Ap Chemistry 2차 - Unit 5

Unit 5: Kinetics — Multiple Choice Questions
Reaction Rate Collision Theory Rate Law Integrated Rate Law Arrhenius Mechanism & RDS

1. Which statement best defines the reaction rate?

  • (A) Change in temperature per unit time
  • (B) Change in concentration per unit time
  • (C) Change in pressure per unit time
  • (D) Change in volume per unit time
View Answer
Correct Answer: (B) Explanation: Rate \(=\Delta[\text{species}]/\Delta t\). Typical unit: M·s\(^{-1}\) (mol·L\(^{-1}\)·s\(^{-1}\)).

2. What is the SI-style unit for reaction rate?

  • (A) s
  • (B) M
  • (C) M·s\(^{-1}\)
  • (D) s\(^{-1}\)
View Answer
Correct Answer: (C) Explanation: Rate is concentration per time: mol·L\(^{-1}\)·s\(^{-1}\) = M·s\(^{-1}\).

3. According to collision theory, a successful reaction requires:

  • (A) Frequent collisions only
  • (B) Proper orientation only
  • (C) Collisions with energy ≥ \(E_a\) only
  • (D) Collision + proper orientation + energy ≥ \(E_a\)
View Answer
Correct Answer: (D) Explanation: All three conditions must be satisfied for reaction to proceed.

4. Which change increases collision frequency most directly for gases?

  • (A) Decreasing pressure
  • (B) Increasing volume
  • (C) Increasing pressure (or concentration)
  • (D) Adding an inert solid
View Answer
Correct Answer: (C) Explanation: Higher pressure (and concentration) means more particles per volume → more collisions.

5. Temperature increases rate because:

  • (A) It lowers the activation energy \(E_a\)
  • (B) It increases the fraction of molecules with \(E \ge E_a\)
  • (C) It changes stoichiometric coefficients
  • (D) It creates new intermediates
View Answer
Correct Answer: (B) Explanation: Higher \(T\) shifts the Maxwell–Boltzmann distribution, increasing high-energy fraction.

6. A catalyst typically:

  • (A) Increases \(E_a\) and decreases \(k\)
  • (B) Lowers \(E_a\) and increases \(k\)
  • (C) Changes \(\Delta H_{\text{rxn}}\)
  • (D) Is consumed stoichiometrically
View Answer
Correct Answer: (B) Explanation: Catalyst provides an alternative pathway with lower \(E_a\); \(k\) increases. It is regenerated.

7. For \(m\text{A} + n\text{B} \to \text{products}\), a general rate law is:

  • (A) \(\text{Rate}=k[\text{A}]^m[\text{B}]^n\) (orders must be coefficients)
  • (B) \(\text{Rate}=k[\text{A}]^x[\text{B}]^y\) (orders are determined experimentally)
  • (C) \(\text{Rate}=k\) only
  • (D) \(\text{Rate}=k[\text{A}][\text{B}]/t\)
View Answer
Correct Answer: (B) Explanation: Reaction orders \(x,y\) are determined by experiment, not necessarily stoichiometry.

8. If doubling \([\text{A}]\) (with \([\text{B}]\) constant) doubles the rate, the order in A is:

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) 1/2
View Answer
Correct Answer: (B) Explanation: Rate \(\propto [\text{A}]^x\). \(2^x=2\Rightarrow x=1\).

9. If doubling \([\text{A}]\) (with \([\text{B}]\) constant) increases the rate by a factor of 4, the order in A is:

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) 3
View Answer
Correct Answer: (C) Explanation: \(2^x=4\Rightarrow x=2\).

10. The units of \(k\) for a zero-order reaction are most likely:

  • (A) M·s\(^{-1}\)
  • (B) s\(^{-1}\)
  • (C) M\(^{-1}\)·s\(^{-1}\)
  • (D) M\(^{1-n}\)·s\(^{-1}\) with \(n=0\)
View Answer
Correct Answer: (A) Explanation: For zero order, rate \(=\) \(k\), so \(k\) has units of rate (M·s\(^{-1}\)).

