1. After five half-lives, 3.125% of the original reactant remains.
2. Yes, if [A]₀ is doubled, t₁/₂ doubles for a first-order reaction because t₁/₂ = ln(2)/k and is independent of initial concentration; for second-order, t₁/₂ = 1/(k[A]₀), so doubling [A]₀ halves t₁/₂.
3. A catalyst provides an alternative reaction pathway with lower activation energy, increasing the rate without being consumed.
4. The rate constant k increases with temperature according to the Arrhenius equation: k = A·e^(-Ea/RT); higher T increases k exponentially.
5. For a first-order reaction, ln([A]₀/[A]) = kt; for second-order, 1/[A] - 1/[A]₀ = kt; for zero-order, [A]₀ - [A] = kt.
6. a) Rate = k[NO]²[Cl₂]; b) Second order in NO, first order in Cl₂, third overall; c) Units: M⁻²s⁻¹; d) Doubling [NO] quadruples rate; e) Tripling [Cl₂] triples rate.
7. The rate law is Rate = k[NO]²[O₂]; it is second order in NO, first order in O₂, third overall.
8. a) First-order; b) k = 0.0231 min⁻¹; c) t₁/₂ = 30.0 min; d) [A] at 90 min = 0.125 M.
9. a) Rate = k[NO]²[Br₂]; b) Second order in NO, first order in Br₂, third overall; c) k = 1.2 × 10⁴ M⁻²s⁻¹.
10. a) Rate = k[NO]²[Cl₂]; b) Second order in NO, first order in Cl₂, third overall; c) k = 0.045 M⁻²s⁻¹.
11. a) Zero-order; b) k = 0.025 M/s; c) [A] at 20 s = 0.50 M; d) t when [A] = 0.10 M is 16 s.
12. a) Rate = k[NO]²[H₂]; b) Second order in NO, first order in H₂, third overall; c) k = 0.012 M⁻²s⁻¹.
13. a) Rate = k[NO]²[Cl₂]; b) Second order in NO, first order in Cl₂, third overall; c) k = 0.030 M⁻²s⁻¹.
14. a) Rate = k[NO]²[O₂]; b) Second order in NO, first order in O₂, third overall; c) k = 0.020 M⁻²s⁻¹.
15. a) Rate = k[NO]²[Br₂]; b) Second order in NO, first order in Br₂, third overall; c) k = 0.015 M⁻²s⁻¹.
Parent Tip: Review the logic above to help your child master the concept of ap chemistry worksheet.
