Chemical Kinetics — Practice Quiz

The concentration of each reactant is unity (1 mol L⁻¹)

Step 1: The rate law is Rate = k[A]^m[B]^n. Step 2: When [A] = [B] = 1 mol L⁻¹, the equation becomes Rate = k × 1^m × 1^n = k × 1 = k. Step 3: Therefore, the rate constant equals the rate of reaction when all reactant concentrations are 1 mol L⁻¹. Step 4: This is an important physical meaning of the rate constant — it represents the intrinsic speed of a reaction at unit concentration. Step 5: Wrong options — zero concentration would give zero rate, temperature of 0°C is unrelated to this definition, and 'reaction complete' means zero reactant remains.

k = (2.303/t) log([A₀]/[A])

Step 1: For a first order reaction, the differential rate law is -d[A]/dt = k[A]. Step 2: Separating variables and integrating: ∫d[A]/[A] = -k∫dt, giving ln[A] = ln[A₀] - kt. Step 3: Rearranging: ln([A₀]/[A]) = kt, which in base-10 log form becomes 2.303 log([A₀]/[A]) = kt. Step 4: Solving for k: k = (2.303/t) log([A₀]/[A]). Step 5: Wrong options — '[A₀]-[A]/t' is for zero order, '(1/t)(1/[A]-1/[A₀])' is for second order, and '[A]=[A₀]-kt' is also the zero order integrated equation.

t₁/₂ = 0.693 / k

Step 1: For a first order reaction, k = (2.303/t) log([A₀]/[A]). Step 2: At half-life (t = t₁/₂), [A] = [A₀]/2. Substituting: k = (2.303/t₁/₂) log([A₀]/([A₀]/2)) = (2.303/t₁/₂) log 2. Step 3: Since log 2 = 0.3010, we get k = (2.303 × 0.3010)/t₁/₂ = 0.693/t₁/₂. Step 4: Rearranging: t₁/₂ = 0.693/k. Step 5: A very important property — for first order reactions, half-life is INDEPENDENT of initial concentration. The option '[A₀]/2k' is for zero order reactions, showing a common confusion between the two orders.

Initial concentration of the reactant

Step 1: For a zero order reaction, the integrated rate law is [A₀] - [A] = kt. Step 2: At t = t₁/₂, [A] = [A₀]/2, so: [A₀] - [A₀]/2 = k × t₁/₂. Step 3: Simplifying: [A₀]/2 = k × t₁/₂, therefore t₁/₂ = [A₀]/(2k). Step 4: This shows t₁/₂ is directly proportional to [A₀] — if you double the initial concentration, the half-life also doubles. Step 5: This is the OPPOSITE of first order reactions where t₁/₂ is independent of concentration. Confusion between these two is a very common exam mistake.