11. The units of \(k\) for a first-order reaction are:

  • (A) M·s\(^{-1}\)
  • (B) s\(^{-1}\)
  • (C) M\(^{-1}\)·s\(^{-1}\)
  • (D) M\(^{-2}\)·s\(^{-1}\)
View Answer
Correct Answer: (B) Explanation: For first order, rate \(=k[\text{A}]\Rightarrow k=\text{rate}/[\text{A}]=\) s\(^{-1}\).

12. The units of \(k\) for a second-order reaction (overall) are:

  • (A) s\(^{-1}\)
  • (B) M·s\(^{-1}\)
  • (C) M\(^{-1}\)·s\(^{-1}\)
  • (D) M\(^{-2}\)·s\(^{-1}\)
View Answer
Correct Answer: (C) Explanation: rate \(=k[\text{A}]^2\) (or \(k[\text{A}][\text{B}]\)) ⇒ \(k\) has units M\(^{-1}\)·s\(^{-1}\).

13. Which plot is linear for a zero-order reaction?

  • (A) \([\text{A}]\) vs \(t\)
  • (B) \(\ln[\text{A}]\) vs \(t\)
  • (C) \(1/[\text{A}]\) vs \(t\)
  • (D) \(\ln k\) vs \(1/T\)
View Answer
Correct Answer: (A) Explanation: Zero-order integrated: \([\text{A}]= -kt+[\text{A}]_0\).

14. Which plot is linear for a first-order reaction?

  • (A) \([\text{A}]\) vs \(t\)
  • (B) \(\ln[\text{A}]\) vs \(t\)
  • (C) \(1/[\text{A}]\) vs \(t\)
  • (D) \([\text{A}]^{1/2}\) vs \(t\)
View Answer
Correct Answer: (B) Explanation: First-order integrated: \(\ln[\text{A}] = -kt + \ln[\text{A}]_0\).

15. Which plot is linear for a second-order reaction (in A only)?

  • (A) \([\text{A}]\) vs \(t\)
  • (B) \(\ln[\text{A}]\) vs \(t\)
  • (C) \(1/[\text{A}]\) vs \(t\)
  • (D) \([\text{A}]^{2}\) vs \(t\)
View Answer
Correct Answer: (C) Explanation: Second-order: \(1/[\text{A}]= kt + 1/[\text{A}]_0\).

16. Which statement about half-life is TRUE?

  • (A) Zero-order: \(t_{1/2}=\frac{0.693}{k}\)
  • (B) First-order: \(t_{1/2}\) is constant and equals \(\frac{0.693}{k}\)
  • (C) Second-order: \(t_{1/2}\) is independent of \([\text{A}]_0\)
  • (D) All orders have constant \(t_{1/2}\)
View Answer
Correct Answer: (B) Explanation: Only first-order half-life is independent of initial concentration.

17. If the measured half-life does not depend on initial concentration, the reaction is most likely:

  • (A) Zero-order
  • (B) First-order
  • (C) Second-order
  • (D) Third-order
View Answer
Correct Answer: (B) Explanation: Signature feature of first-order kinetics.

18. A plot of ln[A] vs. t for a reaction is linear with slope -0.250 s⁻¹. What is the half-life?

  • (A) 1.39 s
  • (B) 2.77 s
  • (C) 3.47 s
  • (D) 4.00 s
View Answer
Correct Answer: (B) Explanation: First-order slope = −k ⇒ k = 0.250 s⁻¹. Half-life t₁/₂ = 0.693/k = 0.693/0.250 ≈ 2.77 s.

19. At constant temperature, which change most directly increases reaction rate by raising collision frequency rather than changing the energy distribution?

  • (A) Increasing temperature
  • (B) Increasing pressure of gaseous reactants
  • (C) Adding an inert solvent at the same concentrations
  • (D) Decreasing surface area of a solid reactant
View Answer
Correct Answer: (B) Explanation: Higher pressure (or concentration) packs more particles per volume, boosting collision frequency without altering the Maxwell–Boltzmann energy distribution at fixed T.

