Effective Collisions
Collisions that lead to product formation (proper energy + orientation).

Temperature Coefficient
Ratio of rate constants at two temperatures differing by 10 K (usually = 2 to 3).

Threshold Energy
E_th = Initial Energy + Activation Energy.

Inversion of Cane Sugar
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (Pseudo 1st order).

Radioactive Decay
All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.

Complex Reactions
Reactions taking place in more than one step.

Rate Determining Step (RDS)
The slowest step in a complex reaction.

Fractional Order
H₂ + Br₂ → 2HBr (Order can be 1.5).

Molecularity vs Order
Order is experimental; Molecularity is theoretical for elementary reactions.

Unit of Pre-exponential factor (A)
Same as the unit of the rate constant (k).

Graphical Representation (zero order)
Plot of [R] vs t is a straight line with slope -k.

Graphical Representation (first order)
Plot of log[R] vs t is a straight line with slope -k/2.303.

Molecularity limit
More than three is rare as simultaneous collision of many molecules is unlikely.

Effect of pressure on gas rate
For gas reactions, [Gas] = P/RT. Increasing pressure increases the rate.

Boltzmann Distribution Curve
Shows shift towards higher energy with increase in temperature.

Frequency Factor
Another name for the Arrhenius factor (A).

Steric Factor (P)
The probability factor in collision theory related to orientation.

Photochemical Reactions
Usually zero order reactions (e.g., H₂ + Cl₂ → 2HCl in sunlight).

Slope of Nernst-like Kinetics Plot
Used to find the order of the reaction graphically.

Rate k and temperature
Rate constant k is independent of the initial concentration of reactants.

Half-life (t_1/2)
Time in which the concentration of reactant is reduced to one half of its initial concentration.

Reaction order and unit
The unit of k tells us the order of the reaction.

Decomposition of N₂O₅
Occurs via first order kinetics.

Elementary Reactions
Reactions taking place in a single step.

Reaction rate and surface area
For heterogeneous reactions, increasing surface area increases the rate.

Role of H⁺ in hydrolysis
Acts as a catalyst, speeding up the breakage of ester bonds.

Arrhenius Plot
Plot of log k vs 1/T is used to find E_a.

Reaction profile
Energy vs reaction coordinate diagram.

Chain Reactions
Complex reactions with initiation, propagation, and termination steps.

Specific Reaction Rate
Another name for the Rate Constant.

Van't Hoff Relationship
Relates equilibrium constant to temperature (similar to Arrhenius for rate).

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