Unit 5 Practice: Hard
Kinetics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Initial-rate data show that when [A] doubles while all other concentrations remain constant, the reaction rate increases by a factor of 8. Determine the order in A.
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Answer: Third order
Work:
2^m = 8
Since:
2³ = 8
m = 3
Question 2
When [B] triples, the rate increases by a factor of 9. Determine the order in B.
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Answer: Second order
Work:
3^n = 9
Since:
3² = 9
n = 2
Question 3
Using Questions 1 and 2, write the rate law and determine the overall reaction order.
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Answer:
Rate = k[A]³[B]²
Overall order:
5
Explanation: Add the exponents:
3 + 2 = 5
Question 4
For:
Rate = k[A]³[B]²
[A] doubles and [B] triples. By what factor does the rate change?
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Answer: 72 times
Work:
Effect of A:
2³ = 8
Effect of B:
3² = 9
Combined:
8 × 9 = 72
Question 5
Why does changing concentration normally change reaction rate without changing k?
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Answer: Concentration appears separately in the rate law, while k depends mainly on temperature, mechanism, and catalyst conditions.
Explanation: For:
Rate = k[A]^m
increasing [A] changes the concentration term.
The identity of the reaction and its activation-energy conditions have not changed, so k stays the same at constant temperature.
Question 6
How could you experimentally distinguish a first-order reaction from a second-order reaction?
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Answer: Test which integrated-rate-law plot is linear.
Explanation:
For first order:
ln[A] vs. t should be linear.
For second order:
1/[A] vs. t should be linear.
The plot with the best straight-line relationship identifies the order.
Question 7
A reaction has the same half-life even when its initial concentration is changed. What reaction order does this strongly suggest?
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Answer: First order
Explanation: The first-order half-life equation is:
t₁/₂ = 0.693/k
Notice that initial concentration is not included.
Question 8
Why is radioactive decay commonly modeled as first-order kinetics?
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Answer: Each radioactive nucleus has a roughly constant probability of decaying during a given time interval.
Explanation: That means a constant fraction decays per unit time rather than a constant amount.
This produces first-order behavior and a constant half-life.
Question 9
Consider this proposed mechanism:
Step 1, slow:
A + B → C
Step 2, fast:
C + A → D
Determine the overall reaction and predicted rate law.
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Answer:
Overall:
2A + B → D
Rate law:
Rate = k[A][B]
Explanation: Add the two steps and cancel intermediate C.
Because Step 1 is slow, its reactants determine the predicted rate law.
Since Step 1 is elementary:
Rate = k[A][B]
Question 10
What two major conditions must a proposed reaction mechanism satisfy?
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Answer:
Its elementary steps must add to the correct overall reaction.
It must produce a rate law consistent with the experimentally observed rate law.
Explanation: A mechanism is not accepted just because its equations add correctly.
It must also agree with kinetic evidence.
Question 11
Why should an experimentally observed rate law usually not contain an intermediate?
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Answer: Intermediates are not initial reactants whose concentrations are independently controlled in the experiment.
Explanation: If an intermediate appears in a mechanism-derived rate law, it usually has to be replaced using another relationship involving measurable reactants.
Question 12
A reaction-energy diagram for a two-step mechanism contains two peaks. What do the peaks represent?
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Answer: Transition states
Explanation: Each elementary step has its own activation-energy barrier.
The highest-energy point along each step is a transition state.
Question 13
What does the valley between two peaks in a multistep energy diagram represent?
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Answer: An intermediate
Explanation: The intermediate is formed after the first transition state and then consumed during the next elementary step.
It is more stable than the neighboring transition states, so it appears as a local minimum.
Question 14
A two-step reaction has a much larger activation barrier for Step 2 than for Step 1. Which step is likely the rate-determining step?
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Answer: Step 2
Explanation: The larger activation barrier usually makes that step slower because fewer particles have enough energy to cross it.
Question 15
A reaction releases a large amount of energy overall but happens very slowly. How is this possible?
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Answer: It can have a large activation energy.
Explanation: Reaction enthalpy or overall energy change tells you the energy difference between reactants and products.
It does not tell you how high the activation barrier is.
A strongly exothermic reaction can still be slow if Ea is large.
Question 16
What happens to a Maxwell-Boltzmann energy distribution when temperature increases?
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Answer: The curve becomes broader and lower, with more particles at higher energies.
Explanation: The total number of particles stays the same, so total area remains constant.
But the fraction of particles above the activation energy increases.
That increases reaction rate.
Question 17
Compare increasing temperature with adding a catalyst using a Maxwell-Boltzmann diagram.
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Answer: Increasing temperature changes the energy distribution; a catalyst lowers the activation-energy threshold.
Explanation: Heating shifts and broadens the particle-energy distribution so more particles exceed Ea.
A catalyst does not change the distribution at the same temperature.
Instead, it creates a lower Ea.
Question 18
Why does a catalyst not change ΔH for the reaction?
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Answer: ΔH depends only on the energies of the reactants and products, not the pathway between them.
Explanation: A catalyst changes the route taken between reactants and products.
It lowers the activation energy but leaves the starting and ending energies unchanged.
Question 19
Why does a catalyst speed up both the forward and reverse reactions?
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Answer: It provides a lower-energy pathway between the same reactant and product states in both directions.
Explanation: The catalyst lowers the activation barriers for both directions.
It therefore helps equilibrium be reached faster but does not change the equilibrium composition.
Question 20
Explain how a heterogeneous catalyst can increase the rate of a reaction occurring at a solid surface.
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Answer: Reactants can adsorb onto the catalyst surface, where they are brought together and may have bonds weakened or orientations improved.
Explanation: A solid catalyst can provide active sites that:
hold reactants near one another
improve collision orientation
weaken existing bonds
provide an alternative mechanism
All of these effects can lower the effective activation energy and increase reaction rate.