20. Which change directly alters the rate constant \(k\)?

  • (A) Changing concentration
  • (B) Changing surface area
  • (C) Changing temperature
  • (D) Diluting with solvent but same \(T\)
View Answer
Correct Answer: (C) Explanation: By Arrhenius, \(k\) depends on \(T\) (and mechanism/catalyst), not on concentration or area.

21. Consider the mechanism:
Step 1: \(2\text{A} + \text{B} \rightleftharpoons \text{C}\) (fast, reversible)
Step 2: \(\text{C} + \text{E} \to \text{D} + \text{A}\) (slow, R.D.S)

  • (A) Intermediate: A
  • (B) Intermediate: C
  • (C) Catalyst: C
  • (D) Catalyst: E
View Answer
Correct Answer: (B) Explanation: C is produced then consumed; A is regenerated (catalyst).

22. In the mechanism above, which species acts as a catalyst?

  • (A) A
  • (B) B
  • (C) C
  • (D) E
View Answer
Correct Answer: (A) Explanation: A appears on both sides overall; it is consumed then regenerated.

23. Using the given mechanism (Step 2 slow), the experimental rate law is most consistent with:

  • (A) \(\text{Rate}=k[\text{A}]^2[\text{B}][\text{E}]\)
  • (B) \(\text{Rate}=k[\text{C}][\text{E}]\)
  • (C) \(\text{Rate}=k[\text{A}]^2[\text{B}]\)
  • (D) \(\text{Rate}=k[\text{E}]^2\)
View Answer
Correct Answer: (A) Explanation: With pre-equilibrium, \([\text{C}]\propto [\text{A}]^2[\text{B}]\); slow step uses C and E.

24. If Step 1 were the slow step instead, the expected rate law would most likely be:

  • (A) \(\text{Rate}=k[\text{A}]^2[\text{B}]\)
  • (B) \(\text{Rate}=k[\text{A}]^2[\text{B}][\text{E}]\)
  • (C) \(\text{Rate}=k[\text{C}][\text{E}]\)
  • (D) \(\text{Rate}=k[\text{E}]\)
View Answer
Correct Answer: (A) Explanation: Rate is governed by the slow elementary step: \(2\text{A}+\text{B}\to\text{C}\).

25. For \(\text{Rate}=k[\text{A}]^2[\text{B}][\text{E}]\), the overall reaction order is:

  • (A) 2
  • (B) 3
  • (C) 4
  • (D) Not defined
View Answer
Correct Answer: (C) Explanation: Sum of exponents \(=2+1+1=4\).

26. Which statement is TRUE regarding \(\Delta H_{\text{rxn}}\) when a catalyst is added?

  • (A) \(\Delta H_{\text{rxn}}\) decreases
  • (B) \(\Delta H_{\text{rxn}}\) increases
  • (C) \(\Delta H_{\text{rxn}}\) is unchanged
  • (D) \(\Delta H_{\text{rxn}}\) becomes zero
View Answer
Correct Answer: (C) Explanation: Catalysts change pathway (and \(E_a\)), not thermodynamic state functions like \(\Delta H\).

27. At constant temperature, increasing volume of a gas-phase mixture generally:

  • (A) Increases rate for positive-order reactions
  • (B) Decreases rate for positive-order reactions
  • (C) Has no effect on rate
  • (D) Always doubles the rate
View Answer
Correct Answer: (B) Explanation: Larger volume → lower concentration → lower rate when order > 0.

28. Which of the following does not change the numerical value of \(k\) at a fixed mechanism?

  • (A) Changing concentration
  • (B) Changing temperature
  • (C) Adding a catalyst
  • (D) Replacing catalyst with a different one
View Answer
Correct Answer: (A) Explanation: \(k\) depends on \(T\) and pathway (catalyst), not on concentrations.

29. You test order by plotting \([\text{A}], \ln[\text{A}], 1/[\text{A}]\) vs \(t\). The only linear plot is \(\ln[\text{A}] \) vs \(t\). Conclude that the reaction is:

  • (A) Zero-order in A
  • (B) First-order in A
  • (C) Second-order in A
  • (D) Third-order in A
View Answer
Correct Answer: (B) Explanation: Linearity of \(\ln[\text{A}] \) vs \(t\) indicates first-order.

