Theory: Chemical Kinetics
1. Integrated Rate Laws — Summary
Zero order: [R] = [R₀] − kt. Graph: [R] vs t = straight line. t½ = [R₀]/2k. First order: ln[R] = ln[R₀] − kt. Graph: ln[R] vs t = straight line. t½ = 0·693/k (constant). Second order: 1/[R] = 1/[R₀] + kt. Graph: 1/[R] vs t = straight line. t½ = 1/(k[R₀]). The GRAPH determines order! Different plots for different orders.
2. Zero Order Reactions
Rate = k[R]⁰ = k. Rate is constant regardless of concentration. [R] = [R₀] − kt. Units of k: mol L⁻¹ s⁻¹. t½ = [R₀]/2k (depends on initial concentration). Examples: decomposition of NH₃ on Pt surface, photochemical reactions (H₂ + Cl₂ in light), enzyme-catalysed reactions at high substrate concentration (Michaelis-Menten kinetics at saturation).
3. First Order Reactions
Rate = k[R]. ln[R] = ln[R₀] − kt. t½ = 0·693/k (INDEPENDENT of initial concentration). Units of k: s⁻¹ (or min⁻¹). Graph: ln[R] vs t = straight line (slope = −k). Examples: radioactive decay, decomposition of N₂O₅, decomposition of H₂O₂, hydrolysis of methyl acetate in acid. All radioactive decay processes are first order.
4. Second Order Reactions
Rate = k[R]². 1/[R] = 1/[R₀] + kt. t½ = 1/(k[R₀]) (depends on initial concentration). Units of k: L mol⁻¹ s⁻¹. Graph: 1/[R] vs t = straight line. Examples: 2NO₂ → 2NO + O₂, 2HI → H₂ + I₂ (thermally).
5. Half-Life Comparison
Zero order: t½ = [R₀]/2k → as [R₀] increases, t½ increases. First order: t½ = 0·693/k → independent of [R₀] (constant). Second order: t½ = 1/(k[R₀]) → as [R₀] increases, t½ decreases. This distinction is important for NEET. First order t½ constant = unique property used in radiocarbon dating.
6. Pseudo First Order Reactions
When one reactant is in large excess, its concentration stays essentially constant → reaction appears first order. Example: CH₃COOC₂H₅ + H₂O →(H⁺) CH₃COOH + C₂H₅OH. Water in large excess → concentration essentially constant → rate = k'[ester] → pseudo first order. k' = k[H₂O]. Many hydrolysis reactions are pseudo first order.
7. Temperature Dependence — Arrhenius Equation
k = Ae^(−Ea/RT). ln k = ln A − Ea/RT. log k = log A − Ea/(2·303RT). Plot of log k vs 1/T: straight line, slope = −Ea/2·303R. Ea from slope: Ea = −slope × 2·303 × R. At two temperatures: log(k₂/k₁) = Ea/2·303R × (1/T₁ − 1/T₂). Rule of thumb: k doubles for every 10°C rise (for reactions with Ea≈50 kJ/mol).
8. Molecularity vs Order
Molecularity = number of molecules participating in elementary step (always integer: 1,2,3). Order = experimentally determined, can be fractional or zero. For elementary reactions: order = molecularity. For complex reactions (multiple steps): order ≠ molecularity. Rate-determining step (slowest step) determines overall order. Example: H₂ + Br₂ → 2HBr has fractional order (complex mechanism).
Frequently Asked Questions
1. Why does a straight line in [R] vs t indicate zero order? ⌄
Zero order integrated rate law: [R] = [R₀] − kt. This is in the form y = mx + c where y=[R], x=t, m=−k(slope), c=[R₀](y-intercept). A straight line with negative slope in [R] vs t graph directly gives zero order. If it were first order, the [R] vs t curve would be exponential decay (curved), not straight.
2. What would the ln[R] vs t graph look like for this reaction? ⌄
Since this is zero order, [R] = [R₀] − kt (linear decrease). Therefore ln[R] = ln([R₀] − kt). This is NOT a straight line — it curves (because ln of a linear function is curved). A straight line in ln[R] vs t would indicate first order. So if you see straight line in [R] vs t → zero order. Straight line in ln[R] vs t → first order.
3. What are examples of zero order reactions? ⌄
Decomposition of NH₃ on Pt/Au surface: 2NH₃ → N₂ + 3H₂ (zero order in NH₃, as surface is saturated). Photochemical reactions: H₂(g) + Cl₂(g) → 2HCl (in light, rate depends on light intensity not reactant concentrations). Enzyme-catalysed reactions at high [substrate]: when all enzyme active sites are occupied, rate = constant = zero order.
4. How do you determine order experimentally? ⌄
Method 1 (Graph): Plot [R] vs t (zero order?), ln[R] vs t (first order?), 1/[R] vs t (second order?). Whichever gives a straight line → that's the order. Method 2 (Initial rate method): vary [R₀], measure initial rate. Rate = k[R]ⁿ → log(rate) vs log[R] gives slope=n. Method 3 (Half-life): if t½ is constant → first order. If t½ ∝ [R₀] → zero order. If t½ ∝ 1/[R₀] → second order.
5. What is the unit of rate constant k for different orders? ⌄
General: units of k = (concentration)^(1-n) × time⁻¹. Zero order (n=0): k = mol L⁻¹ s⁻¹. First order (n=1): k = s⁻¹. Second order (n=2): k = L mol⁻¹ s⁻¹. This can also be used to determine order if k units are given. Easy check: if k has units of s⁻¹ → first order. If k has units of mol L⁻¹ s⁻¹ → zero order. If k has units of L mol⁻¹ s⁻¹ → second order.
6. What is the activation energy and Arrhenius factor? ⌄
Arrhenius equation: k = A × e^(−Ea/RT). A = pre-exponential factor (frequency factor) = total collision frequency × steric factor. Ea = activation energy = minimum energy needed for reaction. e^(−Ea/RT) = fraction of molecules with energy ≥ Ea. Higher T → more molecules have energy ≥ Ea → more successful collisions → higher k. Higher Ea → fewer molecules can react → slower reaction (smaller k).
7. What is the difference between rate and rate constant? ⌄
Rate = change in concentration per unit time = k[A]ⁿ. Rate depends on concentration AND k. Rate changes as reaction proceeds (concentration changes). Rate constant k depends only on temperature (not concentration). k is constant throughout the reaction (at constant T). Units differ: rate = mol L⁻¹ s⁻¹ always; k units depend on order.
8. How to find Ea from two temperature data? ⌄
Arrhenius equation at two temperatures: log(k₂/k₁) = Ea/2·303R × (T₂−T₁)/(T₁×T₂). Ea = 2·303 × R × (T₁×T₂/(T₂−T₁)) × log(k₂/k₁). R = 8·314 J/mol/K = 1·987 cal/mol/K. Example: k doubles from 300K to 310K. log(2) = Ea/(2·303×8·314) × (10/(300×310)). Ea ≈ 53 kJ/mol. This type of calculation appears in NEET.