Unit 3: Energy & Rates — Unit Test

Assessment OF Learning
Graded
Duration: 75 min | Total: /60
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15]
Q1 [3]
Define enthalpy. State the sign of ΔH for endothermic and exothermic reactions.
Q2 [3]
State Hess's law and explain why it works (state function).
Q3 [4]
For rate = k[A]²[B]⁰, what happens to rate if (a) [A] is tripled, (b) [B] doubled, (c) both halved?
Q4 [5]
A 100.0 g sample of water at 25.0 °C absorbs 8.36 kJ. Find final temperature. (c = 4.18 J/g·°C)
Part B: Thinking & Investigation [15]
Q5 [5]
Calculate ΔH° for: 2 NO(g) + O₂(g) → 2 NO₂(g), using ΔHf°: NO = +90.4, NO₂ = +33.2 kJ/mol.
Q6 [5]
From experimental rate data: [A](M) | [B](M) | rate(M/s) → 0.10|0.10|2.0×10⁻³; 0.20|0.10|8.0×10⁻³; 0.10|0.20|2.0×10⁻³. Determine rate law and k.
Q7 [5]
Use Hess's law: (1) C + O₂ → CO₂ ΔH=−393.5; (2) H₂ + ½O₂ → H₂O ΔH=−285.8; (3) CH₄ + 2O₂ → CO₂ + 2H₂O ΔH=−890.4. Find ΔHf° of CH₄.
Part C: Communication [15]
Q8 [5]
Sketch and label a PE diagram for an endothermic reaction; show Ea(forward), Ea(reverse), and ΔH.
Q9 [5]
Explain how a catalyst increases rate using collision theory and a Maxwell–Boltzmann distribution.
Q10 [5]
Define entropy and Gibbs free energy. State conditions for spontaneity.
Part D: Application [15]
Q11 [5]
Hand-warmer chemistry: 4 Fe + 3 O₂ → 2 Fe₂O₃ ΔH=−1648 kJ. How much energy when 5.6 g Fe rusts?
Q12 [5]
A catalyzed industrial process raises the rate 1000-fold without changing yield. Explain whether equilibrium yield can be improved by the catalyst.
Q13 [5]
For the reaction 2H₂O₂ → 2H₂O + O₂, ΔH = −196 kJ. Half-life is 12 h at 25 °C. Discuss why H₂O₂ is stored cold and dark.
Complete Answer Key

Q1. Enthalpy = heat content at constant P. Endo: ΔH > 0; exo: ΔH < 0.

Q2. ΔH for net rxn = sum of ΔH for any path connecting same start/end. H is a state function (independent of pathway).

Q3. (a) ×9; (b) unchanged; (c) ×¼.

Q4. ΔT = 8360/(100·4.18) = 20.0 °C → final 45.0 °C.

Q5. ΔH = 2(33.2) − [2(90.4) + 0] = 66.4 − 180.8 = −114.4 kJ.

Q6. A: doubling A × 4 → second order in A. B: no change → zero order. Rate = k[A]². k = 2.0×10⁻³/(0.10)² = 0.20 M⁻¹s⁻¹.

Q7. CH₄ formation = (1) + 2(2) − (3) = −393.5 + 2(−285.8) − (−890.4) = −74.7 kJ/mol.

Q8. Reactants low → activated complex peak → products higher; Ea(rev) < Ea(fwd); ΔH > 0.

Q9. Catalyst lowers Ea → more molecules in distribution exceed Ea → more effective collisions/sec.

Q10. S = disorder. ΔG = ΔH − TΔS. Spontaneous when ΔG < 0.

Q11. n(Fe) = 5.6/55.85 = 0.100 mol → 0.100 × 1648/4 = 41.2 kJ released.

Q12. No — catalyst speeds both directions equally; equilibrium position (and yield) unchanged. Only conditions (T, P, conc.) change yield.

Q13. Cold lowers k (Arrhenius); dark prevents photochemical decomposition. Combined: extends shelf life.