30. A reaction follows Rate = k[A]²[B][E]. If [B] and [E] are kept in large excess so they remain effectively constant, what is the apparent overall order and which plot is linear?

  • (A) First-order overall; ln[A] vs t linear
  • (B) Second-order in A (apparent overall third-order constant folded into k); 1/[A] vs t linear
  • (C) Zero-order overall; [A] vs t linear
  • (D) Third-order overall; 1/[A] vs t linear with slope k
View Answer
Correct Answer: (B) Explanation: With [B] and [E] constant, Rate = k' [A]² where k' = k[B][E]. The kinetics appear second-order in A (pseudo-conditions), so 1/[A] vs t is linear.

Reference formulas

  • Rate law: Rate = k · [A]^m · [B]^n (orders by experiment)
  • Integrated: Zero \([\text{A}]= -kt + [\text{A}]_0\); First \(\ln[\text{A}]= -kt + \ln[\text{A}]_0\); Second \(1/[\text{A}] = kt + 1/[\text{A}]_0\)
  • Half-life: First \(t_{1/2}=0.693/k\); Zero \(t_{1/2}=[\text{A}]_0/(2k)\); Second \(t_{1/2}=1/(k[\text{A}]_0)\)
  • Arrhenius: \(\ln k = \ln A - E_a/(RT)\)
  • Mechanism example (given): Step 1 fast \(2\text{A}+\text{B}\rightleftharpoons \text{C}\); Step 2 slow \(\text{C}+\text{E}\to\text{D}+\text{A}\) ⇒ \(\text{Rate}\propto [\text{A}]^2[\text{B}][\text{E}]\)

31. From the initial-rates data (Rate in M·s\(^{-1}\)):
Exp 1: [A]=0.10, [B]=0.10 → Rate=0.20
Exp 2: [A]=0.20, [B]=0.10 → Rate=0.80
Exp 3: [A]=0.10, [B]=0.20 → Rate=0.40
For \(\text{Rate}=k[\text{A}]^m[\text{B}]^n\), what are \(m,n\)?

  • (A) \(m=1, n=1\)
  • (B) \(m=2, n=1\)
  • (C) \(m=1, n=2\)
  • (D) \(m=2, n=2\)
View Answer
Correct Answer: (B) Explanation: Doubling A (Exp1→2) quadruples rate ⇒ \(m=2\). Doubling B (Exp1→3) doubles rate ⇒ \(n=1\).

32. Using the data in Q31 and \(\text{Rate}=k[\text{A}]^2[\text{B}]\), what is \(k\) (assume rate in M·s\(^{-1}\))?

  • (A) \(20\ \text{M}^{-2}\text{s}^{-1}\)
  • (B) \(200\ \text{M}^{-2}\text{s}^{-1}\)
  • (C) \(2.0\ \text{M}^{-2}\text{s}^{-1}\)
  • (D) \(0.20\ \text{M}^{-2}\text{s}^{-1}\)
View Answer
Correct Answer: (B) Explanation: From Exp 1: \(k=0.20/(0.10^2\times0.10)=0.20/0.001=200\ \text{M}^{-2}\text{s}^{-1}\).

33. Pseudo-first-order conditions for \(\text{Rate}=k[\text{A}]^m[\text{B}]^n\) are achieved by:

  • (A) Keeping both [A] and [B] very small
  • (B) Keeping [A] \(\gg\) [B] constant and measuring B
  • (C) Keeping [B] \(\gg\) [A] constant and measuring A
  • (D) Varying both [A] and [B] simultaneously
View Answer
Correct Answer: (C) Explanation: Hold B in large excess so \([\text{B}]\) ≈ constant; rate appears first order in A.

34. The instantaneous rate at time \(t\) is best obtained from:

  • (A) Slope of a secant over a large time interval
  • (B) Slope of the tangent to \([\text{A}]\) vs \(t\) at time \(t\)
  • (C) Final minus initial concentration
  • (D) Half of the average rate
View Answer
Correct Answer: (B) Explanation: Instantaneous rate is the derivative at that point, i.e., tangent slope.

