AP Chemistry · Topic 5.5
Collision Model Practice
Part of Kinetics.(TRA-4.B)
Practice questions
22
Sample questions
5 of 22 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 1/5
Two molecules A and B approach each other along a collision trajectory.
For a collision between A and B to lead to reaction, two requirements must be satisfied:
- Acheck_circle
Sufficient KE (≥Ea) and proper orientation
- B
Only sufficient KE; orientation does not matter
- C
Equal velocities and head-on impact only
- D
Same mass and same charge
Why
Effective collisions need (1) energy ≥ Ea and (2) correct geometric orientation.
- A
Sample 2difficulty 1/5
A block of solid reactant and the same mass as a powder are added to acid. Which reacts faster, and why?
- A
Block; less surface tension
- B
Same rate; same mass
- C
Block; more concentrated
- Dcheck_circle
Powder; greater surface area gives more collision sites
Why
Higher surface area in the powder enables more frequent reactant collisions per unit time.
- A
Sample 3difficulty 2/5
Doubling A in k = A·e^(−Eₐ/RT)
- A
Squares k
- Bcheck_circle
Doubles k
- C
k unchanged
- D
Halves k
Why
k is proportional to A; doubling A doubles k.
- A
Sample 4difficulty 2/5
Two requirements for an effective collision are
- A
Equal concentrations
- Bcheck_circle
Sufficient energy and proper orientation
- C
Same molecular mass
- D
Equal speeds
Why
Effective = enough energy (≥ Eₐ) AND correct geometric orientation.
- A
Sample 5difficulty 2/5
Two Arrhenius plots are compared: reaction A has slope −5000 K, reaction B has slope −10000 K.
Which reaction is more sensitive to temperature change, and why?
- A
A, because A starts at higher ln k.
- B
Both are equally sensitive.
- C
A, because shallower slope means stronger T dependence.
- Dcheck_circle
B, because steeper |slope| means larger Eₐ.
Why
Slope = −Eₐ/R; steeper (more negative) slope means larger Eₐ, so k changes more dramatically with T. B is more T-sensitive.
- A