Take timed practice tests on Chemical Kinetics for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
Ready
1. For the reaction 2A + B → C, if [A] decreases at 0.4 mol L⁻¹ s⁻¹, the rate of reaction is:
Explanation: Rate = −(1/2) d[A]/dt = −(1/2)(−0.4) = 0.2 mol L⁻¹ s⁻¹. Each stoichiometric coefficient divides the rate of that species.
2. For a first-order reaction, the rate is:
Explanation: First-order means the rate is proportional to the first power of the reactant concentration: r = k[A].
3. The order of reaction for r = k[A]²[B] is:
Explanation: Order = sum of exponents in the rate law = 2 + 1 = 3 (third order overall).
4. For a first-order reaction, the half-life t½ is:
Explanation: For first-order: t½ = ln2/k = 0.693/k. Note that it is independent of initial concentration.
5. The Arrhenius equation relates rate constant k to temperature by:
Explanation: Arrhenius equation: k = Ae^(−Ea/RT), where A = pre-exponential factor, Ea = activation energy, R = gas constant, T = temperature in K.
6. For a first-order reaction, [A] = [A]₀e^(−kt). If k = 0.1 s⁻¹, time for [A] to reduce to [A]₀/10 is:
Explanation: ln([A]₀/[A]) = kt → ln(10) = 0.1t → t = 2.303/0.1 = 23.03 s.
7. For a zero-order reaction, the rate constant k has units:
Explanation: Zero order: r = k. Units of r = mol L⁻¹ s⁻¹ = units of k. For first order: k in s⁻¹; for second order: k in L mol⁻¹ s⁻¹.
8. If Ea = 0, then the rate constant k:
Explanation: k = Ae^(−Ea/RT). If Ea = 0: k = Ae^0 = A, a constant independent of T. Such reactions have no temperature dependence.
9. The acid hydrolysis of ester in excess water follows pseudo-first order because:
Explanation: In dilute solution, water is in large excess and its concentration barely changes during reaction. The rate law r = k[ester][H₂O] ≈ k'[ester] where k' = k[H₂O] is approximately constant. Hence pseudo-first order.
10. If the rate doubles for every 10°C rise in temperature, increasing temperature from 20°C to 60°C multiplies the rate by:
Explanation: Rate doubles every 10°C. From 20 to 60°C is a 40°C rise = 4 doublings. Rate × 2⁴ = 16.
11. The rate-determining step in a multi-step mechanism is the:
Explanation: The rate-determining step (RDS) is the slowest step in the mechanism. It acts as the bottleneck — the overall reaction rate cannot exceed the rate of the RDS.
12. For a first-order reaction, the time for 75% completion (in terms of t½) is:
Explanation: After 1 half-life: 50% complete. After 2 half-lives: 75% complete (50% of remaining 50% reacts). So t = 2t½.
13. For a second-order reaction, the unit of k is:
Explanation: For nth order: unit of k = (mol L⁻¹)^(1−n) s⁻¹. For n = 2: (mol L⁻¹)^(−1) s⁻¹ = L mol⁻¹ s⁻¹.
14. The Arrhenius equation ln k = ln A − Ea/RT shows that a plot of ln k vs 1/T has slope:
Explanation: Comparing ln k = ln A − (Ea/R)(1/T) with y = mx + c: slope = −Ea/R. A graph of ln k vs 1/T gives a straight line with negative slope −Ea/R.
15. For the reaction 2NO₂ → 2NO + O₂ (second order in NO₂), if initial rate = 4×10⁻³ mol L⁻¹ s⁻¹ at [NO₂] = 0.1 M, then k is:
Explanation: r = k[NO₂]². k = r/[NO₂]² = 4×10⁻³/(0.1)² = 4×10⁻³/10⁻² = 0.4 L mol⁻¹ s⁻¹.
16. A catalyst increases the rate of reaction by:
Explanation: A catalyst provides an alternative reaction pathway with lower activation energy (Ea). It does not change K, temperature, or concentrations.
17. For a zero-order reaction with [A]₀ = 0.1 M and k = 0.01 mol L⁻¹ s⁻¹, the time for complete reaction is:
Explanation: For zero order: [A] = [A]₀ − kt. At completion [A] = 0: t = [A]₀/k = 0.1/0.01 = 10 s.
18. For a reaction with activation energy 100 kJ mol⁻¹, if a catalyst reduces Ea to 70 kJ mol⁻¹ at 300 K, the ratio of catalysed to uncatalysed rate constant (k_cat/k_uncat) is approximately (R = 8.314 J mol⁻¹ K⁻¹):
Explanation: ln(k_cat/k_uncat) = (Ea_uncat − Ea_cat)/(RT) = (100000 − 70000)/(8.314 × 300) = 30000/2494 ≈ 12. So k_cat/k_uncat = e^12 ≈ 1.6×10⁵.