35. Which is TRUE regarding intermediates and transition states (TS)?

  • (A) TS are local minima; intermediates are maxima
  • (B) Intermediates are local minima; TS are maxima on energy profile
  • (C) Both can be isolated routinely
  • (D) Neither affects the rate law
View Answer
Correct Answer: (B) Explanation: Intermediates may be (sometimes) detectable/stable; TS cannot be isolated.

36. The elementary step \(\text{A} + 2\text{B} \to \text{C}\) is:

  • (A) Unimolecular
  • (B) Bimolecular
  • (C) Termolecular
  • (D) Zeromolecular
View Answer
Correct Answer: (C) Explanation: Three reactant particles collide ⇒ termolecular (rare).

37. Which slow (rate-determining) step is consistent with \(\text{Rate}=k[\text{A}]^2[\text{B}]\)?

  • (A) \(2\text{A}+\text{B}\to \text{C}\)
  • (B) \(\text{A}+\text{B}\to \text{C}\)
  • (C) \(\text{A}\to \text{C}\)
  • (D) \(\text{B}\to \text{C}\)
View Answer
Correct Answer: (A) Explanation: For an elementary slow step, orders match molecularity.

38. For a first-order reaction with \(k=0.231\ \text{s}^{-1}\), the half-life \(t_{1/2}\) is:

  • (A) 1.50 s
  • (B) 3.00 s
  • (C) 6.00 s
  • (D) 0.693 s
View Answer
Correct Answer: (B) Explanation: \(t_{1/2}=0.693/k=0.693/0.231=3.00\ \text{s}\).

39. Zero-order: \(k=0.050\ \text{M·s}^{-1}\), \([\text{A}]_0=1.00\ \text{M}\). Time to reach \([\text{A}]=0.25\ \text{M}\)?

  • (A) 5.0 s
  • (B) 10 s
  • (C) 15 s
  • (D) 20 s
View Answer
Correct Answer: (C) Explanation: \([\text{A}]=[\text{A}]_0-kt\Rightarrow t=(1.00-0.25)/0.050=15\ \text{s}\).

40. Second-order in A: \(k=0.200\ \text{M}^{-1}\text{s}^{-1}\), \([\text{A}]_0=0.500\ \text{M}\). \(t_{1/2}\) is:

  • (A) 2.0 s
  • (B) 5.0 s
  • (C) 10 s
  • (D) 20 s
View Answer
Correct Answer: (C) Explanation: \(t_{1/2}=1/(k[\text{A}]_0)=1/(0.200\times0.500)=10\ \text{s}\).

41. At a fixed temperature, adding a catalyst will:

  • (A) Change the equilibrium constant \(K\)
  • (B) Change \(\Delta H^\circ\)
  • (C) Speed up both forward and reverse reactions equally, leaving \(K\) unchanged
  • (D) Stop the reverse reaction
View Answer
Correct Answer: (C) Explanation: Catalysts affect kinetics (rates), not thermodynamic quantities like \(K\) or \(\Delta H^\circ\).

42. A three-step mechanism has activation energies: Step 1 = 10 kJ·mol\(^{-1}\), Step 2 = 25 kJ·mol\(^{-1}\), Step 3 = 5 kJ·mol\(^{-1}\). The RDS is:

  • (A) Step 1
  • (B) Step 2
  • (C) Step 3
  • (D) None (all equal)
View Answer
Correct Answer: (B) Explanation: Highest \(E_a\) step is typically rate-determining.

43. Which best increases the orientation factor in collisions?

  • (A) Using an enzyme active site that binds substrates in a specific geometry
  • (B) Increasing solvent viscosity
  • (C) Decreasing temperature
  • (D) Using an inert diluent gas
View Answer
Correct Answer: (A) Explanation: Enzymes/catalysts orient reactants properly, raising effective collision success.

44. For a plot of \(\ln k\) vs \(1/T\), the slope is \(-1.20\times10^4\ \text{K}\). The activation energy \(E_a\) is approximately:

  • (A) 12.0 kJ·mol\(^{-1}\)
  • (B) 50.0 kJ·mol\(^{-1}\)
  • (C) 100 kJ·mol\(^{-1}\)
  • (D) 1,200 kJ·mol\(^{-1}\)
View Answer
Correct Answer: (C) Explanation: Slope \(=-E_a/R\Rightarrow E_a=1.20\times10^4\times 8.314\approx 9.98\times10^4\ \text{J·mol}^{-1}\approx 100\ \text{kJ·mol}^{-1}\).

45. Overall: \(\text{A}+2\text{B}\to \text{products}\). Measured: \(\text{Rate}=k[\text{A}][\text{B}]\). Which explanation fits?

  • (A) The slow step involves \(\text{A}+\text{B}\); the second B reacts in a fast step
  • (B) The slow step is unimolecular in A only
  • (C) The reaction is zero order overall
  • (D) Orders must equal stoichiometric coefficients
View Answer
Correct Answer: (A) Explanation: Overall stoichiometry need not match the rate law; RDS dictates observed orders.

46. The units of \(k\) for an overall fourth-order reaction are:

  • (A) \(\text{s}^{-1}\)
  • (B) \(\text{M}^{-1}\text{s}^{-1}\)
  • (C) \(\text{M}^{-2}\text{s}^{-1}\)
  • (D) \(\text{M}^{-3}\text{s}^{-1}\)
View Answer
Correct Answer: (D) Explanation: Units \(=\text{M}^{1-n}\text{s}^{-1}\). For \(n=4\): \(\text{M}^{-3}\text{s}^{-1}\).

47. If all reactant concentrations are halved in a third-order reaction overall, the rate becomes:

  • (A) Unchanged
  • (B) Half
  • (C) One-quarter
  • (D) One-eighth
View Answer
Correct Answer: (D) Explanation: Rate \(\propto [\ldots]^3\). \((1/2)^3=1/8\).

48. For \(\text{Rate}=k[\text{A}]^0[\text{B}]^1\), doubling [A] while keeping [B] constant will:

  • (A) Double the rate
  • (B) Halve the rate
  • (C) Not change the rate
  • (D) Quadruple the rate
View Answer
Correct Answer: (C) Explanation: Zero order in A ⇒ rate independent of [A].

49. For \(2\text{A}\to \text{P}\), which relation is correct?

  • (A) \(\text{Rate}=\dfrac{d[\text{A}]}{dt}\)
  • (B) \(\dfrac{d[\text{P}]}{dt}=-\dfrac{d[\text{A}]}{dt}\)
  • (C) \(\dfrac{d[\text{P}]}{dt}=-\dfrac{1}{2}\dfrac{d[\text{A}]}{dt}\)
  • (D) \(\dfrac{d[\text{P}]}{dt}=-2\dfrac{d[\text{A}]}{dt}\)
View Answer
Correct Answer: (C) Explanation: \(-\dfrac{1}{2}\dfrac{d[\text{A}]}{dt}=\dfrac{d[\text{P}]}{dt}\) from stoichiometry.

50. If \(k\) has units \(\text{M}^{-2}\text{s}^{-1}\), the overall reaction order is:

  • (A) 1
  • (B) 2
  • (C) 3
  • (D) 4
View Answer
Correct Answer: (C) Explanation: \(\text{M}^{1-n}=\text{M}^{-2}\Rightarrow 1-n=-2\Rightarrow n=3\).

51. Why don’t experimental rate-law orders necessarily match the coefficients of the overall balanced equation?

  • (A) Measurement error
  • (B) Orders are always integers
  • (C) Overall reaction sums multiple elementary steps; RDS dictates observed orders
  • (D) Because rate is independent of mechanism
View Answer
Correct Answer: (C) Explanation: Only for an elementary (slow) step do exponents equal molecularity.

52. Zero-order with \(k=2.0\times10^{-3}\ \text{M·s}^{-1}\) and \([\text{A}]_0=0.100\ \text{M}\). Time until \([\text{A}]=0\)?

  • (A) 25 s
  • (B) 40 s
  • (C) 50 s
  • (D) 100 s
View Answer
Correct Answer: (C) Explanation: \(t=[\text{A}]_0/k=0.100/0.0020=50\ \text{s}\).

53. First-order with \(k=0.100\ \text{min}^{-1}\), \([\text{A}]_0=0.800\ \text{M}\). Time to reach \([\text{A}]=0.100\ \text{M}\)?

  • (A) 6.93 min
  • (B) 13.9 min
  • (C) 20.8 min
  • (D) 27.7 min
View Answer
Correct Answer: (C) Explanation: \(\ln([\text{A}]/[\text{A}]_0)=-kt\Rightarrow t=\ln(0.100/0.800)/-0.100=\ln(0.125)/-0.100\approx 20.8\ \text{min}\).

54. Second-order in A: \([\text{A}]_0=0.400\ \text{M}\), after 25.0 s \([\text{A}]=0.200\ \text{M}\). Find \(k\).

  • (A) \(0.050\ \text{M}^{-1}\text{s}^{-1}\)
  • (B) \(0.075\ \text{M}^{-1}\text{s}^{-1}\)
  • (C) \(0.100\ \text{M}^{-1}\text{s}^{-1}\)
  • (D) \(0.200\ \text{M}^{-1}\text{s}^{-1}\)
View Answer
Correct Answer: (C) Explanation: \(1/[\text{A}]=kt+1/[\text{A}]_0\Rightarrow 5.0-2.5=kt\Rightarrow k=2.5/25.0=0.100\ \text{M}^{-1}\text{s}^{-1}\).

55. Which experiment most directly indicates first-order behavior?

  • (A) \([\text{A}]\) vs \(t\) is linear
  • (B) \(1/[\text{A}]\) vs \(t\) is linear
  • (C) \(\ln[\text{A}]\) vs \(t\) is linear with slope \(-k\)
  • (D) Rate independent of [A]
View Answer
Correct Answer: (C) Explanation: First-order linearizes \(\ln[\text{A}]\) vs \(t\) with slope \(-k\).

56. The method of isolation (flooding) is used to:

  • (A) Determine activation energy
  • (B) Determine the order with respect to a single reactant by holding others in large excess
  • (C) Balance redox equations
  • (D) Measure equilibrium constants only
View Answer
Correct Answer: (B) Explanation: Excess makes other concentrations effectively constant, simplifying the rate law.

57. Which statement about reaction orders is TRUE?

  • (A) They must be positive integers
  • (B) They equal stoichiometric coefficients
  • (C) They can be zero, fractional, or even negative and are determined experimentally
  • (D) They are independent of the mechanism
View Answer
Correct Answer: (C) Explanation: Experimental determination; non-integer and zero/negative orders are possible.

58. Finely dividing a solid reactant typically:

  • (A) Decreases rate by reducing collisions
  • (B) Increases rate by increasing available surface sites
  • (C) Makes rate zero
  • (D) Only changes equilibrium position
View Answer
Correct Answer: (B) Explanation: Greater surface area → more collision opportunities at the interface.

59. A two-step mechanism has how many transition states and intermediates?

  • (A) 1 TS and 0 intermediates
  • (B) 2 TS and 1 intermediate
  • (C) 2 TS and 2 intermediates
  • (D) 1 TS and 1 intermediate
View Answer
Correct Answer: (B) Explanation: Each step → one TS (peak). Between two peaks lies one intermediate (valley).

60. Which statement about catalysts in rate laws is MOST accurate?

  • (A) Catalysts never appear in rate laws
  • (B) Catalysts always appear squared in rate laws
  • (C) A catalyst can appear in the rate law if it participates in or affects the RDS/pre-equilibrium
  • (D) Catalysts only change \(\Delta H\), not \(E_a\) or \(k\)
View Answer
Correct Answer: (C) Explanation: Catalysts alter mechanism and \(E_a\); if involved before the RDS, their concentration may enter the rate law.

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